I am working on an experiment of rheology and I need to calculate shear stress in order to calculate the viscosity. After some research I found that for the type of viscometer I will be using (cone-plate), the stress is calculated by dividing the torque given by the apparatus by a form factor equal to $\tfrac{2}{3}\pi r^3$. But I wasn't able to find the explanation of how this factor is derived. Any idea?

  • $\begingroup$ Shear stress is not uniformely distributed along the area. $\endgroup$ Commented Apr 7, 2016 at 15:00
  • $\begingroup$ Did you mean $\frac23 \pi r^3$? $\endgroup$
    – Gert
    Commented Apr 7, 2016 at 18:32
  • $\begingroup$ Yes but I am not able to write that in exponential form $\endgroup$
    – Thomson1
    Commented Apr 7, 2016 at 21:31

1 Answer 1


In a cone-plate rheometer the plate is a disc, and the geometry means the strain rate $\dot{\gamma}$ is constant everywhere. The stress is given by:

$$ \tau = \mu\dot{\gamma} $$

Since at equilibrium the viscosity is constant, that means both variables on the right of the equation are constant so the stress is constant everywhere on the plate. Now we just need to work out how the stress relates to the torque.

Let's draw a top view of the plate:

top view

Consider the annulus I've drawn at radius $r$ and with width $dr$. The area of this annulus is:

$$ A = 2\pi r dr $$

Stress is the force per unit area, so force is stress times area giving:

$$ F(r) = \tau 2\pi r dr $$

And torque is just force times (perpendicular) distance so the torque due to the annulus is:

$$ T(r) = \tau 2\pi r^2 dr $$

The total torque is now calculated by integrating:

$$ T = \int_0^R \tau 2\pi r^2 dr $$

where $R$ is the radius of the plate, and this gives:

$$ T = \tau \tfrac{2}{3} \pi R^3 $$

It's conventional to write:

$$ \tau = C_1\,T $$

where $C_1$ is the form factor you mention in your question. So we get the expression for the form factor that you mention:

$$ C_1 = \frac{1}{\tfrac{2}{3} \pi R^3} $$

  • $\begingroup$ Thanks a lot. I think the answer you've provided will be helpful for many as the derivation is not easily available in the internet. $\endgroup$
    – Thomson1
    Commented Apr 7, 2016 at 21:32

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