2. Cosserat linear elastodynamics

#### 2.1 Fundamental equations

The Cosserat linear elasticity balance laws are

$$
\sigma\_{ji,j} = \frac{\partial p\_i}{\partial t},
\tag{1}
$$

$$
\omega\_{ijk}\sigma\_{jk} + \mu\_{ji,j} = \frac{\partial q\_i}{\partial t},\tag{2}
$$

where the <sup>σ</sup>ji is the stress tensor, <sup>μ</sup>ji the couple stress tensor, pi <sup>¼</sup> <sup>ρ</sup> <sup>∂</sup>ui <sup>∂</sup><sup>t</sup> and qi ¼ Jji ∂ϕj <sup>∂</sup><sup>t</sup> are the linear and angular momenta, ρ and Jji are the material density and the rotatory inertia characteristics, εijk is the Levi-Civita tensor.

We will also consider the constitutive equations as in [13]:

$$
\sigma\_{\vec{\mu}} = (\mu + a)\chi\_{\vec{\mu}} + (\mu - a)\chi\_{\vec{\eta}} + \lambda \gamma\_{kk}\delta\_{\vec{\eta}\flat} \tag{3}
$$

$$
\mu\_{ji} = (\chi + \varepsilon)\chi\_{ji} + (\chi - \varepsilon)\chi\_{\vec{\imath}\vec{\jmath}} + \beta \chi\_{kk}\delta\_{\vec{\imath}\vec{\jmath}}.\tag{4}
$$

and the kinematic relations in the form

$$
\chi\_{ji} = u\_{i,j} + \varepsilon\_{ijk} \phi\_k \text{ and } \chi\_{ji} = \phi\_{i,j}. \tag{5}
$$

Here ui and ϕ<sup>i</sup> represent the displacement and microrotation vectors, γji and χji represent the strain and bend-twist tensors, μ, λ are the Lamé parameters and α, β, γ, ε are the Cosserat elasticity parameters.

The constitutive Eqs. (3)–(4) can be written in the reverse form [5].

$$
\gamma\_{\vec{\mu}} = (\mu' + a')\sigma\_{\vec{\mu}} + (\mu' - a')\sigma\_{\vec{\eta}} + \lambda'\sigma\_{kk\nu} \tag{6}
$$

Distinctive Characteristics of Cosserat Plate Free Vibrations DOI: http://dx.doi.org/10.5772/intechopen.87044

$$
\lambda \chi\_{\vec{\mu}} = (\chi' + \varepsilon')\mu\_{\vec{\mu}} + (\chi' - \varepsilon')\mu\_{\vec{\eta}} + \beta' \mu\_{kk\nu} \tag{7}
$$

where <sup>μ</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup><sup>μ</sup>, <sup>α</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup><sup>α</sup>, <sup>γ</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup> 4γ , <sup>ε</sup> <sup>¼</sup> <sup>1</sup> 4ε , <sup>λ</sup><sup>0</sup> <sup>¼</sup> �<sup>λ</sup> <sup>6</sup>μ λþ2<sup>μ</sup> ð Þ<sup>3</sup> and <sup>β</sup><sup>0</sup> <sup>¼</sup> �<sup>β</sup> <sup>6</sup>μ βþ2<sup>γ</sup> ð Þ<sup>3</sup> . We will consider the boundary conditions given in [12].

$$u\_i = u\_i^0, \phi\_i = \phi\_i^0, \text{on } \mathcal{G}\_1^t = \partial B\_0 \wr \partial B\_\sigma \times [t\_0, t], \tag{8}$$

$$
\sigma\_{\vec{\mu}} n\_{\vec{\imath}} = \sigma\_{\vec{\jmath}}^{0}, \mu\_{\vec{\jmath}} n\_{\vec{\imath}} = \mu\_{\vec{\jmath}}^{0} \text{ on } \mathcal{G}\_2^t = \partial B\_{\sigma} \times [t\_0, t], \tag{9}
$$

and initial conditions

$$u\_i(\mathbf{x}, \mathbf{0}) = U\_i^0, \phi\_i(\mathbf{x}, \mathbf{0}) = \Phi\_i^0, \text{in} \, B\_{0i} \tag{10}$$

$$
\dot{\mu}\_i(\mathbf{x}, \mathbf{0}) = \dot{U}\_i^0,\\
\dot{\phi}\_i(\mathbf{x}, \mathbf{0}) = \dot{\Phi}\_i^0,\\
\text{in} \, B\_{0i} \tag{11}
$$

where u<sup>0</sup> <sup>i</sup> and ϕ<sup>0</sup> <sup>i</sup> are prescribed on G1, σ<sup>0</sup> <sup>j</sup> and μ<sup>0</sup> <sup>j</sup> on G2, and ni is the unit vector normal to the boundary ∂B<sup>0</sup> of the elastic body B0.

#### 2.2 Cosserat elastic energy

The strain stored energy UC of the body B<sup>0</sup> is defined by the integral [13]:

$$U\_C = \int\_{B\_0} \mathcal{W}\{\mathbf{y}, \mathbf{y}\} dv,\tag{12}$$

where

$$\begin{split} \mathcal{W}\{\mathbf{y}, \mathbf{y}\} &= \frac{\mu + a}{2} \chi\_{\vec{\mathbf{y}}} \chi\_{\vec{\mathbf{y}}} + \frac{\mu - a}{2} \chi\_{\vec{\mathbf{y}}} \chi\_{\vec{\mathbf{y}}} + \frac{\lambda}{2} \chi\_{kk} \chi\_{nn} \\ &+ \frac{\chi + \varepsilon}{2} \chi\_{\vec{\mathbf{y}}} \chi\_{\vec{\mathbf{y}}} + \frac{\chi - \varepsilon}{2} \chi\_{\vec{\mathbf{y}}} \chi\_{\vec{\mathbf{y}}} + \frac{\beta}{2} \chi\_{kk} \chi\_{nn}, \end{split} \tag{13}$$

is non-negative. The relations Eqs. (3)–(4) can be written in the form [12]:

$$
\sigma = \nabla\_{\mathcal{T}} \mathbf{W} \text{ and } \boldsymbol{\mu} = \nabla\_{\mathcal{X}} \mathbf{W}. \tag{14}
$$

The stress energy is given as

$$U\_K = \int\_{B\_0} \Phi \{ \sigma, \mu \} dv,\tag{15}$$

where

$$\begin{split} \Phi \{ \sigma, \mu \} &= \frac{\mu' + \alpha'}{2} \sigma\_{\vec{\eta}} \sigma\_{\vec{\eta}} + \frac{\mu' - \alpha'}{2} \sigma\_{\vec{\eta}} \sigma\_{\vec{\eta}} + \frac{\lambda'}{2} \sigma\_{kk} \sigma\_{nn} \\ &+ \frac{\chi' + \varepsilon'}{2} \mu\_{\vec{\eta}} \mu\_{\vec{\eta}} + \frac{\chi' - \varepsilon'}{2} \mu\_{\vec{\eta}} \mu\_{\vec{\eta}} + \frac{\beta'}{2} \mu\_{kk} \mu\_{nn}, \end{split} \tag{16}$$

and the relations Eqs. (6)–(7) can be written as [12].

$$\chi = \frac{\partial \Phi}{\partial \sigma}, \text{ and } \chi = \frac{\partial \Phi}{\partial \mu}. \tag{17}$$

We consider the work done by the stresses σ and μ over the strains γ and χ as in [13].

$$U = \int\_{B\_0} [\sigma \cdot \mathfrak{y} + \mathfrak{mu} \cdot \mathfrak{x}] dv \tag{18}$$

and

$$U = U\_K = U\_\mathcal{C} \tag{19}$$

Here σ � γ ¼ σjiγji and μ � χ ¼ μjiχji: The stored kinetic energy TC is defined as

$$T\_C = \int\_{B\_0} \mathbf{Y}\_C dv = \frac{1}{2} \int\_{B\_0} \left( \rho \left( \frac{\partial \mathbf{u}}{\partial t} \right)^2 + \mathbf{J} \left( \frac{\partial \phi}{\partial t} \right)^2 \right) dv,\tag{20}$$

The kinetic energy TK is given as

$$T\_K = \int\_{B\_0} \mathbf{Y}\_K \{ \mathbf{p}, \mathbf{q} \} dv = \frac{1}{2} \int\_{B\_0} (\mathbf{p}^2 \rho^{-1} + \mathbf{q}^2 \mathbf{J}^{-1}) dv,\tag{21}$$

where

$$\mathbf{p} = \frac{\partial \mathbf{Y}\_C}{\partial \dot{\mathbf{u}}} = \rho \frac{\partial \mathbf{u}}{\partial t} \text{ and } \mathbf{q} = \frac{\partial \mathbf{Y}\_C}{\partial \dot{\phi}} = \mathbf{J} \frac{\partial \phi}{\partial t}, \tag{22}$$

and

$$\frac{\partial \mathbf{u}}{\partial t} = \frac{\partial \mathbf{Y}\_K}{\partial \mathbf{p}} = \mathbf{p} \rho^{-1} \text{ and } \frac{\partial \boldsymbol{\Phi}}{\partial t} = \frac{\partial \mathbf{Y}\_K}{\partial \mathbf{q}} = \mathbf{q} \mathbf{J}^{-1}, \tag{23}$$

The work TW done by the inertia forces over displacement and microrotation is given as in [12].

$$T\_W = \int\_{B\_0} \Upsilon\_W dv = \int\_{B\_0} \left(\frac{\partial \mathbf{p}}{\partial t} \cdot \mathbf{u} + \frac{\partial \mathbf{q}}{\partial t} \cdot \boldsymbol{\phi}\right) dv \tag{24}$$

Keeping in mind that the variation of p u, q, ϕ, δu, and δϕ is zero at t<sup>0</sup> and tk we can integrate by parts

$$\int\_{t\_0}^{t\_k} T\_K dt = \int\_{t\_0}^{t\_k} T\_I dt = \frac{1}{2} \int\_{B\_0} (\mathbf{p} \cdot \mathbf{u} + \mathbf{q} \cdot \boldsymbol{\phi}) dv \Big|\_{t\_0}^{t\_k} - \int\_{t\_0}^{t\_k} T\_W dt \tag{25}$$

$$
\delta \int\_{t\_0}^{t\_k} T\_K = -\delta \int\_{t\_0}^{t\_k} T\_W \tag{26}
$$

or

$$
\delta T\_C = \delta T\_K = -\delta T\_W \tag{27}
$$

and therefore

$$\int\_{t\_0}^{t\_k} \int\_{B\_0} \left( \left( \mathbf{p} \cdot \delta \left( \frac{\partial \mathbf{u}}{\partial t} \right) + \mathbf{q} \cdot \delta \left( \frac{\partial \phi}{\partial t} \right) \right) \right) dvdt = - \int\_{t\_0}^{t\_k} \int\_{B\_0} \left( \frac{\partial \mathbf{p}}{\partial t} \cdot \delta \mathbf{u} + \frac{\partial \mathbf{q}}{\partial t} \cdot \delta \phi \right) dvdt \tag{28}$$
