Rheology and non-Newtonian matter

Move one plate, hold the other

Imagine a layer of material between two flat plates. Hold the bottom plate still and move the top plate sideways. The force needed to maintain that motion tells us about the material’s resistance to flow. Two materials can need the same force at one speed, yet very different forces at a higher speed. Rheology studies the deformation and flow of matter; here we focus on steady simple shear.

Use a gap h = 1.00 mm = 0.00100 m and plate area A = 100 cm² = 0.0100 m². Coordinate x follows the sideways motion; +y points from the bottom plate to the top plate. The signed top-plate velocity is V. No slip means fluid touching the bottom has velocity zero and fluid touching the top moves at V. In uniform simple shear, the velocity between them is v_x(y) = V y/h.

The velocity change across the gap gives the engineering shear rate γ̇ = V/h, measured in s⁻¹. At V = 4.00 mm/s, γ̇ = 4.00 s⁻¹. A dot indicates a time rate. Shear stress τ = σ_xy means the x component of traction on the fluid face whose normal points +y. The first index gives the traction direction and the second gives the face-normal direction. Positive τ is a force along +x on that fluid face.

The top plate applies the signed force F = Aτ to the fluid, and the fluid applies −F to that plate. An actuator must supply F to maintain the stated speed against this drag. These equal and opposite forces act on different objects. Stress has unit Pa = N/m²; converting the area before calculating force is essential.

Assume homogeneous incompressible material at fixed temperature, steady laminar fully developed motion, negligible edge effects and no pressure gradient. The five parameter sets below are separate illustrative models, rather than measurements of named fluids. We compare steady states; acceleration during starting or stopping is outside these calculations.

Stress, strain, and Hooke’s law and Stress and strain tensors provide preparation for traction. The definitions needed here are restated. Work, kinetic energy, potential energy, and power helps with the later energy balance. On a first reading, γ̇ = V/h is enough; the velocity-gradient tensor and tangent derivative below provide optional mathematical connections.

Rate tells us how quickly layers slide

After a relative plate displacement Δx, the engineering simple-shear deformation is γ = Δx/h. It is dimensionless. A 1 mm displacement across a 1 mm gap gives γ = 1. If the plate keeps moving, γ continues to grow. The shear rate tells us how quickly this happens: γ̇ = V/h. A fixed speed can therefore give a constant rate while accumulated deformation becomes large.

A shear-rate law is different from a solid’s stress-versus-strain law. The Newtonian relation below uses ongoing motion, not the total distance the plate has already travelled. Do not carry the previous small-strain solid model’s restriction on accumulated strain into this fluid calculation.

An optional tensor connection uses D, the symmetric part of the instantaneous velocity gradient. For this planar motion, D_xy = D_yx = γ̇/2, while its other components vanish. Thus the scalar rate magnitude √(2D:D) equals |γ̇|. Here D:D means the sum of the squares of all components. Both off-diagonal components contribute.

A Newtonian Cauchy shear traction can be written τ = 2ηD_xy = ηγ̇. There is no extra factor of two when using the engineering shear rate. D describes the instantaneous velocity field; it is not an accumulated infinitesimal strain tensor for a specimen that has undergone large shear.

A constant viscosity makes a straight flow curve

Begin with the Newtonian relation τ = ηγ̇. Under the fixed conditions of this model, η is constant and has unit Pa·s. Doubling the shear rate doubles the shear stress. The signed flow curve is a straight line through the origin. Its slope measures the resistance per unit shear rate.

Assign η = 2.00 Pa·s. At γ̇ = 4.00 s⁻¹, τ = (2.00 Pa·s)(4.00 s⁻¹) = 8.00 Pa. Multiply by A = 0.0100 m² to obtain F = 0.0800 N. At γ̇ = 1.00 s⁻¹, the same model needs τ = 2.00 Pa and F = 0.0200 N.

At nonzero rate, define steady-shear viscosity η_ss = τ/γ̇. It is also called a secant viscosity: on a stress-versus-rate plot it is the slope of the line from the origin to the current point. For the Newtonian model this quotient is always 2.00 Pa·s. At exactly zero rate the quotient is 0/0 and is undefined, although its Newtonian limit is η.

A non-Newtonian response need not preserve this constant instantaneous ratio. A steady flow curve may have a rate-dependent viscosity or a yield offset. Other materials also depend on their history. The examples below isolate particular steady-shear behaviours; one scalar law does not cover all non-Newtonian matter.

Thinning and thickening compare viscosities

For three no-yield comparison models, choose a common reference rate γ̇_ref = 1 s⁻¹ and reference stress τ_ref = 2 Pa. Write τ = sgn(γ̇)τ_ref(|γ̇|/γ̇_ref)ⁿ, with n > 0. The sign function sgn gives +1 for a positive argument and −1 for a negative argument; these models give τ = 0 at zero rate. The rate ratio inside the power is dimensionless.

The equivalent form is τ = K sgn(γ̇)|γ̇|ⁿ, where K = τ_ref/γ̇_refⁿ has unit Pa·sⁿ. Use n = 0.5 for power-law thinning, n = 1 for the Newtonian case and n = 1.5 for power-law thickening. All three have numerical SI consistency coefficient 2, but those coefficients have different dimensions when n differs. Comparing them without their units would be misleading.

For nonzero rate, η_ss = K|γ̇|ⁿ⁻¹. When 0 < n < 1, viscosity decreases as rate increases: shear thinning. When n > 1, viscosity increases: shear thickening. At n = 1 it is constant. At 4 s⁻¹ the three comparison viscosities are 1, 2 and 4 Pa·s, giving stresses 4, 8 and 16 Pa respectively.

Thinning does not mean the shear stress must fall. For n = 0.5, stress grows as the square root of rate while viscosity falls as its inverse square root. Increasing rate from 1 to 4 s⁻¹ doubles stress from 2 to 4 Pa, even though viscosity halves from 2 to 1 Pa·s.

The table includes 16 s⁻¹ values beyond the comparison panels’ plotted 0.25–4 s⁻¹ range. Each model cell lists stress first, then viscosity. The panels keep separate vertical units and honest linear scales so a decreasing viscosity cannot be mistaken for a decreasing stress.

A pure power law is a finite-range model choice, not a universal description at every rate. The illustrative calculations here use 0.25–100 s⁻¹ when quoting nonzero-rate predictions. For the n = 0.5 law, viscosity tends to infinity as rate tends to zero, but stress tends to zero: this is not a finite yield stress. The n = 1.5 viscosity tends to zero at zero rate. A measured fluid may instead approach viscosity plateaus that these fits do not contain.

The same force can produce different speeds

So far we prescribed speed, obtained rate from the gap and calculated the required force. A different test prescribes force and asks which steady speed the flow law permits. Keeping the control variable explicit prevents two different experiments from being compared as though their rates were identical.

Apply F = 0.200 N to the area 0.0100 m². The shear stress is τ = F/A = 20.0 Pa. For a no-yield power law, inversion gives γ̇ = sgn(τ)(|τ|/K)¹⁄ⁿ. Equivalently, using the dimensionless reference form, γ̇ = γ̇_ref sgn(τ)(|τ|/τ_ref)¹⁄ⁿ. This last form makes the rate unit explicit.

The assigned Newtonian, thinning and thickening laws give rates 10.000000, 100.000000 and 4.641589 s⁻¹ respectively. With h = 1 mm, V = hγ̇ gives the same numerical speeds in mm/s. A rate and a speed are still different physical quantities; their numerical equality here comes only from the chosen gap.

At the common 1 s⁻¹ reference rate the models required equal forces. At the common 20 Pa stress their rates differ greatly. A fit is useful only under the conditions and rate range for which it has been established. These are conditional mathematical predictions from the supplied parameter sets, not measured operating speeds.

An ideal yield model has a non-flowing stress interval

An ideal yield flow law allows nonzero stress without continuing steady shear. Assign a yield magnitude τ_y = 5.00 Pa. If |τ| ≤ τ_y, the ideal steady rate is γ̇ = 0. Once the stress magnitude exceeds that threshold, a flowing branch relates the excess stress to rate. The statement includes equality at the ideal threshold.

For a Bingham model, choose plastic viscosity η_pl = 2.00 Pa·s. At nonzero rate, τ = sgn(γ̇)[τ_y + η_pl|γ̇|]. For a Herschel–Bulkley model, choose K = 2.00 Pa·s⁰·⁵ and n = 0.5, giving τ = sgn(γ̇)[τ_y + K|γ̇|ⁿ]. The general Herschel–Bulkley flowing law reduces to Bingham when n = 1.

These flowing expressions must not be used to collapse the zero-rate stress to one value by substituting sgn(0) = 0. At zero rate, every stress in [−5,+5] Pa is allowed. On a graph with shear rate horizontal and stress vertical, that complete admissible set is a vertical interval at zero rate. Positive and negative flowing branches approach its upper and lower endpoints.

At γ̇ = 4.00 s⁻¹, Bingham gives τ = 5 + 2 × 4 = 13.0 Pa. Herschel–Bulkley gives τ = 5 + 2 × √4 = 9.00 Pa, with the units supplied by their coefficients. At 1 s⁻¹ both yield models give 7 Pa, not the 2 Pa shared by the three no-yield models.

Under prescribed stress, the flowing inverse is γ̇ = sgn(τ)[(|τ| − τ_y)/K]¹⁄ⁿ for |τ| > τ_y, with K = η_pl and n = 1 for Bingham. First subtract the yield magnitude, then take the power, and finally restore the sign. At τ = 20 Pa the assigned Bingham rate is 7.50 s⁻¹ and the Herschel–Bulkley rate is 56.25 s⁻¹.

The threshold force for this area is Aτ_y = 0.0500 N. A 0.0400 N applied force corresponds to only 4.00 Pa and gives zero steady rate in either ideal yield model. The two-parameter or three-parameter flow laws supply no pre-yield elastic modulus, so they do not determine an elastic displacement during loading. Real yielding can occur over a range; an exact threshold is a feature of these ideal laws.

A viscosity ratio and a tangent slope answer different questions

The Bingham flowing branch is straight, but the full yield law is not the Newtonian line through the origin. At its 4 s⁻¹ point, steady-shear viscosity is η_ss = τ/γ̇ = 13/4 = 3.25 Pa·s. The secant from the origin includes the yield offset. The tangent to the flowing branch instead has slope 2.00 Pa·s, equal to η_pl.

Differential viscosity is η_diff = dτ/dγ̇. It describes the additional stress needed for a small additional rate around a flowing point: Δτ ≈ η_diff Δγ̇. Derivatives, Taylor series, and local approximation provides an optional mathematical explanation of this tangent. Dividing Pa by s⁻¹ gives Pa·s, so slope and secant share units while answering different questions.

On any nonzero Herschel–Bulkley branch, η_ss = τ_y/|γ̇| + K|γ̇|ⁿ⁻¹, whereas η_diff = nK|γ̇|ⁿ⁻¹. At 4 s⁻¹ the assigned Herschel–Bulkley ratio is 9/4 = 2.25 Pa·s and the differential viscosity is 0.500 Pa·s. For the no-yield power law, η_diff = nη_ss. These derivative formulas apply on flowing branches, not across the zero-rate yield interval.

Bingham therefore has a decreasing stress/rate ratio while retaining a constant flowing tangent slope. A viscosity ratio alone does not distinguish every flow-curve family. The vertical zero-rate interval has no single ordinary stress-versus-rate derivative.

Our uniform calculation uses local steady-shear viscosity. An instrument may report an apparent viscosity based on a nominal average shear rate, especially when rates vary across its geometry. That reported ratio need not be a local viscosity at every point. Understand which rate and stress were used before importing a measured coefficient into a flow calculation.

The actuator supplies power in either direction

Power is the rate of mechanical work. The actuator supplies P = FV. No slip makes the fluid’s top-boundary velocity equal to the plate velocity, while the fixed bottom boundary does no work. Substitute F = Aτ and V = hγ̇ to obtain P = Ahτγ̇. The uniform dissipation rate per volume is τγ̇, with unit Pa·s⁻¹ = W/m³.

For all five supplied laws, stress and nonzero shear rate have the same sign, so τγ̇ ≥ 0. Reverse the flow and both τ and γ̇ reverse. The actuator force and velocity then both become negative, but their product remains positive. A negative stress or a negative velocity alone is not evidence of negative dissipation.

At 4 s⁻¹, the main gap gives V = 0.00400 m/s. The Newtonian force is 0.0800 N, so P = 0.000320 W = 0.320 mW. The thinning and thickening powers are 0.160 and 0.640 mW respectively. The sample volume is Ah = 1.00 × 10⁻⁵ m³; multiplying its uniform τγ̇ by that volume gives the same power.

In the ideal zero-rate yield interval, power is zero even when shear stress is nonzero, because the fluid does not undergo ongoing shear. During the steady flow considered here, supplied mechanical work is dissipated; maintaining fixed temperature requires that the resulting heat be removed. The flow laws do not supply a temperature-rise calculation or an elastic energy storage law.

The area under a shear-stress versus shear-rate graph is not an energy density. Its units would be Pa·s⁻¹, whereas energy per volume has unit Pa = J/m³. Dissipated work per volume over a time interval is ∫τγ̇ dt. The previous viscoelastic stress–strain cycle uses strain on its horizontal axis; replacing that axis with rate changes the meaning.

A steady curve does not reveal every form of memory

A steady shear-thinning comparison uses settled responses at different rates. Thixotropy instead concerns a reversible decrease in viscosity with time at a particular fixed shear rate. A useful protocol controls temperature, observes the fixed-rate response and tests recovery after changing or resting the shear history. Merely increasing the rate and finding a smaller viscosity is not a thixotropy test.

The generalized Newtonian idea assigns stress from the current deformation rate and fixed model parameters, without an independent constitutive history variable. A rate-crossover parameter may have units of time, but its reciprocal-rate units alone do not make it a stress-relaxation time. The steady power-law and yield examples above contain no structural recovery law.

Viscoelastic memory concerns dependence on deformation history, as in the held-strain relaxation and held-stress creep tests of Viscoelastic materials. It is not synonymous with thixotropy or with a nonlinear steady flow curve. Different effects can coexist in a material. A single steady curve cannot identify their relative contributions or a unique microscopic mechanism.

Keep the model domain attached to the result. The supplied laws describe homogeneous isothermal scalar simple shear with the declared controls and geometry. They do not establish pipe plug shapes, turbulent mixing, normal stress differences or extensional response. A material may show different regimes outside the measured rate range, so useful fitting requires more than naming the closest-looking curve.

Worked examples

Try it yourself

1. For this exercise use gap 2.00 mm, area 100 cm² and the same Newtonian viscosity 2.00 Pa·s. In a separate steady state the top plate moves at −6.00 mm/s. Find γ̇, D_xy, shear stress, required actuator force and supplied power.

   Show the answer and reasoning

   γ̇ = −3.00 s⁻¹; D_xy = −1.50 s⁻¹; τ = −6.00 Pa; F = −0.0600 N; P = +0.360 mW.

   Rate is V/h, and the symmetric velocity-gradient component is half the engineering rate. Stress and required force reverse with velocity. The fluid applies +0.0600 N to the plate. Supplied power is (−0.0600 N)(−0.00600 m/s) = +0.000360 W, so reversal does not make dissipation negative.

2. In the main geometry, use the assigned no-yield thinning law at +16.0 s⁻¹. Find stress, steady-shear viscosity, required force and power. Has stress fallen compared with its 4 s⁻¹ state?

   Show the answer and reasoning

   τ = 8.00 Pa; η_ss = 0.500 Pa·s; F = +0.0800 N; P = 1.28 mW. Stress has increased.

   The rate ratio is 16, so τ = 2√16 = 8 Pa and η_ss = 8/16 = 0.5 Pa·s. Speed is 16 mm/s = 0.016 m/s, giving P = 0.0800 × 0.016 = 0.00128 W. Relative to 4 s⁻¹, stress doubles from 4 to 8 Pa while viscosity halves from 1 to 0.5 Pa·s.

3. For the assigned Bingham model at +4.00 s⁻¹, a learner reads 3.25 Pa·s from the origin-to-point secant and 2.00 Pa·s from the flowing tangent. Must one value be wrong?

   Show the answer and reasoning

   No. The secant gives steady-shear viscosity τ/γ̇; the tangent gives differential/plastic viscosity dτ/dγ̇.

   The 5 Pa yield offset contributes 5/4 = 1.25 Pa·s to the secant, so 2 + 1.25 = 3.25 Pa·s. It does not change the slope 2 Pa·s of the straight flowing branch.

4. Apply +0.0400 N in the main geometry to either supplied ideal yield model. What steady shear rate follows, and does the flow law give a pre-yield elastic displacement?

   Show the answer and reasoning

   γ̇ = 0. The flow law does not supply a pre-yield elastic displacement.

   The applied stress is 0.0400/0.0100 = +4.00 Pa, inside the inclusive interval [−5,+5] Pa. Nonzero stress is allowed at zero steady rate. No elastic modulus was assigned, so zero steady rate is not a calculation of the loading-stage elastic deformation.

5. At fixed shear rate and controlled temperature, a material’s viscosity decreases with time and recovers under a stated rest-and-retest protocol. Is this the same test as comparing settled viscosities at two different rates? What does a steady power-law curve predict about that recovery?

   Show the answer and reasoning

   The tests differ. The fixed-rate reversible time response is evidence relevant to thixotropy; the steady power-law curve contains no recovery law.

   Steady shear thinning compares rate dependence, while the stated protocol tests reversible history dependence at a particular rate. A current-rate-only algebraic law has no structural state with which to calculate recovery. One test does not identify a unique microscopic mechanism or separate every possible viscoelastic contribution.

What to carry forward

• In uniform simple shear, γ̇ = V/h and D_xy = γ̇/2. The velocity gradient can remain fixed while total shear deformation grows.
• Shear thinning describes decreasing steady-shear viscosity with rate; stress can still increase. A secant ratio and a differential slope share units but need not agree.
• Ideal yield models retain every allowed stress in [−τ_y,+τ_y] at zero steady rate. Their flowing branches alone do not supply a pre-yield elastic modulus.
• Prescribed speed and prescribed force are different controls. For the supplied laws, actuator power FV = Ahτγ̇ is nonnegative in either flow direction.
• Steady rate curves, reversible fixed-rate time tests and viscoelastic loading histories answer different questions. Use fitted laws only within their established conditions and range.