Viscoelastic materials

Hold one quantity and watch the other

Imagine a small specimen pulled in tension. First hold its tensile stress constant and watch its extension. If the extension continues to grow, the specimen creeps. In a separate test, pull it to a chosen extension and hold that extension fixed. If the tensile stress needed to maintain it decreases, the specimen undergoes stress relaxation. The words identify different controlled quantities, so the two tests give different response curves.

Viscoelastic response combines elastic effects with time-dependent motion. Polymers can exhibit this behaviour because deformation includes both relatively rapid changes and slower rearrangements of their molecular structure. A spring–dashpot model represents the measured response without requiring each drawn element to correspond to a particular molecule. The appropriate response depends on the material, temperature and observation time.

Use original length L₀ = 100 mm = 0.100 m and original area A₀ = 100 mm² = 1.00 × 10⁻⁴ m². Engineering strain is ε = ΔL/L₀, and engineering stress is σ = F/A₀, positive in tension. A 0.100 mm extension gives ε = 0.00100 = 0.100% = 1000 microstrain. A 0.100 N tensile force gives σ = 1000 Pa = 1.00 kPa. One microstrain is 10⁻⁶; the percent sign requires a separate factor of 100.

The calculations use a homogeneous one-dimensional, small-strain constitutive model at fixed temperature. Inertia and local grip effects are omitted. Stress steps and strain steps idealize a fast loading stage; real loading takes finite time. We retain the original dimensions and neglect their small changes. The assigned constants are model inputs, not measurements of a named material.

Units, standards, and physical constants and Work, kinetic energy, potential energy, and power supply useful preparation. Read the test descriptions and tables first. Derivatives, Taylor series, and local approximation helps with the dotted rate symbols below; Simple harmonic motion helps with the final cyclic example. The stress and strain definitions needed here are stated in this lesson.

Linear response can still have memory

An ideal linear elastic solid obeys σ = Eε: its current stress follows from its current strain. A viscoelastic material can have different stress at the same current strain if it reached that strain by a different loading history. “Linear” therefore describes a relationship between histories, rather than one fixed instantaneous stress-to-strain ratio.

For a linear model initially at rest, multiplying an entire stress history by two multiplies its strain response by two. Adding two stress histories adds their responses. These scaling and superposition rules apply within the small-strain regime where the model is valid. A slow loading history and a rapid one remain different histories even when their final stresses agree.

A remaining extension immediately after unloading is not enough to establish permanent damage. Some models recover gradually; others retain viscous deformation. A permanent viscous strain need not arise from a yield threshold. Plasticity and viscoplasticity describe additional behaviours that this lesson’s spring–dashpot laws do not include.

Give the spring and dashpot the right units

A constitutive spring obeys σ_s = Eε_s. Its modulus E is measured in Pa. It stores elastic energy and responds immediately in the ideal model. The subscript s labels the spring contribution; ε_s is part of the material strain, not a separately measured specimen extension.

A constitutive dashpot obeys σ_d = η ε̇_d. The dot means a derivative with respect to time, so ε̇_d has unit s⁻¹ and η has unit Pa·s. Under fixed stress, its strain grows at the rate σ_d/η. With zero stress acting on an isolated dashpot, that accumulated strain remains. A spring in a combined network can exert a restoring stress and drive a dashpot in the opposite direction.

These are material analogues written in stress and strain. A laboratory spring written as force = stiffness × extension has stiffness in N/m. The two constants differ: for a uniform axial specimen, the elastic force stiffness is EA₀/L₀. Likewise, a specimen dashpot written in force and extension rate has coefficient ηA₀/L₀ in N·s/m. Here η is an axial constitutive coefficient; it is not automatically the shear viscosity measured in a fluid experiment.

Series and parallel connections give different equations. In a series analogue, the same stress acts on the elements and their strain contributions add. In a parallel analogue, the elements share the same strain and their stress contributions add. The drawings represent these algebraic rules; they do not prescribe the physical lengths of pieces cut from the specimen.

Compare the two simplest connections

The Maxwell model places a spring E and a dashpot η in series. Add their strain rates to obtain ε̇ = σ̇/E + σ/η. Under a stress step σ₀ from an unstressed state, the spring produces the jump σ₀/E, then the dashpot adds σ₀t/η. Thus ε(t) = σ₀/E + σ₀t/η for t ≥ 0. If that stress is removed at time T, the spring recovers σ₀/E immediately and the viscous strain σ₀T/η remains.

For a Maxwell strain step ε₀, the initial stress is Eε₀. Holding the strain fixed gives σ̇ = −(E/η)σ, hence σ(t) = Eε₀ exp(−t/τ), where τ = η/E. Stress relaxes to zero. The Maxwell model describes a material with no nonzero equilibrium tensile modulus; its unbounded creep eventually leaves any small-strain description’s domain.

The Kelvin–Voigt model places the spring and dashpot in parallel: σ = Eε + ηε̇. Under a finite stress step from zero strain, ε(t) = (σ₀/E)[1 − exp(−t/τ)], with τ = η/E. There is no instantaneous strain jump. On removing the stress at T, the strain is continuous and then becomes ε(t) = ε(T) exp[−(t − T)/τ]. The spring drives delayed recovery through the dashpot.

A pure Kelvin–Voigt model has an important limitation in a strain-controlled step test. An instantaneous nonzero strain jump would require an impulsive stress from ηε̇. After a finite loading ramp ends and strain is held, its stress is simply Eε₀; the model gives no gradual relaxation during that hold. It is useful for delayed creep and recovery, but does not reproduce every test with the same two parameters.

Build a solid with an equilibrium spring

Add an equilibrium spring in parallel with a Maxwell arm. The equilibrium spring has modulus E∞, while the Maxwell arm contains spring E₁ and dashpot η in series. Both branches share the specimen strain ε; total stress is σ = E∞ε + σ_m, where σ_m is the Maxwell-arm stress. This three-parameter model is a standard linear solid.

The Maxwell arm obeys σ̇_m + σ_m/τ_R = E₁ε̇, with relaxation time τ_R = η/E₁. Substitute σ_m = σ − E∞ε to obtain σ + τ_Rσ̇ = E∞ε + τ_RE₀ε̇, where E₀ = E∞ + E₁. All terms have unit Pa. E₀ is the instantaneous modulus; E∞ is the long-time equilibrium modulus. The subscripts 0 and ∞ name limiting moduli, not extra springs.

Assign E∞ = 1.00 MPa, E₁ = 2.00 MPa and η = 4.00 MPa·s. Then E₀ = 3.00 MPa and τ_R = 2.00 s. All are positive. Use an initially relaxed specimen: stress, strain and dashpot strain vanish for t < 0. This initial state matters; a specimen still recovering from an earlier load has a different response.

Across an ideal jump the dashpot strain stays continuous. Integrating the differential law across the loading instant gives Δσ = E₀Δε. Once loading is complete, the differential law determines the smooth response. Keeping both the jump and the later response prevents a loading transient from being mistaken for a different material modulus.

Hold strain and calculate relaxation

Apply ε₀ = 0.00100 suddenly at t = 0, then keep that strain fixed. The specimen remains extended by 0.100 mm. The Maxwell-arm stress initially equals E₁ε₀, then decays exponentially as its dashpot changes strain. The equilibrium spring continues to carry E∞ε₀. Therefore σ(t) = ε₀[E∞ + E₁ exp(−t/τ_R)].

Define the relaxation modulus R(t) = σ(t)/ε₀ for this particular step test. For the assigned model, R(t) = [1 + 2 exp(−t/(2 s))] MPa. The initial modulus is 3 MPa and its long-time limit is 1 MPa. One relaxation time reduces the excess R(t) − E∞ to 1/e of its initial excess; it does not reduce the entire modulus to 1/e.

Initially σ = 3.00 kPa and F = σA₀ = 0.300 N. At 2.00 s, σ = 1 + 2/e kPa ≈ 1.735759 kPa and F ≈ 0.173576 N. At long times, σ approaches 1.00 kPa and F approaches 0.100 N. The changing force occurs while extension is fixed; it is not an unloading test.

The relaxation time comes from η/E₁. Increasing η while keeping the two moduli fixed slows the relaxation, while the initial and final stresses stay the same. A time constant describes the rate of a response under stated test conditions; it does not say that every deformation has completely finished after that time.

Hold stress and calculate creep

Now begin a separate test from the same relaxed initial state. Apply σ₀ = 1.00 kPa at t = 0 and hold it fixed. This corresponds to the constant engineering force F = 0.100 N. The instantaneous strain jump is σ₀/E₀ = 1/3000. As the Maxwell arm’s stress decreases, the equilibrium spring carries more of the fixed total stress, and strain increases.

Define creep compliance J(t) = ε(t)/σ₀, measured in Pa⁻¹. Solving the constant-stress equation gives J(t) = (1/E∞)[1 − (E₁/E₀) exp(−t/τ_C)], where τ_C = ηE₀/(E∞E₁) = τ_RE₀/E∞. For this standard linear solid, τ_C = 6.00 s, three times its relaxation time. This ratio follows from this model’s equations; it is not a universal single-time formula for all viscoelastic materials.

Thus J(t) = [1 − (2/3) exp(−t/(6 s))] MPa⁻¹ and ε(t) = 0.00100[1 − (2/3) exp(−t/(6 s))]. At 0⁺ the extension is 0.033333 mm; at 6.00 s it is approximately 0.075475 mm; eventually it approaches 0.100 mm. Here 0⁺ means immediately after loading, rather than the relaxed state just before it.

Creep compliance and relaxation modulus are not reciprocal functions at equal time. At 2 s, R ≈ 1.735759 MPa while J ≈ 0.522312 MPa⁻¹; their product is approximately 0.906608, not 1. Their initial and long-time limiting values are reciprocal for this model. The different intermediate values arise because one test holds strain fixed and the other holds stress fixed.

Remove the load with a second step

Keep the 1.00 kPa stress until T = 6.00 s, then remove it. This is a positive stress step at t = 0 followed by an equal negative step at t = T. Linear superposition gives ε(t) = σ₀J(t) before removal and ε(t) = σ₀[J(t) − J(t − T)] for t ≥ T. Each step uses the time elapsed since that step occurred. The negative step has no response before T.

Just before removal, ε(6⁻) ≈ 754.747039 microstrain. Removal causes the jump −σ₀/E₀ = −333.333333 microstrain, leaving ε(6⁺) ≈ 421.413706 microstrain. For t ≥ 6 s, ε(t) = (2/3) × 0.00100 × (1 − e⁻¹) exp[−(t − 6 s)/(6 s)]. This remaining strain then approaches zero with the creep/recovery time of 6 s.

The dashpot need not erase deformation on its own. With zero external stress after removal, the stretched equilibrium spring carries positive stress E∞ε while the Maxwell arm carries the opposite stress −E∞ε. That internal restoring action drives the dashpot backwards and allows delayed full recovery. Dissipation remains nonnegative: reversing a dashpot’s motion also reverses its stress.

A general history can be built from many small steps. If stress changes by Δσ_j at time t_j, its later contribution is Δσ_jJ(t − t_j), and the contributions add. An optional continuous form, for an initial step σ(0⁺) followed by a smooth history, is ε(t) = σ(0⁺)J(t) + ∫₀ᵗ J(t − s)σ̇(s) ds. For additional jumps, include their separate step contributions rather than treating an ordinary derivative as though it already contained them.

Read phase and loss in a harmonic test

Instead of holding a step, prescribe a repeating small strain ε(t) = ε_a cos(ωt). Here ε_a is strain amplitude and ω is angular frequency in rad/s; ordinary frequency is f = ω/(2π) in Hz. Consider the steady response after initial transients decay. For the passive standard linear solid, stress leads strain: σ(t) = ε_a[E′ cos(ωt) − E″ sin(ωt)] = ε_a|E*| cos(ωt + δ). Equivalently, strain lags stress.

E′ is the storage modulus, the coefficient of stress in phase with strain. E″ is the loss modulus, the coefficient associated with the quadrature component. Their units are Pa. The notation E* = E′ + iE″ is a compact complex-number description; the observed stress is real. Its magnitude is |E*| = √(E′² + E″²), and δ = atan2(E″, E′). The minus sign on the sine term follows from expanding cos(ωt + δ).

For the assigned solid, put x = ωτ_R. Its moduli are E′ = E∞ + E₁x²/(1 + x²) and E″ = E₁x/(1 + x²). At low frequency E′ approaches E∞; at high frequency it approaches E₀. E″ tends to zero in both limits and is largest at x = 1. This is the loss-modulus maximum for this model; it is not a structural resonance or a statement that the phase angle also peaks at x = 1.

Choose ε_a = 0.00100 and ω = 0.500 rad/s, so x = 1, f ≈ 0.079577 Hz and period = 4π s ≈ 12.566371 s. Then E′ = 2.00 MPa, E″ = 1.00 MPa, δ = arctan(1/2) ≈ 26.565051° and stress amplitude = √5 kPa ≈ 2.236068 kPa. Thus σ(t) = [2 cos(ωt) − sin(ωt)] kPa.

Plot signed stress against signed strain during a cycle. At phase 0° the point is (1000 microstrain, 2 kPa); at 90° it is (0, −1 kPa); at 180° it is (−1000 microstrain, −2 kPa); at 270° it is (0, 1 kPa). Returning to 360° closes an ellipse. At zero strain there can be nonzero stress because the internal state remembers the earlier motion.

Work per original volume is ∫σ dε. Over the complete steady cycle the in-phase component contributes zero net work, while the quadrature component gives the positive loss Δw = πE″ε_a². Here Δw = π J/m³ ≈ 3.141593 J/m³. The specimen’s original volume is A₀L₀ = 1.00 × 10⁻⁵ m³, so its cycle loss is 10π microjoules ≈ 31.415927 microjoules. The enclosed loop area represents energy dissipated, not energy that can all be recovered.

Choose the time range and check the model

Temperature can change the rates of molecular rearrangement and therefore the fitted response times and moduli. The constants used above belong to one assigned fixed-temperature model. As a controlled calculation, halving η while keeping E∞ and E₁ fixed halves both τ_R and τ_C: the same endpoint values are reached faster. This calculation is not a prediction of the viscosity change caused by a particular temperature increase.

Some materials permit time–temperature superposition: response curves measured at different temperatures can be shifted along a time or frequency axis to form a master curve. This requires compatible response shapes and a tested shift rule over the intended range. A universal temperature correction should not be assumed for all viscoelastic materials.

One Maxwell arm supplies only one relaxation time. A broader response can be represented by several Maxwell arms in parallel with the equilibrium spring: R(t) = E∞ + Σ E_j exp(−t/τ_j), with positive moduli and times in a passive example. Each arm shares the total strain and adds its stress. These parameters must be fitted to relevant measurements; adding parameters does not itself establish a better physical description.

Check the model against both creep and relaxation, the loading ramp, recovery duration and the frequencies of interest. Confirm that deformation stays small, temperature remains controlled, and effects such as ageing, yielding, damage and large-strain changes do not dominate. A uniaxial model alone does not specify separate bulk and shear relaxation or a time-independent Poisson ratio for a three-dimensional material.

Worked examples

Try it yourself

1. A specimen’s extension is fixed after loading while its tensile force falls. Is this creep, stress relaxation or recovery after unloading?

   Show the answer and reasoning

   Stress relaxation.

   Strain is held fixed and the stress needed to maintain it decreases. Creep holds stress fixed; recovery after unloading begins when the applied stress is removed.

2. For the assigned standard linear solid, does the 2.00 s relaxation time also give the time for the difference between creep strain and its final value to fall to 1/e? Find the correct time.

   Show the answer and reasoning

   No. The creep time is τ_C = τ_RE₀/E∞ = 2.00 × 3.00/1.00 = 6.00 s.

   Holding stress rather than strain gives a different differential equation. The creep strain’s deficit below its long-time limit decays as exp(−t/τ_C).

3. Repeat the six-second stress pulse from a relaxed state using 0.200 N instead of 0.100 N. What strain remains immediately after removal? Assume the same linear model remains valid.

   Show the answer and reasoning

   Approximately 842.827412 microstrain.

   The entire applied stress history doubles, so every strain value doubles. The original 6⁺ strain is 421.413706 microstrain. The eventual recovered strain is still zero.

4. A pure Kelvin–Voigt specimen is initially unstrained. Can a finite stress step produce an instantaneous nonzero strain jump? What happens after a finite strain-loading ramp ends and strain is held?

   Show the answer and reasoning

   No instantaneous strain jump occurs under a finite stress step. After the strain ramp ends, the held-strain stress is Eε₀, with no gradual relaxation in this model.

   An ideal strain jump requires an impulsive ηε̇ contribution. Under finite stress the strain is continuous; under a subsequent strain hold ε̇ = 0 and only the spring stress remains.

5. In the same steady harmonic test, halve strain amplitude from 0.00100 to 0.000500 while retaining ω = 0.500 rad/s. What is the specimen’s cycle loss, and does a negative instantaneous stress imply negative dissipation?

   Show the answer and reasoning

   Cycle loss is (5π/2) microjoules ≈ 7.853982 microjoules. A negative instantaneous stress does not imply negative dissipation.

   Cycle loss scales with ε_a², so halving amplitude divides the loss by four. The dashpot’s dissipation rate is σ_m²/η ≥ 0: its stress and strain rate reverse together. Net cycle work remains positive.

What to carry forward

• Creep holds stress fixed; stress relaxation holds strain fixed. Their response functions answer different questions.
• Linear viscoelasticity combines scaling and superposition with memory of the loading history.
• The standard linear solid has immediate elastic response, finite equilibrium stiffness and delayed full recovery; the assigned relaxation and creep times are 2 s and 6 s.
• Storage and loss moduli describe steady harmonic response. The signed stress–strain loop has positive cycle loss πE″ε_a² per volume.
• The usable time, temperature and strain range must be tested. One scalar model does not determine every three-dimensional material response.