Suppose I have a tank filled and there is no slip at the walls. If the tank is filled with a Newtonian fluid and is in static equilibrium, we know that the pressure is defined as $p = \rho g z$.

But what if the tank is filled with a yield stress fluid. For example, the Bingham plastic model defines viscous stress as:

$$\tau = \tau_0 + \mu \gamma$$

(or if it is more clear, the definition of shear rate) $$ \gamma = \begin{cases}0 & |\tau|\leq\tau_0 \\ \frac{1}{\mu}\left(\tau-\tau_0\right) & |\tau|>\tau_0\end{cases} $$

where $\tau_0$ is the yield stress of the fluid, $\mu$ is the plastic viscosity, and $\gamma$ is the shear rate. The material acts as a solid below the yield stress, and then as a Newtonian fluid above it. Intuitively, I feel as though the static pressure is reduced relative to the yield stress and that the pressure distribution is not a continuous function. Is my thinking correct? How do I calculate the static pressure in a yield stress fluid?


2 Answers 2


Below the yield stress your fluid is behaving like an elastic solid. Imagine putting your tank in zero g, so there are no forces, and then removing the base. The result would look like the left hand figure in the diagram below:

Yield stress

Now turn on gravity, or apply an external force and the result will be the middle diagram. As long as the stress is below the yield stress you simply elastically deform the fluid downwards so it bulges out the bottom of the tank. The downwards force $F$ is balanced by the upwards forces $F_s$ on the sides of the tank so $F = F_s$.

Now imagine bringing the bottom of the tank up from below to push the bulge back up again. This is going to require a force $F_b$ so there will be a force $F_b$ on the bottom of the tank so now $F = F_s + F_b$.

If your fluid is incompressible then when you push the bulge back up you reduce the strain in the fluid to zero, so the force on the sides gets reduced to zero and you simply have $F = F_b$. In practice I suspect there would be some mechanical compliance and the forces on the side would be much reduced but not reduced to zero.


John's answer is pretty good - I will just add my own to give some alternative insights ( I had made a comment to John's answer, but comments tend to get deleted after a while).

First - a Bingham plastic acts as a solid under low shear; this means that in the static case (which you explicitly ask for when you ask for "static pressure"), it is a solid.

The pressure distribution in a solid depends on the elastic stresses in that solid - as any elastic stresses need to be taken into account when computing the pressure. For an incompressible liquid that is allowed to set without stresses, the pressure profile will be exactly that of an ordinary liquid of the same density. It is only if there are residual stresses in the solid that you get a change in the pressure profile.

Residual stresses might arise in many different ways, depending on how you filled your container (for example, if you spun the container after filling it with a liquid, you might end up with a parabolic surface). You need to consider how those stresses "carry" some of the mass above them when trying to compute the static pressure. The problem would be further complicated if you had a compressible fluid.

But whichever way you look at it, if you are looking at the static case, the entire substance should be a solid.


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