# Shear stress of a suspended sphere in a viscoelastic fluid

What I am trying to solve right now is that I have a magnetic particle (nanoparticle to be exact) floating in liquid that is viscoelastic and apply sinusoidal magnetic field. The applied magnetic field will create torque that aligns the magnetic moment of the particle to the field. Also, the particle is large enough to be thermally blocked meaning that when the particle is rotated, it will physically rotate the particle not just the internal magnetic moment.

So, when the particle is rotated, it will drag along the fluid at the surface creating shear stress.

As of now I am interested in maximum force that the particle drags the fluid so I will assume that the internal magnetic moment and applied magnetic field are perpendicular to each other. My goal is calculate shear stress so I start with calculating force. Suppose that the applied torque is in +y-axis so we have only forces in x- and z-axis direction. $\vec{F} = F_x\hat{a_x}+F_z\hat{a_z}$ and $\vec{r} = r_x\hat{a_x} + r_y\hat{a_y} + r_z\hat{a_z}$. And then I apply $\vec{\tau} = \vec{r} \times \vec{F}$ and get $\vec{\tau} = -(r_xF_z - r_zF_x)\hat{a_y}$.

Decomposing components, and calculate only on y-axis we get

$$\tau_y = -|\vec{r}|\cos{\theta}\cos{\phi}\cdot |\vec{F}|\cos{\theta} + |\vec{r}|\sin{\theta}\cdot |\vec{F}|\sin{\theta}$$

Integrating from $\theta = -\frac{\pi}{2}$ to $\theta = \frac{\pi}{2}$ and from $\phi = 0$ to $\phi = 2\pi$,

$$\tau_y = |\vec{r}||\vec{F}|\int_{\theta = -\frac{\pi}{2}}^{\theta = \frac{\pi}{2}}\int_{\phi = 0}^{\phi = 2\pi} -\cos^2{\theta}\cos{\phi}\cdot + \sin{\theta}\cdot \sin{\theta} d\phi d\theta$$

I got this

$$\tau_y = |\vec{r}||\vec{F}|\pi^2$$

Now comes where I get stuck when I want to apply $\sigma = G\gamma$ linear relationship (in reality, this is not always the case but I want to test it first with linear assumption) where $\sigma$ is shear stress, $\gamma$ is shear strain, and $G$ is shear modulus.

From my understanding, shear stress is $\frac{F}{A}$, but the force is not constant so I have to calculate force acted on infinitsimal surface area $dA = r^2\sin{\theta}d\phi d\theta$. However, I feel something is wrong here and cannot proceed forward. I found the result $d\sigma = \frac{F}{dA}$ very strange and wrong but I have no idea what is the better way.

I also welcome other solution or equations with more complicated or rigorous approach if there is any.

• Have you tried solving this for a purely viscous fluid? – Chet Miller Jan 22 '18 at 13:46
• For purely viscous fluid, we already have brownian relaxation time $\tau_b = \frac{4\pi r_{hydro}^3 \eta}{K_B T}$ where $\eta$ viscosity. But we do not have this for viscoelastic media – Nabs Jan 22 '18 at 15:14
• You would like to know the shear stress distribution on the surface of the spherical particle as a result of the angular rotation history for small angular displacements, correct? – Chet Miller Jan 22 '18 at 15:20
• Actually, the shear stress distribution is nice, but it would be nice if I can just calculate the maximum shear stress right now. So, I take it when the torque is max and try to calculate what is supposed to be max shear stress. The idea is that the fluid is linear only when the force acting on it is low. So, I am trying to confirm that the applied torque still results in linear region – Nabs Jan 22 '18 at 22:10
• Wouldn't you rather be able to predict the torque exerted by the fluid on the sphere as a result of the fluid deformation, from knowledge of the fluid rheological properties? – Chet Miller Jan 22 '18 at 23:07

For the purely elastic solid problem, you would express the torque as a function of the angular displacement and the shear modulus: $$\tau=kGa^3\Delta \theta$$where a is the radius of the sphere and k is a dimensionless constant that arises from the solution to the solid mechanics problem. You would then represent the angular displacement as: $$\Delta \theta=\frac{\tau}{kGa^3}=J\frac{\tau}{ka^3}$$where J is the shear compliance (=1/G). For a viscoelasic material, the shear compliance J is a function of time J(t), and the angular displacement at time t is related to the history of the torque variation by: $$\Delta \theta(t)=\frac{1}{ka^3}\int_{-\infty}^t{J(t-\xi)}\frac{d\tau}{d\xi}d\xi$$ I leave it up to you to work out how this plays our for a case in which the torque is varying sinusoidally.
• I arrived at this using dimensional analysis. It is the only form that is dimensionally consistent. I haven't tried to solve the Navier Stokes equations (neglecting inertial terms) for this problem by myself yet, although I am confident that I could do it. But, I've Googled some links that seem to address it. I just don't want to put in too much work on this personally. In one reference, I saw mention of Landau and Lifshitz, 1984, which gives the solution for Newtonian fluid, and determines the constant k as $8\pi$. – Chet Miller Jan 26 '18 at 15:29