# Well-posedness of linear elasticity boundary conditions

I have several questions regarding suitable boundary conditions for linear elasticity.

I have read that in order for modeling linear elasticity to be well-posed, the entire boundary cannot be traction/Neumann boundary conditions, and that at least part of it needs to be specified as a fixed/Dirichlet boundary condition.

My first question is: Does at least part of the boundary need to be "fixed," as in "zero" displacement, or could the boundary have a specified non-zero displacement Dirichlet condition and still have the problem be well-posed?

My second question is: Physically, couldn't you have applied loadings to all parts of a boundary? For example you could have a rod and apply a compression/load on all surfaces? Computationally, to model this, all of the boundaries would then be a Neumann traction condition?

That's a really nice question :)

The main equation that is being solved here is: $$\rho \frac{\partial^2 \vec{u}}{\partial t^2}-\textrm{div}(\sigma(\vec{u}))=\vec{f}\tag{1}$$ with $\sigma(\vec{u})$ the elastic tensor that describes the behaviour of the material that is being considerated. This is given by a constitutive law. And $\vec{f}$ the internal forces.

This equation has (as is natural) two kind of BC: $$\vec{u}=\vec{u}_D \quad \forall \vec{x}\in\partial \Omega\qquad \sigma\vec{n}=\vec{t}\quad \forall \vec{x}\in\partial\Omega\tag{BC}$$

that are imposed in the Dirichlet boudary where the displacements are fixed and in the Neumann boundary where tractions are known, respectively. There are as well two initial conditions, but they are not so important here...

In dynamics (equation $(1)$ you can impose whatever, because it is not a requirements that the body is at rest... but what happens if the body does? This is statics... and its behaviour is described by the next equation: $$-\textrm{div}(\sigma(\vec{u}))=\vec{f}\tag{2}$$ subjected to the same BC as before $(BC)$.

Now everything change...

Physics

Could a body be at rest without moving? The answer is yes! If the resultant of the applied forces adds to 0, then the body is in equilibrium. For this to happen, there is no need to specify any Dirichlet BC but in one node (to specify the spatial reference).

Therefore, equilibrium of forces must be present:

$$\vec{F}=\int_{\Omega}\vec{f}\,dV=-\int_{\partial\Omega}\vec{t}\,dS=\vec{T}\tag{*}$$ This means that the resultant of the body forces $\vec{F}$ must exert a force equal but in oposite direction respect to the external forces that are applied to the body (as a boundary condition).

If this occurs we don't have anything to be afraid of.

For example, a rope, whose extrema are being equally pulled does not remains at equilibrium even the fact that in their boundaries two traction BC are imposed. Of course we must specify in which spatial point the rope is, otherwise there would be infinite solutions to the problem in which the rope would be translated in space. In this example, if the rope is hung under gravity, in the lower extremum the BC would be $\vec{t}_l=\sigma \vec{n}=\vec{0}$ (no applied force), while in the upper $\vec{T}=\vec{t}_u-\vec{t}_l=\vec{F}=\int_0^{L}\rho \vec{g}\,dz$ (force exerted by the whole mass that it supports). This problem will have a unique solution if one point of the rope is known, otherwise the rope could be in equilibrium wherever.

Mathematics

This is so easy:

Apply the Gauss Th. to $(2)$ and apply the $(BC)$. If to Dirichlet BC are imposed, we obtain the restriction: $$\int_{\Omega}\left[-\textrm{div}(\sigma(\vec{u}))\right]dV=-\int_{\partial\Omega}\sigma\vec{n}\,dS=-\int_{\partial\Omega}\vec{t}\,dS=\vec{T}=\int_{\Omega}\vec{f}\,dV=\vec{F}$$ Which is the same equation deduced above.

Edit

Thanks to @origimbo to remember me the rotations!

Of course, apart from the equilibrium of forces, there must be an equilibrium of torques! Therefore, the provided boundary conditions must be consistent with this equilibrium: $$\int_{\Omega}\vec{x}\times\vec{f}\,dV=-\int_{\partial\Omega}\vec{x}\times\vec{t}\,dS\tag{**}$$ with respect to an arbitrary origin of coordinates.

Also, this last statement may be deduced taking the cross product of $(2)$ with an arbitrary vector $\vec{x}$, integrating by parts and noting that $\sigma = \sigma^{T}$.

If the provided $BC$ (neumann only) given by $(BC)$ fulfills $(*)$ and $(**)$ then your problem is well posed once fixed one node.

• Ah this is interesting. When you say "For this to happen, there is no need to specify any Dirichlet BC but in one node (to specify the spatial reference)," The Dirichlet BC must specify ZERO displacement right? Or could the the Dirichlet BC specify a non-zero displacement at this one node and still lead to a well-posed problem? – user27504 Apr 30 '18 at 0:11
• You can fix whatever displacement you want, but you have to specify a number. Tha value of 0 is set for convenience because all displacements are referenced to this node, otherwise you should subtract this quantity to the displacement of all nodes. This leads to a well posed problem. – HBR Apr 30 '18 at 8:28
• @user27504 Note that "fix one node" for elasticity is often only sufficient in 1d. The class of displacements in the null space of the operator are usually solid body motions, and you need more prescribed points to describe rotation in higher dimensions. – origimbo Apr 30 '18 at 10:38
• @origimbo , true.. but you still must fix always one node to avoid solid displacements. Then the provided boundary conditions and $\vec{f}$ must be compatible with each other for the problem $(P)$ to be well posed... I've added an edit. – HBR Apr 30 '18 at 11:14
• there is no need to fix any point. you can also use a integral condition to remove displacement from the system. – Bort May 27 at 12:20