Solid mechanics

Strong formulation

The strong form of deformable solids is given by the Cauchy momentum equation

$$\begin{equation} \nabla \cdot \boldsymbol{\sigma} + \boldsymbol{f} = \rho \boldsymbol{\ddot{u}}, \end{equation}$$

where

• $\boldsymbol{\sigma}$ is the stress tensor
• $\boldsymbol{f}$ is the volumetric body force vector
• $\rho$ is the mass density
• $\boldsymbol{u}$ is the displacement field vector
• $\boldsymbol{\ddot{u}}$ is the acceleration vector

Constitutive equation

Generalized Hooke’s law defines the relation between the stress and strain components

$$\boldsymbol{\sigma} = \boldsymbol{C} \boldsymbol{\varepsilon},$$

where $\boldsymbol{C}$ is a fourth-order elasticity tensor and $\boldsymbol{\varepsilon}$ is the strain tensor.

Strain tensor

• Small displacement theory: $\qquad$

$$\boldsymbol{\varepsilon} = \dfrac{1}{2} \left( \nabla \boldsymbol{u} + \nabla \boldsymbol{u}^T \right)$$

• Large displacement theory: $\qquad$

$$\boldsymbol{\varepsilon} = \dfrac{1}{2} \left( \nabla \boldsymbol{u} + \nabla \boldsymbol{u}^T + \nabla \boldsymbol{u}^T \nabla \boldsymbol{u} \right)$$

Voigt notation

In Voigt notation, the stress and strain tensors are represented as vectors. Consequently, the constitutive relation takes the form

$$\boldsymbol{\sigma}_v = \boldsymbol{H} \boldsymbol{\varepsilon}_v,$$

where $\boldsymbol{H}$ is the elasticity matrix which is a symmetric $6 \times 6$ matrix.

Weak formulation

The weak form of solid mechanics is obtained by multiplying the partial differential equation in $(1)$ with the test function of displacement field $\boldsymbol{u}^\prime$ and integrating over the whole domain $\Omega$.

$$\int_{\Omega}\; -\rho \boldsymbol{\ddot{u}} \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; (\nabla \cdot \boldsymbol{\sigma}) \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; \boldsymbol{f} \cdot \boldsymbol{u}^\prime \; d\Omega = 0$$

Applying Leibniz rule on the divergence term, we get

$$\int_{\Omega}\; -\rho \boldsymbol{\ddot{u}} \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; -\boldsymbol{\sigma} \colon \nabla \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; \nabla \cdot (\boldsymbol{\sigma} \cdot \boldsymbol{u}^\prime) \; d\Omega + \int_{\Omega}\; \boldsymbol{f} \cdot \boldsymbol{u}^\prime \; d\Omega = 0.$$

Applying Divergence theorem on the divergence term, we get

$$\int_{\Omega}\; -\rho \boldsymbol{\ddot{u}} \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; -\boldsymbol{\sigma} \colon \nabla \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; \boldsymbol{u}^\prime \cdot (\boldsymbol{\sigma} \cdot \boldsymbol{n}) \; d\Omega + \int_{\Omega}\; \boldsymbol{f} \cdot \boldsymbol{u}^\prime \; d\Omega = 0.$$

Since the stress tensor is symmetric, the following relation holds.

$$\begin{align*} \boldsymbol{\sigma} \colon \nabla \boldsymbol{u}^\prime &= \boldsymbol{\sigma} \colon \nabla {\boldsymbol{u}^\prime}^T \\ &= \boldsymbol{\sigma} \colon \dfrac{1}{2} \left(\nabla {\boldsymbol{u}^\prime}^T + \nabla {\boldsymbol{u}^\prime}^T \right) \\ &= \boldsymbol{\sigma} \colon \dfrac{1}{2} \left(\nabla {\boldsymbol{u}^\prime} + \nabla {\boldsymbol{u}^\prime}^T \right) \\ &= \boldsymbol{\sigma} \colon \boldsymbol{\varepsilon}^\prime \end{align*}$$

Substituting in the above relation $\boldsymbol{\sigma} \colon \nabla \boldsymbol{u}^\prime = \boldsymbol{\sigma} \colon \boldsymbol{\varepsilon}^\prime$ and following the definition of traction vector $\boldsymbol{t} = \boldsymbol{\sigma \cdot n}$, we get the final form of our weak formulation:

$$\int_{\Omega}\; -\rho \boldsymbol{\ddot{u}} \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Omega}\; -\boldsymbol{\sigma} \colon \boldsymbol{\varepsilon}^\prime \; d\Omega + \int_{\Omega}\; \boldsymbol{f} \cdot \boldsymbol{u}^\prime \; d\Omega + \int_{\Gamma}\; \boldsymbol{t} \cdot \boldsymbol{u}^\prime \; d\Gamma = 0$$