advanced28 min studyLesson 170 in Physics

Viscoelastic Response and Relaxation

Derive Maxwell stress relaxation, distinguish it from creep, and interpret a relaxation time without confusing it with oscillator damping.

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What you’ll understand

  • Build a Maxwell constitutive equation from a series spring and dashpot.
  • Calculate stress relaxation and constant-stress creep with correct units.
  • Distinguish a constitutive relaxation time from a mechanical oscillator’s damping ratio.

Hold a stretched polymer at the same length and the force can decrease with time. Hold the force instead and the extension can increase. These are different experiments: stress relaxation fixes strain, whereas creep fixes stress. A material that stores energy elastically and dissipates energy through delayed rearrangement can exhibit both. The Maxwell spring–dashpot model gives a precise first example, with deliberately limited realism, and lets us derive the time dependence rather than describe every fading signal as the same kind of damping.

A series connection imposes a constitutive equation

Consider a one-dimensional elastic spring and a Newtonian dashpot connected in series. Here their variables are stress and strain, so the spring stiffness parameter is a modulus in pascals rather than an ordinary spring constant in newtons per metre. Both elements transmit the same stress, while their strains add. The spring obeys a linear stress–strain law and the dashpot obeys a linear stress–strain-rate law. The dashpot viscosity consequently has units of pascal seconds.

Differentiate the sum of strains and replace each element’s strain rate using its own law. The result is the Maxwell differential constitutive equation. It specifies the relation between a material point’s stress history and strain history; it is not Newton’s equation for an attached moving mass. The characteristic relaxation time is viscosity divided by modulus. Positive values of both parameters permit elastic storage and positive viscous dissipation. A single pair of constants is an approximation over a chosen strain, temperature, and time range, not a universal description of every polymer.

ε˙=σ˙E+ση,σ˙+στ=Eε˙,τ=ηE\dot\varepsilon=\frac{\dot\sigma}{E}+\frac{\sigma}{\eta},\qquad \dot\sigma+\frac{\sigma}{\tau}=E\dot\varepsilon,\qquad \tau=\frac{\eta}{E}
E is a modulus in Pa, η is a viscosity in Pa·s, and τ is a time in seconds. The model is linear, one-dimensional, and uses a series spring–dashpot connection.↔ Scroll wide equations horizontally.

Relaxation and creep impose different boundary histories

Apply an ideal instantaneous strain step and then keep the strain fixed. The dashpot cannot acquire a finite instantaneous strain under finite stress, so the initial strain is carried by the spring. After the step, total strain rate is zero, and the constitutive equation becomes a first-order decay equation for stress. Stress falls exponentially toward zero because strain transfers from the spring into the dashpot. The relaxation modulus is stress divided by the imposed strain; it has units of pascals and is not a dimensionless fraction.

Now impose constant stress instead. After the spring’s initial extension, the dashpot continues to extend at a constant rate. This Maxwell creep strain grows without bound for an indefinitely sustained nonzero stress. That is appropriate for a simple viscoelastic fluid model, but many viscoelastic solids retain a nonzero equilibrium stiffness. A parallel equilibrium spring added to a Maxwell branch creates a standard linear solid and a nonzero long-time relaxation plateau. Do not fit a decaying-to-zero model to such a plateau and interpret the discrepancy as mere measurement noise.

σ(t)=Eε0e−t/τ after a strain step,ε(t)=σ0E+σ0ηt after a stress step\sigma(t)=E\varepsilon_0e^{-t/\tau}\ \text{after a strain step},\qquad \varepsilon(t)=\frac{\sigma_0}{E}+\frac{\sigma_0}{\eta}t\ \text{after a stress step}
Both expressions apply for t ≥ 0 after an ideal step from the unstressed reference state. The first holds strain fixed; the second holds stress fixed.↔ Scroll wide equations horizontally.

Compare time scales without mixing physical models

A relaxation time only becomes informative relative to an observation or loading time. Watch a Maxwell material for a duration much shorter than its relaxation time and little stress has relaxed. Wait many relaxation times and the stress has nearly disappeared under fixed strain. Temperature and molecular structure can change the relevant parameters substantially, and real materials often need a spectrum of relaxation times. Linear viscoelasticity also permits superposition of small loading histories, but that does not justify extending the same constants through yielding or very large strains.

Use the oscillator lab to investigate an analogy with an explicit boundary. At fixed natural frequency, predict that increasing damping ratio from 0.2 to 0.5 will make the underdamped envelope decay faster, then compare the histories. The lab solves a second-order unit-mass oscillator released from unit displacement and zero velocity. Its position can overshoot and oscillate, whereas Maxwell stress under the specified strain step decays monotonically. The damping-ratio slider is not a viscosity control, and its natural-frequency slider is not a modulus measurement. The shared lesson is that a decay needs a governing equation and a time scale, not that the models are interchangeable.

Stress remaining after one relaxation time

An illustrative Maxwell material has E = 8 MPa and η = 40 MPa·s. Apply strain ε₀ = 0.02 instantaneously, then hold it fixed. Find its relaxation time and its stress at 5 s and 10 s.

  1. Divide viscosity by modulus: τ = 40/8 = 5 s. The MPa units cancel, leaving seconds.

  2. The initial stress is Eε₀ = 8 × 0.02 = 0.16 MPa. At 5 s, t/τ = 1, so stress is 0.16/e ≈ 0.0589 MPa.

  3. At 10 s, t/τ = 2. Stress is 0.16e⁻² ≈ 0.0217 MPa, or 21.7 kPa. The imposed total strain remains 0.02 throughout the hold.

The result

After one relaxation time, about 36.8% of the initial stress remains; after two, about 13.5% remains. These percentages describe the specified fixed-strain Maxwell experiment.

Explore the model

Interactive lab

Interactive lab · oscillator

Compare a decay with its governing model

At fixed natural frequency, compare damping ratios 0.2 and 0.5 and predict which underdamped envelope decays faster. The displayed motion is a unit-mass oscillator with fixed initial conditions, not Maxwell stress relaxation.

Compare a decay with its governing modelUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDissipation can appear in several equations. An oscillator’s damping ratio and a Maxwell material’s relaxation time must be interpreted within their own models.

Check your understanding

For the stated Maxwell model with positive parameters, which experiment produces σ(t) = Eε₀ exp(−t/τ) after the loading step?

Try it yourself

Use the worked material under a constant applied stress of 0.08 MPa. Find its strain immediately after the stress step and after a 10 s hold. Does this contradict the relaxation calculation?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For creep, add the instantaneous spring strain σ₀/E to the dashpot strain σ₀t/η. The controlled variable has changed.

Compare with the solution

The initial spring strain is 0.08/8 = 0.01. The dashpot accumulates 0.08 × 10/40 = 0.02 over 10 s, so total strain is 0.03. There is no contradiction: this calculation holds stress fixed, while the earlier one held strain fixed. A material law must be combined with the actual loading history to predict either experiment.

Take it with you

  • Stress relaxation fixes strain; creep fixes stress. They are different loading histories for the same constitutive law.
  • For a Maxwell element, τ = η/E has units of time and fixed-strain stress decays to zero.
  • A solid with a finite long-time stiffness needs a richer model; oscillator damping is a separate dynamical problem.

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