弹性波动力学笔记(五) 应变张量简介下

2.5 Proof that strain tensor is a tensor

To prove that $\varepsilon_{ij}$ is a tensor it is necessary to show that under a rotation of axis the components transform according to

\begin{matrix} (2.5.1) & ε_{k l}^{^{'}} = a_{k i} a_{l j} ε_{i j} \end{matrix}

$\varepsilon_{kl}^{'}=a_{ki}a_{lj}\varepsilon_{ij} \tag{2.5.1}$

To prove equation (2.5.1) let us translate the coordinate axes in such a way that the new origin is at the end of $\mathbf{R}$ . Then $dX_i$ can be replaced by $X_i$ , where $X_i$ indicates (temporarily) local coordinates measured with respect to the new origin. Then, equation (2.4.6) can be written as

\begin{matrix} (2.5.2) & k = d S (d s - d S) = ε_{i j} X_{i} X_{j} \end{matrix}

$k=dS(ds-dS)=\varepsilon_{ij}X_iX_j \tag{2.5.2}$

Because $k$ is a quantity related to the length of vectors, it is independent of the coordinate system used to describe the deformation. Equation (2.5.2) represents a quadratic function, and is known as the strain quadric. After a rotation of the local coordinates the following relations as satisfied:

\begin{matrix} (2.5.3) & X_{i}^{^{'}} = a_{i j} X_{j} \end{matrix}

$X_{i}^{'}=a_{ij}X_j \tag{2.5.3}$

\begin{matrix} (2.5.4) & X_{i} = a_{j i} X_{j}^{^{'}} \end{matrix}

$X_i=a_{ji}X_{j}^{'} \tag{2.5.4}$

In the rotated system the relation equivalent to (2.4.24) is

\begin{matrix} (2.5.5) & k = ε_{i j}^{^{'}} X_{i}^{^{'}} X_{j}^{^{'}} \end{matrix}

$k=\varepsilon_{ij}^{'}X_i^{'}X_j^{'} \tag{2.5.5}$

Therefore, from equation (2.5.2) and (2.5.5) we obtain

\begin{matrix} (2.5.6) & ε_{i j} X_{i} X_{j} = ε_{i j}^{^{'}} X_{i}^{^{'}} X_{j}^{^{'}} \end{matrix}

$\varepsilon_{ij}X_iX_j=\varepsilon_{ij}^{'}X_i^{'}X_j^{'} \tag{2.5.6}$

Introducing (2.5.3) and (2.5.4) gives

\begin{matrix} (2.5.7) & ε_{i j} a_{k i} X_{k}^{^{'}} a_{l j} X_{l}^{^{'}} = ε_{k l}^{^{'}} X_{k}^{^{'}} X_{l}^{^{'}} \end{matrix}

$\varepsilon_{ij}a_{ki}X_k^{'}a_{lj}X_l^{'}=\varepsilon_{kl}^{'}X_k^{'}X_l^{'} \tag{2.5.7}$

which can be written as

\begin{matrix} (2.5.8) & (a_{k i} a_{l j} ε_{i j} - ε_{k l}^{^{'}}) X_{k}^{^{'}} X_{l}^{^{'}} = 0 \end{matrix}

$(a_{ki}a_{lj}\varepsilon_{ij}-\varepsilon_{kl}^{'})X_{k}^{'}X_l^{'}=0 \tag{2.5.8}$

This equation is valid for arbitrary $X_k^{'}X_{l}^{'}$ , which implies that the term in parentheses has to be equal to zero. Therefore,

\begin{matrix} (2.5.9) & ε_{k l}^{^{'}} = a_{k i} a_{l j} ε_{i j} \end{matrix}

$\varepsilon_{kl}^{'}=a_{ki}a_{lj}\varepsilon_{ij} \tag{2.5.9}$

which shows that $\varepsilon_{ij}$ is a tensor.

2.6 The rotation tensor and strain tensor

The strain tensor was introduced by analyzing the change in length of line elements, but it does not represent the whole effect of the deformation. To see this refer to Fig. 2.2. The displacement vector $\mathbf{u}$ is a function of the coordinates of the point in the body being considered. To make this dependence explicit we will write $\mathbf{u}(\mathbf{R})$ . In general, the displacement will be different at different points. Therefore, the difference $\mathbf{u}(\mathbf{R}+d\mathbf{R})-\mathbf{u}(\mathbf{R})$ fully describes the deformation in the vicinity of $\mathbf{R}$ . Expanding this difference in a Taylor series under the assumption of small deformations we obtain

\begin{matrix} (2.6.1) & d u_{i} = u_{i} (R + d R) - u_{i} (R) = \frac{\partial u_{i}}{\partial X_{j}} d X_{j} = u_{i, j} d X_{j} \end{matrix}

$du_i=u_i(\mathbf{R}+d\mathbf{R})-u_i(\mathbf{R})=\frac{\part{u_i}}{\part{X_j}}d{X_j}=u_{i,j}d{X_j}\tag{2.6.1}$

Because of the small-deformation hypothesis the higher order terms in $dX_j$ can be neglected.

We know that $u_{i,j}$ is a tensor, which can be written as the sum of two tensors, one symmetric and one anti-symmetric. Adding and subtracting $u_{j,i}dX_j/2$ to the last term of equation (2.6.1) we obtain

\begin{matrix} (2.6.2) & d u_{i} = \frac{1}{2} (u_{i, j} + u_{j, i}) d X_{j} + \frac{1}{2} (u_{i, j} - u_{j, i}) d X_{j} \end{matrix}

$d{u_i}=\frac{1}{2}(u_{i,j}+u_{j,i})d{X_j}+\frac{1}{2}(u_{i,j}-u_{j,i})d{X_j} \tag{2.6.2}$

where

\begin{matrix} (2.6.3) & w_{i j} = \frac{1}{2} (u_{i, j} - u_{j, i}) \end{matrix}

$w_{ij}=\frac{1}{2}(u_{i,j}-u_{j,i}) \tag{2.6.3}$

The tensor $w_{ij}$ is anti-symmetric, and it is likely to be related some infinitesimal rotation. To show that this is indeed the case note that

\begin{matrix} (2.6.4) & d r = d R + d u \end{matrix}

$d{\mathbf{r}}=d{\mathbf{R}}+d{\mathbf{u}} \tag{2.6.4}$

wrriten as

\begin{matrix} (2.6.5) & d x_{i} = d X_{i} + d u_{i} \end{matrix}

$d{x_i}=d{X_i}+d{u_i} \tag{2.6.5}$

Then using (2.6.1) and (2.6.5) and writing $dX_i=\delta_{ij}dX_j$ , we find

\begin{matrix} (2.6.6) & d x_{i} = d X_{i} + [ε_{i j} + (δ_{i j} + w_{i j})] d X_{j} \end{matrix}

$d{x_i}=d{X_i}+[\varepsilon_{ij}+(\delta_{ij}+w_{ij})]dX_j \tag{2.6.6}$

we know that $\delta_{ij}+w_{ij}$ represents an infinitesimal rotation. For this reason $w_{ij}$ is known as the rotation tensor. Therefore, the deformation of the line element $d{\mathbf{R}}$ consists of two terms，one involving the strain tensor already described， and the other being an infinitesimal rotation of $d{\bf{R}}$ . It is important to recognize that this is a local rotation, associated with a particular $d{\bf{R}}$ , not a whole-body rotation.

The vector $w_{k}$ associated with the anti-symmetric tensor $w_{ij}$ is

\begin{matrix} (2.6.7) & ω_{i} = \frac{1}{2} ϵ_{i j k} ω_{j k} = \frac{1}{4} (ϵ_{i j k} u_{j, k} - ϵ_{i j k} u_{k, j}) = \frac{1}{4} (ϵ_{i j k} u_{j, k} - ϵ_{i k j} u_{j, k}) = - \frac{1}{2} ϵ_{i k j} u_{j, k} = - \frac{1}{2} {(\nabla \times u)}_{i} \end{matrix}

$\omega_{i}=\frac{1}{2}\epsilon_{ijk}\omega_{jk}=\frac{1}{4}(\epsilon_{ijk}u_{j,k}-\epsilon_{ijk}u_{k,j})=\frac{1}{4}(\epsilon_{ijk}u_{j,k}-\epsilon_{ikj}u_{j,k})=-\frac{1}{2}\epsilon_{ikj}u_{j,k}=-\frac{1}{2}\left(\nabla \times \bf{u} \right)_{i} \tag{2.6.7}$

Here we have used the following facts: $\epsilon_{ijk}u_{k,j}=\epsilon_{ikj}u_{j,k}$ because $j$ and $k$ are dummy indices, and $\epsilon_{ijk}=-\epsilon_{ikj}$ .

Equations (2.6.7) and (2.6.6) show that the term $\delta_{ij}+\omega_{ij}$ corresponds to the rotation of the vector element $d{\mathbf{R}}$ through a small angle $\frac{1}{2}|\nabla\times{\mathbf{u}} |$ about an axis parallel to $\nabla\times\mathbf{u}$ .