Boundary condition of limiting friction in continuum mechanics

[hunt_mat]

TL;DR Summary

What is the correct boundary condition for limiting friction?

Suppose I have a block of deformable material on a rough surface. I want to have the boundary condition for the stress tensor that takes into account of friction. If the mass of my block is $m$, and of density $\rho$ and the coefficient of friction is $\mu$ as well as gravity $g$. The resultant force is given by $R=mg=\rho gdV$, so the frictional force per unit mass is [itex]\mu\rho g$. Linking to the tangential part of the stress tensor, this yields: $$\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}\cdot\hat{\mathbf{t}}=\mu\rho g$$

Does this reasoning seem okay?

 

[anuttarasammyak]

I assume that force per area or pressure on the surface matters but I do not find it in your formula where volume matters.

 

[hunt_mat]

Pressure is part of the stress tensor.

 

[anuttarasammyak]

Yea, but I mentioned about RHS $\mu \rho g$ which has physical dimension of ML^-2 T^-2 force per volume.

 

[hunt_mat]

Surely that's correct, as when you use this as a boundary condition, you require the force per unit volume in 3D. Navier's equations in differential form require force per unit volume. That was my thinking here.

 

[anuttarasammyak]

Your LHS and RHS have same dimension ?

 

[Chestermiller]

The $\rho g$ is not present in the boundary condition; it is only in the stress-equilibrium differential equation.

If the usual convention is followed, where tensile stresses are considered positive, then $-n \centerdot \sigma \centerdot n$ is the normal compressive stress exerted by the surface on the block.

 

[hunt_mat]

Your LHS and RHS have same dimension ?

I'm pretty sure they are. I thought about this. The differential equation uses per unit volume. This is the RHS.

 

[hunt_mat]

The $\rho g$ is not present in the boundary condition; it is only in the stress-equilibrium differential equation.

If the usual convention is followed, where tensile stresses are considered positive, then $-n \centerdot \sigma \centerdot n$ is the normal compressive stress exerted by the surface on the block.

I think that $\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}\cdot\hat{\mathbf{t}}$ is the correct component for the frictional force here. One can also generalise this slightly by the equation:
$$\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}-(\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}\cdot\hat{\mathbf{n}})\hat{\mathbf{n}}=f(|\mathbf{u}|)\hat{\mathbf{u}}$$
The motion is very slight, so static friction should be a good approximation I think.

 

[Frabjous]

On the boundary, I would expect to see something like
$\sigma_{shear}\leq\mu\sigma_{normal}$
So you are making approximations on the rhs.
Friction opposes motion, so there is also going to be a need for some additional logic to determine which direction the shear points. There is also going to be a need for logic to handle the inequality.

 

[hunt_mat]

On the boundary, I would expect to see something like
$\sigma_{shear}\leq\mu\sigma_{normal}$
So you are making approximations on the rhs.
Friction opposes motion, so there is also going to be a need for some additional logic to determine which direction the shear points. There is also going to be a need for logic to handle the inequality.

Friction will oppose motion obviously. I am making an approximation; that approximation is that the velocity is small enough that static friction will be enough to accurately capture this. I did explain that it was limiting friction.

 

[Frabjous]

Friction will oppose motion obviously. I am making an approximation; that approximation is that the velocity is small enough that static friction will be enough to accurately capture this. I did explain that it was limiting friction.

What happens if you double the height of the block?

 

[Chestermiller]

What happens if you double the height of the block?

It doesn’t change the boundary condition expression.

 

[hunt_mat]

What happens if you double the height of the block?

It's scaled with volume.

 

[Frabjous]

If you double the height of the block, you double it’s mass. One would expect the friction to increase. This is not reflected in $\mu\rho g$.

 

[hunt_mat]

It's scaled with volume. The governing equations are per unit volume.

 

[Chestermiller]

It's scaled with volume. The governing equations are per unit volume.

Not the boundary conditions.

 

[hunt_mat]

Not the boundary conditions.

The boundary conditions have to fit the governing equations. So I don't understand what you're saying here.

 

[Chestermiller]

The boundary conditions have to fit the governing equations. So I don't understand what you're saying here.

The boundary conditions are per. unit area, not unit volume.