# Boundary condition of limiting friction in continuum mechanics

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• hunt_mat
hunt_mat
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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 $\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? I assume that force per area or pressure on the surface matters but I do not find it in your formula where volume matters. Pressure is part of the stress tensor. Yea, but I mentioned about RHS ##\mu \rho g## which has physical dimension of ML^-2 T^-2 force per volume. 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. Your LHS and RHS have same dimension ? 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. anuttarasammyak said: 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. Chestermiller said: 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 [itex]\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.

Chestermiller
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.

Last edited:
Frabjous said:
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.

hunt_mat said:
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?

Frabjous said:
What happens if you double the height of the block?
It doesn’t change the boundary condition expression.

hunt_mat
Frabjous said:
What happens if you double the height of the block?
It's scaled with volume.

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##.

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

hunt_mat said:
It's scaled with volume. The governing equations are per unit volume.
Not the boundary conditions.

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

hunt_mat said:
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.

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