Viscoelasticity, temperature and time

Response depends on the clock

A polymer may deform gradually under a constant load: creep. If held at a fixed deformation, its stress can decrease: stress relaxation. Both arise from time-dependent rearrangements; they are different experiments with different controlled quantities.

A stretched polymer strip can recover some length immediately when released and more over the following minute. The length still present just after release is therefore not automatically permanent deformation. Conversely, a strip may flow or become damaged and fail to recover fully. Observe the load, deformation and recovery history before deciding which behaviour occurred.

Engineering stress here is σ = F/A₀ and engineering strain is ε = u/L₀, using original area A₀, original gauge length L₀ and change u between the gauge marks. In a small-deformation creep test, hold the applied stress σ₀ and record ε(t). With fixed A₀ that is also fixed force. If the actual area changes appreciably, constant force is not constant true stress F/A(t). In relaxation, hold a chosen strain ε₀ and record σ(t).

Creep compliance J(t) = ε(t)/σ₀ belongs to the specified stress-step test and has inverse-stress units. Relaxation modulus E_r(t) = σ(t)/ε₀ belongs to the strain-step test and has stress units. They are different measured functions, not automatically reciprocals at each time. “At 10 seconds” has meaning only with a defined start and history.

Fast distortion and slower rearrangement share the deformation

A small change in bond lengths and angles can give a fast elastic response. Polymer segments can also rotate around permitted backbone bonds and rearrange their conformations, the spatial shapes of the chains. Such motion requires suitable local configurations and can involve neighbouring segments moving cooperatively. It takes time to explore those configurations; a sustained mechanical bias changes how the rearrangements contribute to strain or stress.

Chains do not have to break for a delayed response. Temporary entanglements restrict how a chain can move past other chains. Stable chemical crosslinks connect chains more persistently; crystalline regions can also constrain motion while they remain intact. These constraints can support a remaining elastic response even after faster rearrangements have occurred. If constraints can eventually be escaped, long-time flow can instead become possible. The relevant times can span a broad range.

Higher temperature often makes a given rearrangement faster by increasing molecular mobility. A rapid test or a colder test can leave a slow process effectively unable to respond; a longer or warmer test can allow it to contribute. This is a comparison of a process time with the observation time, not a rule that all bonds stop moving below a glass transition.

Near a glass-transition range, segmental motion in amorphous regions becomes consequential on the measurement clock. The observed transition depends on the response and protocol. A crosslinked network need not become a freely flowing liquid above that range; a semicrystalline polymer also has crystalline constraints until they change or melt. Glasses, disorder and glass transition develops the thermal/structural observation problem. Here the question is mechanical response under a controlled load or deformation.

Viscoelastic dissipation means that some mechanical work is not recovered on the same mechanical cycle. It can become heat through molecular rearrangement. That does not imply that every delayed strain is permanent. Conversely, a damaging or nonlinear experiment need not obey the small, linear models developed next.

A model with an elastic load path and a relaxing load path

Use a constitutive analogue, not a literal drawing of molecular hardware. An elastic spring with modulus E∞ acts in parallel with a second branch containing a spring E₁ and a viscous dashpot in series. Parallel branches have the same total strain, and their stress contributions add. Within the series branch the spring and dashpot share branch stress s, while their strains add. This is the Maxwell form of the standard linear solid.

The equilibrium branch contributes E∞ε. In the series branch, spring strain is s/E₁ and dashpot strain rate is s/η, where η is a positive viscosity parameter in Pa·s or MPa·s. Thus total stress is σ = E∞ε + s, and the common strain rate satisfies dε/dt = (1/E₁)(ds/dt) + s/η. The symbol d/dt means rate of change per time; no new measurement is implied by writing the equation.

For the original teaching model, set E∞ = 20.0 MPa, E₁ = 80.0 MPa and η = 800 MPa·s at a reference temperature T_ref = 300 K. The rapid-response modulus is E₀ = E∞ + E₁ = 100 MPa. The series branch has relaxation time τ_r = η/E₁ = 10.0 s. The units matter: 800 MPa·s = 8.00 × 10⁸ Pa·s, so either consistent unit system gives the same time.

All values are fictional, not measured properties of a named polymer. Assume homogeneous small uniaxial strain, a previously relaxed state, constant parameters at each stated temperature, and no damage, ageing, crystallisation, mass uptake or thermal expansion. Linearity is stipulated over the illustrated strains up to 1%; it is not established for real polymers by calling a strain small. Rapid load steps are ideal limits compared with the stated times, not instantaneous real-machine instructions.

The nonzero E∞ branch makes this model a solid with bounded creep and eventual recovery in the absence of damage. A thermoplastic that flows on longer times would need another description. The dashpot dissipates rather than creates energy: its stress times its strain rate is s²/η ≥ 0. Neither spring represents one identifiable population of actual bonds, and one fitted time does not identify a unique molecular mechanism.

At fixed strain, stress transfers out of the relaxing branch

Impose ε₀ = 0.0100, or 1.00%, and hold it. During the ideal strain step the dashpot cannot acquire a finite strain in zero time under finite stress, so both springs initially carry load. The initial stress is E₀ε₀ = 1.00 MPa. Subsequently the dashpot deforms, the series spring relaxes, and that branch’s stress falls. The E∞ spring remains at the held strain, so its 0.200 MPa contribution remains.

Setting dε/dt = 0 in the series equation gives ds/dt = −s/τ_r. Therefore s(t) = E₁ε₀ exp(−t/τ_r), and σ(t) = [E∞ + E₁ exp(−t/τ_r)]ε₀. Here exp means the exponential function. After each τ_r the remaining series stress is multiplied by exp(−1) ≈ 0.368. The exponent t/τ_r is a dimensionless time ratio.

At 10.0 s, σ = 0.200 + 0.800 exp(−1) ≈ 0.49430 MPa. At 50.0 s it is about 0.20539 MPa, approaching the model’s 0.200 MPa limit. One relaxation time removes about 63.2% of the transient 0.800 MPa contribution, not 63.2% of the entire initial stress. Stress has not fallen to zero and the material has not been assigned a failure time.

For a uniform model coupon with A₀ = 10.0 mm² and L₀ = 50.0 mm, the held gauge extension is 0.500 mm. Force falls from 10.0 N initially to 4.943 N at 10 s, tending to 2.00 N. The original seal case describes the same type of control, but an actual confined seal has a different geometry and stress state; these uniaxial numbers are not seal-contact-pressure predictions.

At fixed stress, strain grows and can recover later

Now use a separate initially relaxed specimen of the same model. Rapidly apply σ₀ = 0.200 MPa and hold it for 50.0 s, then unload. This is not the same history as holding 1% strain. At the stress step, the rapid modulus gives ε(0+) = σ₀/E₀ = 0.00200, or 0.200%. The plus sign means just after the step.

During the constant-stress hold, s = σ₀ − E∞ε. Substitute its rate ds/dt = −E∞ dε/dt into the branch equation: dε/dt = −(E∞/E₁)dε/dt + (σ₀ − E∞ε)/η. Collecting terms gives dε/dt = (σ₀/E∞ − ε)/τ_c, with τ_c = η(E∞ + E₁)/(E∞E₁) = τ_r E₀/E∞ = 50.0 s. Creep and relaxation have different characteristic times even in this single three-element model.

The resulting strain is ε(t) = σ₀/E∞ − σ₀(1/E∞ − 1/E₀) exp(−t/τ_c). It starts at σ₀/E₀ and tends to σ₀/E∞ = 0.0100. At 50.0 s, ε ≈ 0.00705696 = 0.705696%. A reader can use these endpoint checks and the exponential without repeating the rate-equation derivation.

On rapid unloading at t_h = 50.0 s, strain drops by σ₀/E₀ = 0.00200. Thus ε(t_h+) ≈ 0.00505696 = 0.505696%. The dashpot strain is continuous, but stress redistribution changes both spring strains. Do not subtract the long-time 1% strain or assume that all of the strain disappears immediately.

After unloading the applied stress is zero, but the internal stresses need not be: s = −E∞ε. The equilibrium spring drives the series branch back while the dashpot dissipates stored spring energy. The same equation now gives ε(t) = ε(t_h+) exp[−(t − t_h)/τ_c]. At total time 100 s, 50 s after release, the remaining strain is about 0.00186035 = 0.186035%. It tends to zero in this model. The strain at 50 s just after release is delayed recovery, not a permanent set.

For the 50.0 mm gauge, extension is 0.100 mm initially, 0.352848 mm just before release, 0.252848 mm just after, and 0.093018 mm at total time 100 s. The held force was 2.00 N, then zero. In a real test, finite ramp times, fixture movement, temperature and recovery duration belong in the record; an unrecovered short record alone cannot decide between very slow recovery, flow and damage.

Temperature changes accessible motion

Higher temperature often speeds molecular relaxation, but phase transitions or degradation can change mechanisms. A short-time modulus cannot automatically describe a long-duration load. Time–temperature shifting is useful only where its assumptions hold; it is not permission to merge all temperatures into one curve.

A restricted time–temperature superposition assumes that temperature changes the clock without changing the relevant curve shape. In a multi-time model, all relevant times must shift by the same factor for a simple horizontal shift to work. End moduli may also need a separate justified vertical correction in real data. One cannot hide a changed plateau or another mechanism by moving the time labels.

For a deliberately simple second model condition at T = 320 K, keep E∞ and E₁ exactly unchanged but set η = 160 MPa·s. Define a_T = η(T)/η(300 K) = 0.200. Both τ_r and τ_c are then multiplied by 0.200: they become 2.00 s and 10.0 s. This fivefold speed change is stipulated, not a universal effect of a 20 K warming and not an Arrhenius or WLF fit for an actual polymer.

Write reference-equivalent time θ = t/a_T. A response after 10.0 s at 320 K equals this model’s response after 50.0 s at 300 K under the corresponding same control and initial state. For fixed 1% strain the warm stress at 10 s is 0.20539 MPa. For a still-loaded 0.200 MPa creep test, warm strain at 10 s is 0.705696%. These remain different experiments; shifting time does not exchange their controls.

The third figure uses relaxation at fixed strain. A warm point at 2 s maps to a reference point at 10 s: both have stress about 0.49430 MPa. Moving the warm record to the reference-time axis increases its time coordinate by five. Stating a_T and θ explicitly avoids the ambiguous instruction “shift left” without naming the axis or reference.

If this same ideal model changes only viscosity during a constant-stress history, add the reference-equivalent durations of the stages. Ten seconds at 300 K followed by 8 s at 320 K gives θ = 10/1 + 8/0.200 = 50 s. The resulting strain is 0.705696%, even though only 18 s of laboratory time elapsed. This optional clock calculation excludes thermal expansion, gradients, structural change and changes in moduli; it is not a heating prescription.

To use shifting on actual data, test overlap under matched strain/stress range, prior history, state, geometry and temperature control. Changes in crystallinity, ageing, absorbed species, damage or chemical degradation can change the response rather than merely its clock. Different relaxation processes can also shift differently. A successful overlap over a finite window is evidence for that window, not a lifetime or unlimited-temperature guarantee.

Match the response to the actual use

A strap held under nearly constant force needs a creep question; a gasket held at fixed compression needs a relaxation question. A clamp whose separation and force both change needs the stiffness and motion of the rest of the assembly as well. A free material curve does not determine every assembled part’s history.

Even this model demonstrates why a single “modulus at 10 seconds” is insufficient. In relaxation, E_r(10) ≈ 49.4304 MPa; in creep, J(10) ≈ 0.0172508 MPa⁻¹. J(10) is not 1/E_r(10): the latter is about 0.0202305 MPa⁻¹. Each value arises from a different controlled history. They agree as reciprocal limits at the instantaneous and equilibrium endpoints here, but not at intermediate times.

A linear-viscoelastic comparison should also test whether multiplying a small load multiplies the whole strain response without changing its shape or characteristic times. If it does not, a linear model may be inadequate even though each trace is smooth. Repeated loading can introduce additional history, heating or damage; the equations above do not establish cyclic life.

Report composition and conditioning, relevant structure and direction, specimen region/geometry, ramp and hold histories, actual temperature, recovery duration and measurement uncertainty. Polymer chains and molecular architecture and Crystallinity, crosslinks and network formation explain structural constraints; Sustained loads and creep mechanisms later compares polymer response with other time-dependent deformation routes.

Worked examples

Try the reasoning

1. Which quantity is fixed in ideal creep?

   Show the answer and reasoning

   Applied stress or load as specified.

   Deformation is then monitored over time. For engineering stress F/A₀ a fixed force keeps that nominal quantity fixed. A changing current area can still change true stress.

2. Can a single loading speed describe every use?

   Show the answer and reasoning

   No.

   Different relaxation times become relevant. A short test may leave a delayed mode unexpressed; a different temperature or longer hold can reveal it. No single loading speed establishes the full response or eventual recovery.

3. Use the same 300 K model and 0.200 MPa creep hold, but A₀ = 4.00 mm² and L₀ = 80.0 mm. Find force and extensions at 0+ and 50−. Did changing the gauge alter τ_c?

   Show the answer and reasoning

   Force is 0.800 N; extensions are 0.160 mm initially and about 0.564557 mm at 50−. τ_c stays 50.0 s.

   F = σA₀ = 0.200 × 4.00. The material strains stay 0.00200 and 0.00705696; multiplying by 80.0 mm changes extension. The original model moduli and viscosity did not change. Geometry does not by itself alter this constitutive time.

4. A worksheet subtracts the long-time 0.0100 strain from the 50− creep strain and reports negative strain immediately after release. What response should it use?

   Show the answer and reasoning

   Use the rapid unloading jump σ₀/E₀ = 0.00200, leaving positive strain 0.00505696.

   A finite stress step cannot instantly change the dashpot strain. The 1/E∞ compliance is the model’s long-time limit, not the instantaneous unloading compliance. Subsequent zero-load recovery supplies the further decrease.

5. At 10 s, the model has E_r ≈ 49.4304 MPa in relaxation and J ≈ 0.0172508 MPa⁻¹ in creep. A learner sets J = 1/E_r. Calculate that proposed inverse and compare.

   Show the answer and reasoning

   1/E_r ≈ 0.0202305 MPa⁻¹, which differs from the creep compliance 0.0172508 MPa⁻¹.

   The tests have different controlled histories and τ_c = 50 s versus τ_r = 10 s. In this model reciprocal endpoint limits do not imply a pointwise reciprocal function throughout the changing response.

6. At 300 K, keep the 0.200 MPa creep load for only 10 s before removing it. Find the strain immediately before and after release. Why is it smaller after release than the 50 s-hold case?

   Show the answer and reasoning

   Before release, ε ≈ 0.00345015 (0.345015%); just after, ε ≈ 0.00145015 (0.145015%).

   Use 0.0100 − 0.00800 exp(−10/50), then subtract the same immediate 0.00200 unloading jump. The shorter hold allowed less delayed deformation. Subsequent recovery remains governed by τ_c in this model; the remaining value is not automatically permanent.

7. Two real relaxation records begin near 100 MPa modulus, but their established long-time plateaux are 20 and 10 MPa. Can a horizontal rescaling of time alone make both entire curves identical?

   Show the answer and reasoning

   No. A horizontal shift changes when a response occurs, not its endpoint values.

   A justified vertical correction or a different state/model might be needed. Heating may change density or structure as well as mobility; overlap over one small interval does not erase incompatible full-curve limits.

8. A different polymer is tested 20 K warmer. May you automatically use this lesson’s a_T = 0.200, or interpret a few seconds of remaining strain as damage?

   Show the answer and reasoning

   Neither conclusion follows. The shift factor and recovery interpretation need material- and history-specific evidence.

   The fivefold clock change was an invented input. Real polymers can have different processes, phase changes, physical ageing, uptake or degradation. A finite residual can reflect slow recoverable response, flow or damage; use the actual control and recovery record.

9. Within the model’s stated linear range, halve the creep stress to 0.100 MPa while keeping the 50 s hold and all parameters. Predict the 50− strain and τ_c. What would a changed curve shape in actual data suggest?

   Show the answer and reasoning

   The 50− strain halves to about 0.00352848 (0.352848%); τ_c stays 50.0 s. A changed normalized shape would challenge this linear fixed-parameter description.

   The entire strain response is proportional to the imposed stress under the model. Halving stress does not halve η/E or the characteristic time. A real comparison must control temperature and prior state before attributing the difference to nonlinear response.

What to carry forward

• State what is held fixed.
• Match test duration and temperature to use.
• Fast elastic distortion and slower constrained molecular rearrangements can contribute to one mechanical response.
• A recoverable delayed response and a permanent set are different claims; observe unloading and sufficient recovery.
• The three-element model has τ_r = η/E₁ and τ_c = τ_r E₀/E∞. One clock label cannot be exchanged between different tests.
• A time–temperature shift must state its convention and its unchanged-state/shape assumptions; the fictional factor here is not a universal polymer law.