Skip to content
Open
Show file tree
Hide file tree
Changes from 10 commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
5 changes: 4 additions & 1 deletion _toc.yml
Original file line number Diff line number Diff line change
Expand Up @@ -58,7 +58,10 @@ parts:
- file: chapter2/stokes-v2
- file: chapter2/elliptic-curl
- file: chapter2/elliptic-div
- file: chapter2/linear-elasticity
- file: chapter2/linear-elasticity-intro
sections:
- file: chapter2/linear-elasticity-v1
- file: chapter2/linear-elasticity-v2
- caption: Nonlinear Problems
chapters:
- file: chapter3/poisson
Expand Down
137 changes: 137 additions & 0 deletions chapter2/linear-elasticity-intro.md
Original file line number Diff line number Diff line change
@@ -0,0 +1,137 @@
# Linear Elasticity

## The PDE problem
The governing equations for small elastic deformations of a body $\Omega$ can be expressed as:

$$
\begin{aligned}
-\nabla \cdot \sigma(u) &= f & \text{in } \Omega \\
\sigma(u) &= C : \epsilon(u) & \text{in } \Omega \\
\end{aligned}
$$

where :
- $\Omega$ is the domain of interest,
- $u$ is the displacement vector field,
- $\sigma(u)$ is a second-order symmetric tensor, called stress tensor,
- $f$ represents the load vector field (force per unit volume),
<!-- $\kappa$ and $\mu$ are Lamé's elasticity parameters for the material, $I$ denotes the identity tensor, -->
- $\epsilon(u)$ is a second-order symmetric tensor, called strain tensor and by definition $\epsilon(u) := \frac{1}{2}(\nabla u + (\nabla u)^T)$.
- $C$ is the elasticity tensor, which relates stress and strain.

Then, the strong formulation of linear elasticity is :
Find $u \in V$ such that

$$
\begin{aligned}
-\nabla \cdot \sigma (u) &= f & \text{in } & \Omega \\
u &= 0 & \text{on } & \partial \Omega_D \\
\sigma(u) \cdot n &= g_T & \text{on } & \partial \Omega_T
\end{aligned}
$$

where :
- $V = \{ v \in (H^1(\Omega))^3 : v = 0 \text{ on } \partial \Omega_D \}$,

- $g_T$ is the traction vector on the part $\partial \Omega_T$ of the boundary,

- $n$ is the outward normal vector on the boundary $\partial \Omega$,

- $\partial \Omega_D \cap \partial \Omega_T = \emptyset$ and $\partial \Omega = \partial \Omega_D \cup \partial \Omega_T$.
- $\partial \Omega_D$ is the Dirichlet boundary condition where the displacement is fixed to zero, and $\partial \Omega_T$ is the Neumann boundary condition where the traction is prescribed.

## The Variational Formulation
The variational formulation of the linear elasticity equations involves forming the inner product of the PDE with a vector test function $ v \in V $ and integrating over the domain $ \Omega $. This yields:

$$
\int_{\Omega} - \nabla \cdot \sigma(u) \cdot v \, \mathrm{d} x = \int_{\Omega} f \cdot v \, \mathrm{d} x
$$

Integrating the term $ \nabla \cdot \sigma(u) \cdot v $ by parts, considering boundary conditions, we obtain:

$$
\int_{\Omega} \sigma(u) : \nabla v \, \mathrm{d} x = \int_{\Omega} f \cdot v \, \mathrm{d} x + \int_{\partial \Omega_T} g_T \cdot v \, \mathrm{d} s
$$

By using the symmetry of the stress tensor $ \sigma $ and its definition from $(2)$, we can notice that :

$$
\begin{aligned}
\int_{\Omega} \sigma(u) : \nabla v \, \mathrm{d} x &= \int_{\Omega} \sigma(u) : \epsilon(v) \, \mathrm{d} x = \int_{\Omega} C : \epsilon(u) : \epsilon(v) \, \mathrm{d} x \\ &= \int_{\Omega} \epsilon(u) : C : \epsilon(v) \, \mathrm{d} x
\end{aligned}
$$

This leads to the following variational formulation:

$$
\boxed{
\begin{aligned}
&\text{Find } u \in V \text{ such that:} \\
&\qquad a(u, v) = L(v) \quad \forall v \in V
\end{aligned}
}
$$

with

$$
\begin{aligned}
&a :
\begin{cases}
V \times V \rightarrow \mathbb{R} \\
(u, v) \longmapsto \int_{\Omega} \epsilon(u) : C : \epsilon(v) \, \mathrm{d} x
\end{cases} \\[0.3cm]
&L :
\begin{cases}
V \rightarrow \mathbb{R} \\
v \longmapsto \int_{\Omega} f \cdot v \, \mathrm{d} x + \int_{\partial \Omega_T} g_T \cdot v \, \mathrm{d} s
\end{cases}
\end{aligned}
$$

## Isotropic Materials
For isotropic materials, the elasticity tensor $C$ can be expressed in terms of the Lamé parameters $\lambda$ and $\mu$ as follows:

$$
C := \lambda (\nabla \cdot u) I_3 + 2\mu \epsilon(u)
$$
Then, the stress tensor can be expressed as:
$$\sigma(u) = \lambda (\nabla \cdot u) I_3 + 2\mu \epsilon(u)$$


This leads to the variational formulation:
$$
\boxed{
\begin{aligned}
&\text{Find } u \in V \text{ such that:}
\\
&\qquad a(u, v) = L(v) \quad \forall v \in V
\end{aligned}
}
$$

with
$$
\begin{aligned}
&a :\begin{cases}
V \times V \rightarrow \mathbb{R} \\
(u, v) \longmapsto \int_{\Omega} \sigma(u) : \epsilon(v) \, \mathrm{d} x
\end{cases} \\[0.3cm]
&L :\begin{cases}
V \rightarrow \mathbb{R} \\
v \longmapsto \int_{\Omega} f \cdot v \, \mathrm{d} x + \int_{\partial \Omega_T} g_T \cdot v \, \mathrm{d} s
\end{cases}
\end{aligned}
$$

With this formulation, the problem is well-posed under the assumption that the material is isotropic and the boundary conditions are properly defined. While $\frac{\lambda}{\mu}$ is not too large (typically $\frac{\lambda}{\mu} \leq 10^4$), the problem remains well-posed numerically. However, as $\frac{\lambda}{\mu}$ increases, the problem can become ill-posed, leading to numerical difficulties in finding a solution. The first notebook of this chapter illustrates the case of isotropic materials with $\frac{\lambda}{\mu} \leq 10^4$ and the second notebook is trying to illustrate the case of isotropic materials with $\frac{\lambda}{\mu} > 10^4$.

<p align="center">

| Material | $\lambda$ (GPa) | $\mu$ (GPa) |
|-----------|:--------------:|:-----------:|
| Steel | 120 | 80 |
| Concrete | 17 | 14 |
| Rubber | 0.16 | 0.00033 |

</p>
Loading