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# 物理代写|弹性力学代写Elasticity代考|EML6653 Strain transformation

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## 物理代写|弹性力学代写Elasticity代考|Strain transformation

Because the strains are components of a second-order tensor, the transformation theory discussed in applying this to the strain gives
$$e_{i j}^{\prime}=Q_{i p} Q_{j q} e_{p q}$$
where the rotation matrix $Q_{i j}=\cos \left(x_i^{\prime}, x_j\right)$. Thus, given the strain in one coordinate system, we can determine the new components in any other rotated system. For the general three-dimensional case, define the rotation matrix as
$$Q_{i j}=\left[\begin{array}{lll} l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 \ l_3 & m_3 & n_3 \end{array}\right]$$
Using this notational scheme, the specific transformation relations from Eq. (2.3.1) become
\begin{aligned} e_x^{\prime} &=e_x l_1^2+e_y m_1^2+e_z n_1^2+2\left(e_{x y} l_1 m_1+e_{y z} m_1 n_1+e_{z x} n_1 l_1\right) \ e_y^{\prime} &=e_x l_2^2+e_y m_2^2+e_z n_2^2+2\left(e_{x y} l_2 m_2+e_{y z} m_2 n_2+e_{z x} n_2 l_2\right) \ e_z^{\prime} &=e_x l_3^2+e_y m_3^2+e_z n_3^2+2\left(e_{x y} l_3 m_3+e_{y z} m_3 n_3+e_{z x} n_3 l_3\right) \ e_{x y}^{\prime} &=e_x l_1 l_2+e_y m_1 m_2+e_z n_1 n_2+e_{x y}\left(l_1 m_2+m_1 l_2\right)+e_{y z}\left(m_1 n_2+n_1 m_2\right)+e_{z x}\left(n_1 l_2+l_1 n_2\right) \ e_{y z}^{\prime} &=e_x l_2 l_3+e_y m_2 m_3+e_z n_2 n_3+e_{x y}\left(l_2 m_3+m_2 l_3\right)+e_{y z}\left(m_2 n_3+n_2 m_3\right)+e_{z x}\left(n_2 l_3+l_2 n_3\right) \ e_{z x}^{\prime} &=e_x l_3 l_1+e_y m_3 m_1+e_z n_3 n_1+e_{x y}\left(l_3 m_1+m_3 l_1\right)+e_{y z}\left(m_3 n_1+n_3 m_1\right)+e_{z x}\left(n_3 l_1+l_3 n_1\right) \end{aligned}
For the two-dimensional case shown in Fig. 2.7, the transformation matrix can be expressed as
$$Q_{i j}=\left[\begin{array}{cll} \cos \theta & \sin \theta & 0 \ -\sin \theta & \cos \theta & 0 \ 0 & 0 & 1 \end{array}\right]$$

## 物理代写|弹性力学代写Elasticity代考|Principal strains

From the previous discussion in Section 1.6, it follows that because the strain is a symmetric secondorder tensor, we can identify and determine its principal axes and values. According to this theory, for any given strain tensor we can establish the principal value problem and solve the characteristic equation to explicitly determine the principal values and directions. The general characteristic equation for the strain tensor can be written as
$$\operatorname{det}\left[e_{i j}-e \delta_{i j}\right]=-e^3+\vartheta_1 e^2-\vartheta_2 e+\vartheta_3=0$$
where $e$ is the principal strain and the fundamental invariants of the strain tensor can be expressed in terms of the three principal strains $e_1, e_2, e_3$ as
\begin{aligned} &\vartheta_1=e_1+e_2+e_3 \ &\vartheta_2=e_1 e_2+e_2 e_3+e_3 e_1 \ &\vartheta_3=e_1 e_2 e_3 \end{aligned}
The first invariant $\vartheta_1=\vartheta$ is normally called the cubical dilatation, because it is related to the change in volume of material elements (see Exercise 2.11).
The strain matrix in the principal coordinate system takes the special diagonal form
$$e_{i j}=\left[\begin{array}{lll} e_1 & 0 & 0 \ 0 & e_2 & 0 \ 0 & 0 & e_3 \end{array}\right]$$
Notice that for this principal coordinate system, the deformation does not produce any shearing and thus is only extensional. Therefore, a rectangular element oriented along principal axes of strain will retain its orthogonal shape and undergo only extensional deformation of its sides.

## 物理代写|弹性力学代写elastic代考|应变转换

.

$$e_{i j}^{\prime}=Q_{i p} Q_{j q} e_{p q}$$

$$Q_{i j}=\left[\begin{array}{lll} l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 \ l_3 & m_3 & n_3 \end{array}\right]$$

\begin{aligned} e_x^{\prime} &=e_x l_1^2+e_y m_1^2+e_z n_1^2+2\left(e_{x y} l_1 m_1+e_{y z} m_1 n_1+e_{z x} n_1 l_1\right) \ e_y^{\prime} &=e_x l_2^2+e_y m_2^2+e_z n_2^2+2\left(e_{x y} l_2 m_2+e_{y z} m_2 n_2+e_{z x} n_2 l_2\right) \ e_z^{\prime} &=e_x l_3^2+e_y m_3^2+e_z n_3^2+2\left(e_{x y} l_3 m_3+e_{y z} m_3 n_3+e_{z x} n_3 l_3\right) \ e_{x y}^{\prime} &=e_x l_1 l_2+e_y m_1 m_2+e_z n_1 n_2+e_{x y}\left(l_1 m_2+m_1 l_2\right)+e_{y z}\left(m_1 n_2+n_1 m_2\right)+e_{z x}\left(n_1 l_2+l_1 n_2\right) \ e_{y z}^{\prime} &=e_x l_2 l_3+e_y m_2 m_3+e_z n_2 n_3+e_{x y}\left(l_2 m_3+m_2 l_3\right)+e_{y z}\left(m_2 n_3+n_2 m_3\right)+e_{z x}\left(n_2 l_3+l_2 n_3\right) \ e_{z x}^{\prime} &=e_x l_3 l_1+e_y m_3 m_1+e_z n_3 n_1+e_{x y}\left(l_3 m_1+m_3 l_1\right)+e_{y z}\left(m_3 n_1+n_3 m_1\right)+e_{z x}\left(n_3 l_1+l_3 n_1\right) \end{aligned}

$$Q_{i j}=\left[\begin{array}{cll} \cos \theta & \sin \theta & 0 \ -\sin \theta & \cos \theta & 0 \ 0 & 0 & 1 \end{array}\right]$$

## 物理代写|弹性力学代写Elasticity代考|主应变

.

$$\operatorname{det}\left[e_{i j}-e \delta_{i j}\right]=-e^3+\vartheta_1 e^2-\vartheta_2 e+\vartheta_3=0$$
，其中$e$是主应变，应变张量的基本不变量可以用三个主应变$e_1, e_2, e_3$表示为
\begin{aligned} &\vartheta_1=e_1+e_2+e_3 \ &\vartheta_2=e_1 e_2+e_2 e_3+e_3 e_1 \ &\vartheta_3=e_1 e_2 e_3 \end{aligned}
。第一个不变量$\vartheta_1=\vartheta$通常被称为立方膨胀，因为它与材料元素的体积变化有关(见练习2.11)。

$$e_{i j}=\left[\begin{array}{lll} e_1 & 0 & 0 \ 0 & e_2 & 0 \ 0 & 0 & e_3 \end{array}\right]$$

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