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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Derive and calculate
Chapters 3–5. Connect torque, power-law stress and a worked specimen.
Check the experiment
Chapters 9–10. Inspect geometry, boundaries and common errors.
Practise and continue
Chapters 11–12. Change rate and model slope, then follow specialist guides.
Full chapter index · Practice and worked checks · Rheology subject root
A parallel-plate rheometer reports one torque, although material near the centre moves differently from material near the rim. Converting that integrated signal into a local stress requires a model of the geometry and the fluid response.
This guide derives the ideal torque integral, works an original shear-thinning example and shows why the familiar Newtonian conversion can overestimate rim stress. The calculation is for steady, homogeneous, isothermal, no-slip flow in a narrow uniform gap. It is a scientific interpretation exercise, not an instrument setup or operating procedure.
Chapter index
Define the geometry · Chapters 1–2
1. Define the local shear rate
2. Add the torque of each circular ring
Derive and calculate · Chapters 3–5
3. Introduce a power-law fluid within a stated range
4. Compare with the Newtonian conversion
5. Work an original specimen calculation
Read the correction · Chapters 6–8
6. Obtain the correction from a local torque slope
7. See why outer material contributes so much
8. Compare different material slopes
Check the experiment · Chapters 9–10
9. Check assumptions before trusting the corrected number
10. Questions readers ask
Practise and continue · Chapters 11–12
11. Transfer checks with worked answers
12. Continue with the quantity you need
Take circular plates of radius R separated by gap h. The lower plate is stationary and the upper rotates at angular speed Ω in radians per second. At radius r, the upper surface speed is Ωr.
Under the ideal narrow-gap, no-slip velocity approximation, local shear rate is γ̇(r) = Ωr/h. Its rim value is γ̇_R = ΩR/h. The centre has zero local rate in this model; the rest of the radius spans a continuous range of rates.
Angular speed is not shear rate. They share an inverse-time dimension, but geometry supplies the factor r/h. Nor should a speed in revolutions per minute be inserted numerically as radians per second without conversion.
A thin ring between r and r + dr has area dA = 2πr dr. If its tangential shear stress is σ(r), its tangential force is σ(r)dA. Torque multiplies that force by the moment arm r.
Therefore dM = 2πσ(r)r² dr, and total torque is M = 2π∫₀ᴿ σ(r)r² dr. The r² factor includes both the larger circumference and the longer lever arm of outer rings.
One torque is consequently a weighted integral of many local stresses. Dividing it by total plate area does not produce a shear stress: torque per area has the wrong units. The geometry conversion must account for the moment arm.
CHAPTER 3 OF 12 · Derive and calculate
3. Introduce a power-law fluid within a stated range
Suppose the local steady relation is σ = Kγ̇ⁿ, with positive rate. K is a consistency parameter and n is the flow index. K has units Pa·sⁿ, so it is a viscosity in Pa·s only when n = 1.
Substitute γ̇(r) = γ̇_Rr/R into the torque integral. This gives M = 2πKγ̇_RⁿR³/(n + 3), for the usual positive n considered here. The integral uses the power law throughout the radius, not just at its endpoint.
Since rim stress is σ_R = Kγ̇_Rⁿ, rearrangement yields σ_R = (n + 3)M/(2πR³). This local quantity is inferred from the integrated torque through the assumed response law.
For n = 1, the expression becomes σ_N = 2M/(πR³), the ideal Newtonian rim-stress conversion. Applying that same conversion to every fluid silently retains a Newtonian radial stress profile.
For a power-law fluid, the true-to-Newtonian ratio is σ_R/σ_N = (n + 3)/4. With n = 0.50, the factor is 0.875. The Newtonian conversion is then about 14.29% above the true model rim stress, measured relative to the true value.
The 12.5% reduction from the uncorrected value and the 14.29% overestimate relative to the corrected value use different denominators. Both describe the same pair of numbers. State which reference a percentage uses.
Choose R = 0.020 m, h = 0.0010 m, Ω = 1.0 rad/s, K = 2 Pa·s^0.5 and n = 0.50. These are hypothetical model inputs, not specifications for a real material.
The rim rate is 1.0 × 0.020/0.0010 = 20 s⁻¹. The local constitutive law gives σ_R = 2√20 ≈ 8.9443 Pa. The predicted torque is 2π × 0.020³ × 8.9443/3.5 ≈ 0.000128454 N·m.
| Quantity | Correct model value | Using the Newtonian rim conversion |
|---|---|---|
| Measured/modelled torque | 0.000128454 N·m | Same torque |
| Rim shear rate | 20 s⁻¹ | Same geometric rate |
| Inferred rim stress | 8.9443 Pa | 10.2220 Pa |
| Rim apparent viscosity | 0.44721 Pa·s | 0.51110 Pa·s |
The disagreement comes from interpreting the same torque differently. It does not require a change in the sample. The corrected apparent viscosity remains tied to the stated rim rate; it is not a universal viscosity for the whole plate.
CHAPTER 6 OF 12 · Read the correction
6. Obtain the correction from a local torque slope
For a more general steady local relation σ(γ̇), substitute x = γ̇_Rr/R into the integral. Then Mγ̇_R³ = 2πR³∫₀^γ̇_R σ(x)x² dx. Differentiate with respect to rim rate while keeping geometry fixed.
Rearranging gives σ_R = [3M + γ̇_R dM/dγ̇_R]/(2πR³), or σ_R = M[3 + d ln M/d ln γ̇_R]/(2πR³). Since γ̇_R is proportional to Ω at fixed gap and radius, the logarithmic slope can also use angular speed.
For an exact power law, d ln M/d ln γ̇_R = n, recovering the earlier result. For data, the derivative is local and uncertain. It should not be replaced by one global slope across a range where the response changes.
The primary Liu and colleagues rim-stress study addresses non-Newtonian correction using torque-versus-rim-rate data and compares estimation approaches. Our derivation and numerical specimen are separate educational cases.
CHAPTER 7 OF 12 · Read the correction
7. See why outer material contributes so much
In the power-law model, the fraction of torque contributed by radii below aR is a^(n + 3). Thus the outer half-radius region, from R/2 to R, supplies 1 − 0.5^(n + 3) of total torque.
For n = 0.50, that contribution is about 91.16%. For a Newtonian fluid it is 93.75%. The outer half-radius region contains 75% of the plate area, illustrating why area fraction and torque fraction are different weights.
This makes edge conditions important, but it does not mean that only the rim was measured. The entire ideal flow field enters the integral. Lost contact or a changed sample radius can substantially affect the inferred quantity.
| Power-law index n | True/uncorrected rim-stress factor | Interpretation within this model |
|---|---|---|
| 0.20 | 0.800 | Stronger shear thinning |
| 0.50 | 0.875 | Main worked case |
| 1.00 | 1.000 | Newtonian response |
| 1.50 | 1.125 | Shear thickening |
For n above one, the Newtonian conversion instead underestimates the model rim stress. The correction therefore cannot be described as a universal fixed reduction applied to every non-Newtonian sample.
Differentiating noisy data can amplify scatter. Smoothing or fitting may help estimate a slope, but the fitting range, uncertainty and residuals need scrutiny. A precise-looking corrected curve can inherit an unstable derivative.
CHAPTER 9 OF 12 · Check the experiment
9. Check assumptions before trusting the corrected number
The derivation assumes that local stress is a single-valued function of the local steady shear rate under the stated conditions. Time-dependent structure, elastic transients, heating or significant secondary flow can require a different treatment.
Wall slip breaks the assumed relation between plate motion and bulk rate. Edge fracture, underfilling or evaporation changes the effective geometry or sample. Correcting the radial non-Newtonian weighting does not automatically correct any of these effects.
The rotational-rheometry guide explains the wider evidence chain. Record actual radius, gap, raw torque, speed, temperature, test history, instrument window and the software’s conversion. Check whether a correction has already been applied before applying it again.
Is the displayed rate an average across the plate?
Its convention must be checked. The rate used here is explicitly the rim value ΩR/h. The local rate elsewhere is smaller. A rate specified at another representative radius needs its corresponding conversion.
Is multiplying by 0.875 always the right correction?
Only for the stipulated power-law index 0.50 under the ideal assumptions. A different local slope gives a different factor. A transient, slipping or heterogeneous experiment may not satisfy this formula at all.
Can this formula be used directly for oscillatory moduli?
Not as a general substitution. We derived a steady torsional-flow relation. Oscillatory and nonlinear transient measurements require analysis appropriate to their forcing and response, rather than copying a steady-flow correction.
Double Ω in the main example while keeping geometry and material parameters unchanged. Rim rate doubles to 40 s⁻¹. Stress and torque increase by √2, because n = 0.50. Torque becomes approximately 0.000181661 N·m and rim apparent viscosity decreases by 1/√2.
Next halve the gap at the original speed within the same ideal model. Rim rate again doubles, giving the same √2 torque factor. This prediction does not establish no slip in an actual gap series; deviations could arise from slip, geometry errors or other violated assumptions.
Finally suppose a justified local torque-versus-speed logarithmic slope is 0.80. The correction factor relative to Newtonian rim stress is (3 + 0.80)/4 = 0.95. Use that local slope, not the n = 0.50 from the earlier specimen. Report how the slope and its uncertainty were estimated.
Geometry and measurement
Return to the Rheology root for the subject map, rotational rheometry for the instrument chain and rheology terminology for quantity and protocol distinctions.
Compare different loading and flow models
Stress relaxation after a finite ramp concerns time history. Yield-stress pipe flow and the moving plug concerns another spatial flow field. Neither equation can be transferred to rotating plates without deriving the relevant geometry and assumptions.
Sources and transferable learning
The primary 2020 rim-stress correction study provides a research route into estimating corrected stress from torsional data. The torque integral, constructed power-law example and transfer checks above show the reasoning explicitly and do not reproduce that paper’s experimental results.
Continue through How X Works and the Gold Standard of Education. The transferable skill is to identify how a whole-system signal was weighted before interpreting it as a local property.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.