Viscoelasticity, Mechanical Models, Dynamic Testing and Cure Rheometry

This chapter answers the second half of Section XE5.6, Polymer Rheology, of XE5 Polymer Science and Engineering, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The syllabus reads: “viscoelasticity, creep and stress relaxation, oscillatory or dynamic testing, mechanical models, control of rheological characteristics through compounding, rubber curing in parallel plate viscometer, ODR and MDR.” The chapter teaches viscoelastic behaviour, creep and stress relaxation, the storage and loss moduli and tan δ, the Maxwell, Kelvin–Voigt and standard linear solid models with their responses, how compounding changes rheology, and the torque–time curve of the Mooney viscometer, ODR and MDR. The numericals are Maxwell stress decay and relaxation time, Kelvin–Voigt creep strain, standard-linear-solid modulus, and the loss modulus from the phase angle.

1. Viscoelasticity: between the solid and the liquid

An ideal elastic solid (Hooke) stores all the work done on it: σ = E ε, and the strain follows the stress instantly. An ideal viscous liquid (Newton) dissipates it all: σ = η dε/dt. A polymer does both at once, so its response depends on time: it stores part of the energy, like a spring, and dissipates part, like a dashpot. The effect is largest near Tg and in melts. Whether a polymer looks solid or liquid depends on the Deborah number De = λ/t_obs, the ratio of its relaxation time to the observation time: for De ≫ 1 it behaves as a solid (rubber under a fast impact), for De ≪ 1 as a liquid (the same rubber under slow loading, or a glass on a geological time scale).

Visible signs in melts are die swell (the extrudate swells on leaving the die because stored elastic strain recovers), rod climbing (the Weissenberg effect, the melt climbing a rotating rod because of normal stresses) and elastic recoil. A Newtonian liquid shows none of them. A crosslinked rubber, on the other hand, stores energy at any time scale and so creeps to a finite strain and relaxes to a finite stress, not to zero.

2. Creep and stress relaxation

Creep: a constant stress σ₀ is applied at t = 0 and the strain ε(t) is followed; the creep compliance is J(t) = ε(t)/σ₀. Recovery: after unloading, the elastic part returns at once, the viscoelastic part returns slowly, and the viscous (flow) part never returns, leaving a permanent set. Stress relaxation: a constant strain ε₀ is imposed and the stress σ(t) is followed; the relaxation modulus is E(t) = σ(t)/ε₀ (G(t) in shear). Creep is the practical test for load-bearing plastics, stress relaxation for seals, gaskets and rubber bands, which lose their clamping force.

Responses of the three kinds of material
MaterialCreep under constant stressRelaxation at constant strain
Elastic solidInstant strain, then constantStress stays constant
Viscous liquidStrain grows linearly for everStress falls to zero at once
Uncrosslinked polymerElastic jump, delayed elasticity, then steady flowStress decays to zero over a time of the order of λ
Crosslinked polymerStrain levels off at a finite valueStress levels off at a finite value

In the linear range (small strain) the response is proportional to the load, and the Boltzmann superposition principle holds: the response to a sequence of loads is the sum of the responses to each. This is what makes recovery after creep predictable from the creep curve itself, and it is lost at large strains. Creep and relaxation increase with temperature, and also increase at large stress.

3. Mechanical models: Maxwell, Kelvin–Voigt and the standard linear solid

The models combine a spring (modulus E, σ = E ε) and a dashpot (viscosity η, σ = η dε/dt). Maxwell model: spring and dashpot in series; the strains add and the stress is shared, giving dε/dt = σ̇/E + σ/η, or σ + λ σ̇ = η ε̇ with relaxation time λ = η/E. Under constant strain, σ(t) = σ₀ e^(−t/λ): the stress decays to 1/e (36.8 %) in a time λ. Under constant stress, ε(t) = σ₀/E + σ₀ t/η: an instant jump and then unbounded linear flow, so it is a viscoelastic liquid. Kelvin–Voigt model: spring and dashpot in parallel; the stress adds and the strain is shared, σ = E ε + η dε/dt. Under constant stress σ₀, ε(t) = (σ₀/E)(1 − e^(−t/τ)) with retardation time τ = η/E; the strain approaches σ₀/E, and after unloading it recovers as e^(−t/τ). It is a viscoelastic solid, but it cannot relax stress, since under fixed strain the stress stays at E ε₀.

The standard linear solid (Zener) cures both defects. In the form used here a spring E₁ is in parallel with a Maxwell arm (spring E₂ and dashpot η in series). Relaxation: σ(t) = ε₀ [E₁ + E₂ e^(−t/τ)] with τ = η/E₂: the stress falls from ε₀(E₁ + E₂) at t = 0 to a finite value ε₀ E₁, so it is a solid. Creep: ε(t) = σ₀ {1/E₁ − [E₂/(E₁(E₁ + E₂))] e^(−t/τ_c)} with τ_c = η (E₁ + E₂)/(E₁ E₂); the instantaneous strain is σ₀/(E₁ + E₂) and the final strain σ₀/E₁. Worked examples. (i) Maxwell, σ₀ = 100 kPa, λ = 20 s: at t = 30 s, σ = 100 e^(−1.5) = 22.3 kPa. (ii) Kelvin–Voigt, σ₀ = 2 MPa, E = 10 MPa, η = 200 MPa·s: τ = 20 s, ε(40 s) = 0.2 (1 − e^(−2)) = 0.173. (iii) SLS with E₁ = 2 MPa, E₂ = 3 MPa, η = 60 MPa·s: τ = 20 s and E(t = τ) = 2 + 3 e^(−1) = 3.10 MPa.

⚠️ Which model for which test
Maxwell describes stress relaxation well but creeps without limit, so it cannot describe a solid; Kelvin–Voigt describes creep and recovery of a solid but cannot relax stress. Real polymers have a spectrum of relaxation times, which is represented by many Maxwell elements in parallel, and a single time constant is only a first approximation.

4. Oscillatory (dynamic) testing: G′, G″ and tan δ

A sinusoidal strain γ = γ₀ sin ωt is applied in the linear range, and the stress is σ = σ₀ sin(ωt + δ), where δ is the phase angle. For a purely elastic solid δ = 0°, for a purely viscous liquid δ = 90°, and a polymer lies between. The stress splits into an in-phase and an out-of-phase part. Storage modulus G′ = (σ₀/γ₀) cos δ (elastic energy stored and returned); loss modulus G″ = (σ₀/γ₀) sin δ (energy lost as heat); complex modulus magnitude |G*| = σ₀/γ₀ = √(G′² + G″²); and loss tangent tan δ = G″/G′. The dynamic viscosity is η′ = G″/ω. The energy dissipated per unit volume per cycle is π G″ γ₀².

For the Maxwell model, G′ = G (ωλ)²/[1 + (ωλ)²] and G″ = G ωλ/[1 + (ωλ)²]. At low frequency (the terminal region) G′ ∝ ω² and G″ ∝ ω, the liquid signature; at high frequency G′ → G, a solid. G′ and G″ cross at ω = 1/λ, where tan δ = 1/(ωλ) = 1: with G = 1 MPa and λ = 2 s at ω = 0.5 rad/s, G′ = G″ = 0.5 MPa. A temperature sweep at fixed frequency (DMTA) shows G′ falling by about three decades through the glass transition while G″ and tan δ pass through peaks, the tan δ peak being commonly quoted as Tg; a crosslinked rubber keeps a flat G′ plateau above it. A frequency sweep gives the same picture by the time–temperature equivalence, and the Cox–Merz rule says that the complex viscosity |η*|(ω) is often close to the steady viscosity η(γ̇) at γ̇ = ω.

Worked example. A specimen is tested at a strain amplitude γ₀ = 0.001 and a stress amplitude σ₀ = 1.2 MPa with a phase lag δ = 20°. |G*| = 1.2/0.001 = 1200 MPa, G′ = 1200 cos 20° = 1127.6 MPa, G″ = 1200 sin 20° = 410.4 MPa and tan δ = tan 20° = 0.364. If G′ = 4.0 MPa and G″ = 1.0 MPa instead, tan δ = 0.25 and δ = 14.0°. In rubbers, a low tan δ at 60 °C goes with low rolling resistance, and a high tan δ at 0 °C with good wet grip.

5. Controlling rheological behaviour through compounding

Compounding is the route by which a rubber or plastic is made to flow as the process needs. Filler: rigid particles raise the viscosity, as Einstein showed for dilute spheres, η = η₀ (1 + 2.5 φ), and the Guth–Gold extension η = η₀ (1 + 2.5 φ + 14.1 φ²) for higher volume fractions φ; at φ = 0.2 it gives a factor of 1 + 0.5 + 0.564 = 2.06. Carbon black of high structure and fine size raises viscosity more and reduces die swell. At high loading a filler network forms and gives a yield-like behaviour and the Payne effect, the fall of G′ with rising strain amplitude as the network breaks. Plasticiser and process oil lower the viscosity and the Tg. Lubricants and processing aids lower the melt viscosity or promote slip and fusion. Molar mass and its distribution: a lower molar mass gives lower viscosity and lower die swell, while a broader distribution lowers the viscosity at processing rates (thinning starts earlier) but usually increases elasticity and die swell; peptisers and mastication cut chains to reduce Mooney viscosity; branching raises elasticity and melt strength.

ℹ️ Mooney viscosity
Mooney viscosity is reported as ML 1+4 (100 °C): the large rotor turns at 2 rpm in the heated, closed cavity of raw rubber, after 1 minute of preheating and a reading after 4 minutes, in Mooney units. A higher reading means a stiffer, higher molar mass or more filled stock. It is a single-point viscosity at a low rate and does not capture shear thinning.

6. Rubber curing in the parallel-plate viscometer, ODR and MDR

The Mooney viscometer is a parallel-plate (shearing-disc) instrument: a disc rotor turns slowly in a closed, heated cavity of rubber between two plates. Held at a vulcanising temperature (for example 120 to 135 °C), the viscosity first falls as the rubber warms, passes through a minimum and then rises when crosslinking starts. The scorch time t₅ is the time for the reading to rise 5 Mooney units above the minimum; it is a measure of processing safety, but the instrument does not follow the cure to completion. The ODR (oscillating disc rheometer) holds the sample in a heated biconical cavity and oscillates a disc embedded in it through a small arc (commonly 1°) at about 100 cycles/min; the MDR (moving die rheometer) has a sealed, pressurised biconical cavity with no rotor, the lower die oscillating (commonly 0.5° at about 1.7 Hz). The MDR needs a smaller sample, reaches the test temperature faster, and gives better repeatability, and it can run frequency and strain sweeps; the ODR is the older instrument. Both record torque (dN·m) against time, and the torque is proportional to the shear modulus of the sample.

Reading the torque–time curve
ParameterMeaningUse
MLMinimum torqueViscosity of the compound at test temperature, before cure
ts2Time to ML + 2 unitsScorch safety, time available for moulding flow
MHMaximum (plateau) torqueStiffness, MH − ML ∝ crosslink density
t90Time to ML + 0.9 (MH − ML)Optimum cure time, press time

Worked example. A curve has ML = 1.5 dN·m, MH = 18.5 dN·m, ts2 = 3.0 min and t90 = 11.0 min. The torque at t90 is 1.5 + 0.9 × 17.0 = 16.8 dN·m, the cure rate index is 100/(11.0 − 3.0) = 12.5 min⁻¹, and the torque rise is 17.0 dN·m. A compound with the same ML but MH = 24 dN·m has a higher crosslink density, and since modulus is proportional to crosslink density for an ideal rubber, G ≈ ν R T, its stiffness is higher in the same ratio, 22.5/17.0 = 1.32. A falling torque after MH is reversion, and a plateau that keeps rising is a marching cure.

Key takeaways

  • Polymers are viscoelastic: they store energy like a spring and dissipate it like a dashpot, so the response depends on time; the Deborah number De = λ/t_obs tells solid-like (large) from liquid-like (small). Die swell and rod climbing are signs of stored elasticity; a crosslinked rubber relaxes to a finite stress.
  • Creep compliance J(t) = ε/σ₀ and relaxation modulus E(t) = σ/ε₀. Maxwell (series): λ = η/E, σ = σ₀ e^(−t/λ), creeps without limit. Kelvin–Voigt (parallel): τ = η/E, ε = (σ₀/E)(1 − e^(−t/τ)), cannot relax. Standard linear solid: E(t) = E₁ + E₂ e^(−t/τ), a solid that relaxes to E₁.
  • Dynamic testing: G′ = (σ₀/γ₀) cos δ is the stored energy, G″ = (σ₀/γ₀) sin δ the lost energy, tan δ = G″/G′, and the energy lost per cycle is π G″ γ₀². The Maxwell model gives G′ = G (ωλ)²/(1 + (ωλ)²) and G″ = G ωλ/(1 + (ωλ)²), crossing at ω = 1/λ.
  • Fillers raise viscosity (Einstein 1 + 2.5φ, Guth–Gold with 14.1φ²) and carbon black structure cuts die swell; plasticisers, oils and processing aids lower viscosity; a lower or broader molar mass distribution gives lower viscosity at processing rates; peptisers and mastication reduce Mooney viscosity.
  • The Mooney viscometer (parallel-plate disc) gives the raw viscosity ML 1+4 and the scorch time t5; the ODR and MDR record torque against time at the cure temperature, and the MDR, with a sealed cavity and small sample, is quicker and more repeatable.
  • Cure curve: ML, ts2, MH, t90 (torque at t90 = ML + 0.9 (MH − ML)), CRI = 100/(t90 − ts2); torque is proportional to modulus, so MH − ML tracks crosslink density; a falling torque after MH is reversion, a rising one is a marching cure.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The Deborah number is the ratio of a material’s relaxation time to the observation time. For a polymer tested at a very high De, how does it behave?

    1. Like a viscous liquid, since it has no time to store energy in the spring
    2. Like an elastic solid, since it cannot relax during the observation
    3. Like a Newtonian liquid with a lower viscosity than at low De
    4. Like a perfectly plastic solid with a fixed yield stress
    Show answer

    Answer: B — Like an elastic solid, since it cannot relax during the observation

    A high De means the relaxation time is long compared with the observation time, so the material has no chance to flow and responds elastically, like rubber under a fast impact. The liquid-like response belongs to low De.

  2. In the Maxwell model, a spring of modulus E and a dashpot of viscosity η are joined in series. What is the relaxation time?

    1. λ = E η
    2. λ = E/η
    3. λ = η/E
    4. λ = 1/(E η)
    Show answer

    Answer: C — λ = η/E

    In the Maxwell model σ(t) = σ₀ e^(−t/λ) with λ = η/E, which has the dimension of time (Pa·s divided by Pa). The inverted ratio E/η has the unit of a rate, and the products do not have the dimension of time.

  3. A strip of uncrosslinked polymer and a strip of lightly crosslinked rubber are each held at a fixed strain for a very long time. What happens to the stress in each?

    1. The stress in both decays to zero, because all polymers eventually flow
    2. The stress in both stays unchanged, because strain is held constant
    3. The stress in the uncrosslinked strip levels off at a finite value and in the rubber it decays to zero
    4. The uncrosslinked strip relaxes to zero; the rubber levels off at a finite stress
    Show answer

    Answer: D — The uncrosslinked strip relaxes to zero; the rubber levels off at a finite stress

    Without crosslinks the chains can slip through entanglements, so the stress relaxes to zero; the permanent network of the rubber holds an equilibrium stress. The claim that both decay to zero ignores crosslinks, and constant stress at constant strain would mean no relaxation at all.

  4. Which test is the Kelvin–Voigt model unable to describe, even qualitatively?

    1. Creep under a constant stress
    2. Recovery of strain after the load is removed
    3. Stress relaxation under a constant strain
    4. Delayed approach to an equilibrium strain
    Show answer

    Answer: C — Stress relaxation under a constant strain

    In the parallel arrangement, a held strain ε₀ gives a constant stress E ε₀ once the dashpot stops moving, so stress does not decay as the model has no way to relax. It does describe creep as ε = (σ₀/E)(1 − e^(−t/τ)), the delayed approach and the recovery.

  5. A polymer is tested in oscillatory shear and gives a storage modulus G′ = 4.0 MPa and a loss modulus G″ = 1.0 MPa. What is tan δ?

    1. 4.0
    2. 0.25
    3. 0.80
    4. 0.20
    Show answer

    Answer: B — 0.25

    tan δ = G″/G′ = 1.0/4.0 = 0.25, so δ is about 14°. The value 4.0 is the inverse ratio G′/G″, and 0.20 would be G″/(G′ + G″), a wrong combination.

  6. For a single-element Maxwell fluid with plateau modulus G and relaxation time λ, at the angular frequency ω = 1/λ the storage modulus G′ and the loss modulus G″ are

    1. G′ = G and G″ = 0, as a solid at high frequency
    2. G′ = 0 and G″ = G, as a liquid
    3. both equal to G/2, so tan δ = 1
    4. G′ = G/2 and G″ = G, so tan δ = 2
    Show answer

    Answer: C — both equal to G/2, so tan δ = 1

    With x = ωλ = 1, G′ = G x²/(1 + x²) = G/2 and G″ = G x/(1 + x²) = G/2, so tan δ = 1; this is the crossover. The limits G′ → G and G″ → 0 apply only for x ≫ 1.

  7. The viscosity of a filled rubber compound is estimated by the Guth–Gold relation η = η₀ (1 + 2.5 φ + 14.1 φ²), where φ is the filler volume fraction. By what factor does the viscosity rise for φ = 0.20?

    1. 2.06
    2. 1.50
    3. 2.50
    4. 3.82
    Show answer

    Answer: A — 2.06

    The factor is 1 + 2.5 × 0.20 + 14.1 × 0.04 = 1 + 0.50 + 0.564 = 2.06. Using only the Einstein term gives 1.50, which ignores the particle interactions that dominate at this loading.

  8. What is the main advantage of a moving die rheometer (MDR) over an oscillating disc rheometer (ODR)?

    1. It tests a much larger sample, so the result is an average over a bigger volume
    2. It has no heated cavity, so the sample stays at room temperature during the cure
    3. It measures the Mooney viscosity of raw rubber directly in Mooney units
    4. It has a sealed cavity and a small sample, so it heats fast and repeats better
    Show answer

    Answer: D — It has a sealed cavity and a small sample, so it heats fast and repeats better

    The MDR's closed, pressurised cavity, small sample volume and rotor-free design give fast thermal equilibrium and better repeatability. Both instruments use a heated cavity and measure torque, and Mooney units belong to the Mooney viscometer.

  9. An NR compound cured with a conventional sulphur system shows a torque that falls steadily after the maximum, MH, at a high test temperature. What is this, and what is its cause?

    1. Reversion, caused by breakdown of heat-labile polysulphidic crosslinks
    2. Scorch, caused by too much accelerator
    3. Marching cure, caused by crosslinks still being formed
    4. A wall-slip artefact of an unfilled compound, which has no chemical cause
    Show answer

    Answer: A — Reversion, caused by breakdown of heat-labile polysulphidic crosslinks

    A falling torque after MH is reversion, the loss of crosslink density when long polysulphidic bonds break down in heat; EV or semi-EV systems reduce it. A marching cure is a torque that keeps rising, and scorch refers to premature crosslinking at the start.

  10. Which statements about creep, relaxation and the simple models are correct?

    1. The creep compliance is J(t) = ε(t)/σ₀ for a constant applied stress σ₀
    2. A crosslinked polymer relaxes to a finite stress under a constant strain
    3. A Maxwell element under constant stress approaches a finite limiting strain σ₀/E
    4. A Kelvin–Voigt element under constant strain relaxes its stress to zero
    Show answer

    Answer: A — The creep compliance is J(t) = ε(t)/σ₀ for a constant applied stress σ₀; B — A crosslinked polymer relaxes to a finite stress under a constant strain

    The first two statements are standard definitions and behaviour. A Maxwell element under constant stress keeps flowing: ε = σ₀/E + σ₀ t/η grows without limit. A Kelvin–Voigt element holds the stress E ε₀ under fixed strain and does not relax it.

  11. Which statements about oscillatory testing are correct?

    1. The phase angle of a perfectly elastic solid is 90°
    2. The loss modulus is G″ = (σ₀/γ₀) sin δ
    3. The energy dissipated per cycle is proportional to the storage modulus G′
    4. tan δ is the ratio G″/G′
    Show answer

    Answer: B — The loss modulus is G″ = (σ₀/γ₀) sin δ; D — tan δ is the ratio G″/G′

    G″ is the out-of-phase component and tan δ = G″/G′. A perfectly elastic solid has δ = 0°, and the energy lost per cycle, π G″ γ₀², goes with the loss modulus, not the storage modulus, which measures the energy returned.

  12. Which statements about a rheometer cure curve are correct?

    1. ML reflects the viscosity of the uncured compound at the test temperature
    2. t90 is the time to reach ML + 0.9 (MH − ML)
    3. The scorch time ts2 is measured from the maximum torque MH
    4. MH − ML is a relative measure of the crosslink density
    Show answer

    Answer: A — ML reflects the viscosity of the uncured compound at the test temperature; B — t90 is the time to reach ML + 0.9 (MH − ML); D — MH − ML is a relative measure of the crosslink density

    These three are the definitions of the curve parameters. The scorch time is measured from the start of the test, as the time for the torque to rise 2 units above ML, long before MH is reached.

  13. A Maxwell element is held at a constant strain so that the initial stress is σ₀ = 100 kPa. Its relaxation time is λ = 20 s and σ(t) = σ₀ exp(−t/λ). Find the stress at t = 30 s, in kPa, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 22.3

    σ = 100 × e^(−30/20) = 100 × e^(−1.5) = 100 × 0.2231 = 22.3 kPa. Taking the 20 s as the time to zero stress, or using e^(−0.5), are the common errors.

  14. In a stress-relaxation test on a Maxwell material the stress falls from 80 kPa to 30 kPa in 12 s. Using σ(t) = σ₀ exp(−t/λ), find the relaxation time λ in seconds, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 12.2

    ln(σ₀/σ) = ln(80/30) = ln 2.667 = 0.9808, so λ = t/ln(σ₀/σ) = 12/0.9808 = 12.23 s, which rounds to 12.2 s. The decay to 1/e (36.8 %) takes exactly λ, and 30/80 = 37.5 % is close to that, which is a good check.

  15. A Kelvin–Voigt element has E = 10 MPa and η = 200 MPa·s and is loaded with a constant stress σ₀ = 2 MPa. Its creep strain is ε(t) = (σ₀/E)(1 − e^(−t/τ)) with τ = η/E. Find the strain at t = 40 s as a percentage, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 17.29

    τ = 200/10 = 20 s, so t/τ = 2. σ₀/E = 0.2, ε = 0.2 × (1 − e^(−2)) = 0.2 × 0.8647 = 0.1729, which is 17.29 %. The limiting strain is 20 %, and the element has reached 86.5 % of it.

  16. In an oscillatory shear test the stress amplitude is σ₀ = 1.2 MPa, the strain amplitude is γ₀ = 0.001 and the stress leads the strain by a phase angle δ = 20°. Using G″ = (σ₀/γ₀) sin δ, find the loss modulus in MPa, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 410.4

    |G*| = σ₀/γ₀ = 1.2 MPa/0.001 = 1200 MPa, so G″ = 1200 × sin 20° = 1200 × 0.3420 = 410.4 MPa. Using cos 20° gives the storage modulus, 1127.6 MPa.

  17. A standard linear solid has a spring E₁ = 2 MPa in parallel with a Maxwell arm made of a spring E₂ = 3 MPa and a dashpot η = 60 MPa·s in series. Its relaxation modulus is E(t) = E₁ + E₂ exp(−t/τ), with τ = η/E₂. Find E(t) at t = τ, in MPa, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 3.10

    τ = 60/3 = 20 s, so at t = τ the exponential is e^(−1) = 0.3679 and E = 2 + 3 × 0.3679 = 3.10 MPa. The modulus starts at E₁ + E₂ = 5 MPa and tends to E₁ = 2 MPa, never to zero.