弹性波动力学笔记(六) 应力张量简介中

3.4 The Equation Of Motion and Symmetry of the Stress Tensor

Here we will combine several results from the previous sections into the so-called equation, which constitutes one of the most important results of this chapter. we will see that this equation leads to the elastic wave equation and in this section it will be used to prove that the stress tensor \(\tau_{ij}\) is symmetric. From equation (3.2.3) and (3.3.6) and Gauss' theorem for tensors we find

\[\int_{S}n_j\tau_{ji}d{S}+\int_{V}\rho f_i d{V}=\int_{V}(\tau_{ji,j}+\rho f_i)d{V}=\int_{V}{\rho\frac{D{v_i}}{D{t}}}d{V} \tag{3.4.1} \]

where \(\tau_{ji,j}\) is the divergence of \(\tau_{ji}\) , which is a vector. A good discussion of \(\tau_{ji,j}\) and its expression in cylindrical and spherical is provided by Auld(1990).

The last equality in (3.4.1) can be written as

\[ \int_{V}(\tau_{ji,j}+\rho f_i-\rho\frac{D{v_i}}{D{t}})dV=0 \tag{3.4.2} \]

Since (3.4.2) is valid for any arbitrary volume V inside the body, and continuity of the integrand is assumed, we obtain

\[ \tau_{ji,j}+\rho f_i=\rho\frac{Dv_i}{Dt} \tag{3.4.3} \]

The proof that equation(3.4.3) follows equation (3.4.2) is by contradiction. If the integrand were different from zero. Equation (3.4.3) is known as Euler's equation of motion. In dyadic form (3.4.3) becomes

\[\nabla \cdot{T}+\rho\mathbf{f}=\rho\frac{D{\mathbf{v}}}{D{t}} \tag{3.4.4} \]

The symmetry of the stress tensor will be proved using the principle of angular momentum, which should be written as follows when body forces are present:

\[\frac{d}{dt}\int_V(\mathbf{r}\times\rho\mathbf{v})dV=\int_{S}\mathbf{r}\times\mathbf{T}d{S}+\int_{V}(\mathbf{r}\times\rho\mathbf{f})d{V} \tag{3.4.5} \]

Writing in component form, collecting terms in one side, introducing \(\tau_{ij}\) , and using Gauss' theorem and

\[\frac{d}{d{t}}\int_V\rho\phi d{V}=\int_{V}\rho\frac{D{\phi}}{D{t}}d{V} \tag{3.4.6} \]

where \(\phi\) is any scalar, vector, or tensor property we obtain

\[\begin{equation} \begin{aligned} 0 &=\int_V\epsilon_{ijk}\Bigg\{ (x_j\tau_{rk})_{,r}+\rho x_jf_k-\frac{D}{Dt}(x_jv_k)\Bigg\}dV \\ &=\int_V\epsilon_{ijk}\Bigg\{ (\tau_{jk}+x_j\tau_{rk,r})+\rho x_{j}f_{k}-\rho v_k v_j-\rho x_j \frac{Dv_k}{Dt} \Bigg\}d{V} \end{aligned} \tag{3.4.7} \end{equation} \]

where the relation \(x_{j,r}=\delta_{jr}\) was used. Now using equation (3.4.3) and symmetry of \(v_jv_k\), equation (3.4.7) becomes

\[ \int_V\epsilon_{ijk}\tau_{jk}dV=0 \tag{3.4.8} \]

Because the integration volume is arbitrary, if the integrand in equation (3.4.8) is continuous it must be equal to zero:

\[ \epsilon_{ijk}\tau_{jk}=0, \tag{3.4.9} \]

which means that \(\tau_{jk}\) is symmetric,

\[\tau_{jk}=\tau_{kj} \tag{3.4.10} \]

Equation (3.4.10) can be used to rewrite (3.3.14) as

\[ T_j(\mathbf{n})=\tau_{ji}n_i \space j=1,2,3, \tag{3.4.11} \]

so that the matrix equation (3.3.15) becomes

\[\left( {\begin{array}{*{20}{c}} {{T_1}}&\\{{T_2}}&\\{{T_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{\tau _{11}}}&{{\tau _{12}}}&{{\tau _{13}}}\\ {{\tau _{21}}}&{{\tau _{22}}}&{{\tau _{23}}}\\ {{\tau _{31}}}&{{\tau _{32}}}&{{\tau _{33}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{n_1}}&\\{{n_2}}&\\{{n_3}} \end{array}} \right) \tag{3.4.12} \]

Equation (3.4.12) can also be written in dyadic form:

\[ \mathbf{T}=T\cdot{\mathbf{n}} \tag{3.4.13} \]