Posted on Categories:Elasticity, 弹性力学, 物理代写

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## 物理代写|弹性力学代写Elasticity代考|Spherical and deviatoric strains

In particular applications it is convenient to decompose the strain tensor into two parts called spherical and deviatoric strain tensors. The spherical strain is defined by
$$\tilde{e}{i j}=\frac{1}{3} e{k k} \delta_{i j}=\frac{1}{3} \vartheta \delta_{i j}$$
while the deviatoric strain is specified as
$$\widehat{e}{i j}=e{i j}-\frac{1}{3} e_{k k} \delta_{i j}$$
Note that the total strain is then simply the sum
$$e_{i j}=\tilde{e}{i j}+\widehat{e}{i j}$$
The spherical strain represents only volumetric deformation and is an isotropic tensor, being the same in all coordinate systems (as per the discussion in Section 1.5). The deviatoric strain tensor then accounts for changes in shape of material elements. It can be shown that the principal directions of the deviatoric strain are the same as those of the strain tensor.

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

We now investigate in more detail the nature of the strain-displacement relations $(2.2 .5)$, and this will lead to the development of some additional relations necessary to ensure continuous, single-valued displacement field solutions. Relations (2.2.5), or the index notation form (2.2.6), represent six equations for the six strain components in terms of three displacements. If we specify continuous, single-valued displacements $u, v, w$, then through differentiation the resulting strain field will be equally well behaved. However, the converse is not necessarily true; given the six strain components, integration of the strain-displacement relations (2.2.5) does not necessarily produce continuous, single-valued displacements. This should not be totally surprising since we are trying to solve six equations for only three unknown displacement components. In order to ensure continuous, singlevalued displacements, the strains must satisfy additional relations called integrability or compatibility equations.

Before we proceed with the mathematics to develop these equations, it is instructive to consider a geometric interpretation of this concept. A two-dimensional example is shown in Fig. $2.8$ whereby an elastic solid is first divided into a series of elements in case (a). For simple visualization, consider only four such elements. In the undeformed configuration shown in case (b), these elements of course fit together perfectly. Next, let us arbitrarily specify the strain of each of the four elements and attempt to reconstruct the solid. For case (c), the elements have been carefully strained, taking into consideration neighboring elements so that the system fits together yielding continuous, single-valued displacements. However, for case (d), the elements have been individually deformed without any concern for neighboring deformations. It is observed for this case that the system will not fit together without voids and gaps, and this situation produces a discontinuous displacement field. So, we again conclude that the strain components must be somehow related to yield continuous, single-valued displacements. We now nursue these narticular relations.

## 物理代写|弹性力学代写弹性代考|球形和偏应变

$$\tilde{e}{i j}=\frac{1}{3} e{k k} \delta_{i j}=\frac{1}{3} \vartheta \delta_{i j}$$
，而偏应变指定为
$$\widehat{e}{i j}=e{i j}-\frac{1}{3} e_{k k} \delta_{i j}$$

$$e_{i j}=\tilde{e}{i j}+\widehat{e}{i j}$$

.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Elasticity, 弹性力学, 物理代写

## avatest™帮您通过考试

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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)。

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

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Elasticity, 弹性力学, 物理代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 物理代写|弹性力学代写Elasticity代考|Cartesian tensors

Scalars, vectors, matrices, and higher-order quantities can be represented by a general index notational scheme. Using this approach, all quantities may then be referred to as tensors of different orders. The previously presented transformation properties of a vector can be used to establish the general transformation properties of these tensors. Restricting the transformations to those only between Cartesian coordinate systems, the general set of transformation relations for various orders can be written as
\begin{aligned} a^{\prime} &=a, \text { zero order (scalar) } \ a_i^{\prime} &=Q_{i p} a_p, \text { first order (vector) } \ a_{i j}^{\prime} &=Q_{i p} Q_{j q} a_{p q}, \text { second order (matrix) } \ a_{i j k}^{\prime} &=Q_{i p} Q_{j p} Q_{k r} a_{p q r}, \text { third order } \ a_{i j k l}^{\prime} &=Q_{i p} Q_{j q} Q_{k r} Q_{l s} a_{p q r s}, \text { fourth order } \ \vdots \ a_{i j k k m}^{\prime} &=Q_{i p} Q_{j q} Q_{k r} \cdots Q_{m t} a_{p q r \ldots t}, \text { general order } \end{aligned}
Note that, according to these definitions, a scalar is a zero-order tensor, a vector is a tensor of order one, and a matrix is a tensor of order two. Relations (1.5.1) then specify the transformation rules for the components of Cartesian tensors of any order under the rotation $Q_{i j}$. This transformation theory proves to be very valuable in determining the displacement, stress, and strain in different coordinate directions. Some tensors are of a special form in which their components remain the same under all transformations, and these are referred to as isotropic tensors. It can be easily verified (see Exercise 1.8) that the Kronecker delta $\delta_{i j}$ has such a property and is therefore a second-order isotropic tensor. The alternating symbol $\varepsilon_{i j k}$ is found to be the third-order isotropic form. The fourth-order case (Exercise 1.9) can be expressed in terms of products of Kronecker deltas, and this has important applications in formulating isotropic elastic constitutive relations in Section $4.2$.

## 物理代写|弹性力学代写Elasticity代考|Principal values and directions for symmetric second-order tensors

Considering the tensor transformation concept previously discussed, it should be apparent that there might exist particular coordinate systems in which the components of a tensor take on maximum or minimum values. This concept is easily visualized when we consider the components of a vector as shown in Fig. 1.1. If we choose a particular coordinate system that has been rotated so that the $x_3$-axis lies along the direction of the vector, then the vector will have components $v={0,0,|v|}$. For this case, two of the components have been reduced to zero, while the remaining component becomes the largest possible (the total magnitude).

This situation is most useful for symmetric second-order tensors that eventually represent the stress and/or strain at a point in an elastic solid. The direction determined by the unit vector $\boldsymbol{n}$ is said to be a principal direction or eigenvector of the symmetric second-order tensor $a_{i j}$ if there exists a parameter $\lambda$ such that
$$\begin{gathered} a_{i j} n_j=\lambda n_i \ a_{11} n_1+a_{12} n_2+a_{13} n_3=\lambda n_1 \ a_{21} n_1+a_{22} n_2+a_{23} n_3=\lambda n_2 \ a_{31} n_1+a_{32} n_2+a_{33} n_3=\lambda n_3 \end{gathered}$$

## 物理代写|弹性力学代写弹性代考|笛卡尔张量

\begin{aligned} a^{\prime} &=a, \text { zero order (scalar) } \ a_i^{\prime} &=Q_{i p} a_p, \text { first order (vector) } \ a_{i j}^{\prime} &=Q_{i p} Q_{j q} a_{p q}, \text { second order (matrix) } \ a_{i j k}^{\prime} &=Q_{i p} Q_{j p} Q_{k r} a_{p q r}, \text { third order } \ a_{i j k l}^{\prime} &=Q_{i p} Q_{j q} Q_{k r} Q_{l s} a_{p q r s}, \text { fourth order } \ \vdots \ a_{i j k k m}^{\prime} &=Q_{i p} Q_{j q} Q_{k r} \cdots Q_{m t} a_{p q r \ldots t}, \text { general order } \end{aligned}

## 物理代写|弹性力学代写弹性代考|对称二阶张量的主值和方向

$$\begin{gathered} a_{i j} n_j=\lambda n_i \ a_{11} n_1+a_{12} n_2+a_{13} n_3=\lambda n_1 \ a_{21} n_1+a_{22} n_2+a_{23} n_3=\lambda n_2 \ a_{31} n_1+a_{32} n_2+a_{33} n_3=\lambda n_3 \end{gathered}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。