# 数学代写|有限元方法代写Finite Element Method代考|ENG4118

#### Doug I. Jones

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## 数学代写|有限元方法代写Finite Element Method代考|Stress transformations1

The state of stress of the point $P^{\prime}$ is expressed by the stress tensor in the global, Cartesian, coordinates $(x, y, z)$ by Eq. (2.5). However, the choice of this coordinate system and the small volume $(d V=d x . d y . d z)$ is arbitrary. We would like to be able to transform the stress state between different orientations.

In order to develop these transformations, we consider an oblique crosssection of the hexahedron by a plane of arbitrary, but known orientation. This results in the tetrahedral volume on one side of the plane as shown in Fig. 2.3. The traction components $\vec{T}_x, \vec{T}_y, \vec{T}_z$, and $\vec{T}_n$, acting on the small triangular areas $\Delta A_x, \Delta A_y, \Delta A_z$ and $\Delta A_n$, respectively, must be in static equilibrium in order to keep the continuum whole. Let us look into the arbitrary plane in more detail before we state this equilibrium condition.

The orientation of the plane is identified by its outward unit normal $\vec{n}$, as shown in the figure. The unit normal is defined as follows:
$$\vec{n}=n_x \hat{i}+n_y \hat{j}+n_z \hat{k}$$
in Cartesian coordinate system. In vector notation, the unit normal is expressed as follows:
$${n}=\left{\begin{array}{lll} n_x & n_y & n_z \end{array}\right}^T$$
The components $n_x, n_y$, and $n_z$ are the direction cosines of $\vec{n}$ with respect to the $(x, y, z)$ axes, and have the property,
$$n_x^2+n_y^2+n_z^2=1$$
It can be shown that the direction cosines can be expressed as the ratios $\Delta A_i$ to $\Delta A_n$ as follows [7]:
$$n_i=\frac{\Delta A_i}{\Delta A_n} \text { for } i=x, y, z$$
Let’s consider traction $\vec{T}n$ acting on the oblique plane in Fig. 2.3. Since the traction represents the intensity of force per unit area, it can be expressed by the vector representation as follows: $$\vec{T}_n=T{n x} \hat{i}+T_{n y} \hat{j}+T_{n z} \hat{k}$$
Next, consider the static equilibrium of the forces acting on the pyramid, as follows:
$$\vec{T}n \Delta A_n=\vec{T}_x \Delta A_x+\vec{T}_y \Delta A_y+\vec{T}_z \Delta A_z$$ where $\vec{T}_x, \vec{T}_y$ and $\vec{T}_z$ are the tractions acting on the other faces $\left(\Delta A_x, \Delta A_y\right.$, and $\Delta A_z$ ) of the tetrahedron. By using Eq. (2.10), this expression becomes, $$\vec{T}_n=\vec{T}_x n_x+\vec{T}_y n_y+\vec{T}_z n_z$$ On each lateral face of the tetrahedron, we can express the tractions in terms of their components, by using Eqs. (2.1) and (2.2) as follows: \begin{aligned} & \vec{T}_x=\sigma{x x} \hat{i}+\tau_{x y} \hat{j}+\tau_{x z} \hat{k} \ & \vec{T}y=\tau{y x} \hat{i}+\sigma_{y y} \hat{j}+\tau_{y z} \hat{k} \ & \vec{T}z=\tau{z x} \hat{i}+\tau_{z y} \hat{j}+\sigma_{z z} \hat{k} \end{aligned}

## 数学代写|有限元方法代写Finite Element Method代考|Principal stresses and directions

Infinitely many planes can pass through a given point in a continuum, such as the plane that gives rise to the tetrahedron shown in Fig. 2.3. The normal and shear traction components $\sigma_{n n}$ and $\sigma_{n t}$ will vary according to the orientation of the cutting-plane. It is reasonable to assume that among these planes, there is one on which the shear tractions vanish, i.e., $\sigma_{n t}=0$. This plane is called the principal plane, and its orientation is called the principal orientation $\vec{n}p$. The normal component of the traction $\sigma{n n}$ acting on the principle plane has the magnitude $\lambda$ and it is named the principal stress.

The traction acting on the principal plane can be expressed by a simple vector argument and by using Eq. (2.17a) as follows:
\begin{aligned} & \vec{T}{n_p}=\lambda \vec{n}_p \ & \vec{T}{n_p}=[\sigma]^T\left{n_p\right} \end{aligned}
By combining these two relationships, we find,
\begin{aligned} \lambda\left(n_{p_x} \hat{i}+n_{p_y} \hat{j}+n_{p_z} \hat{k}\right)= & {\left[\left(\sigma_{x x} n_{p_x}+\tau_{y x} n_{p_y}+\tau_{z x} n_{p_z}\right) \hat{i}+\left(\tau_{x y} n_{p_x}+\sigma_{y y} n_{p_y}+\tau_{z y} n_{p_z}\right) \hat{j}\right.} \ & \left.+\left(\tau_{x z} n_{p_x}+\tau_{y z} n_{p_y}+\sigma_{z z} n_{p_z}\right) \hat{k}\right] \end{aligned}
The following system of equations can be obtained for the unknown vector components of the normal vector of the principle plane from this relationship,
\begin{aligned} & \left(\sigma_{x x} n_{p_x}+\tau_{y x} n_{p_y}+\tau_{z x} n_{p_z}\right)=\lambda n_{p_x} \ & \left(\tau_{x y} n_{p_x}+\sigma_{y y} n_{p_y}+\tau_{z y} n_{p_z}\right)=\lambda n_{p_y} \ & \left(\tau_{x z} n_{p_x}+\tau_{y z} n_{p_y}+\sigma_{z z} n_{p_z}\right)=\lambda n_{p_z} \end{aligned}
This equation can be represented in matrix form as follows:
$$\left[\begin{array}{ccc} \left(\sigma_{x x}-\lambda\right) & \tau_{y x} & \tau_{z x} \ \tau_{x y} & \left(\sigma_{y y}-\lambda\right) & \tau_{z y} \ \tau_{x z} & \tau_{y z} & \left(\sigma_{z z}-\lambda\right) \end{array}\right]\left{\begin{array}{l} n_{p_x} \ n_{p_y} \ n_{p_z} \end{array}\right}=0$$
or,
$$([\sigma]-\lambda[I])\left{n_p\right}=0$$
where $[I]$ is the identity matrix. This relationship represents the equilibrium of the principal stress $\lambda$ acting on the principal plane $\vec{n}_p$.

# 有限元方法代考

## 数学代写|有限元方法代写Finite Element Method代考|Stress transformations1

$$\vec{n}=n_x \hat{i}+n_y \hat{j}+n_z \hat{k}$$

$${n}=\left{\begin{array}{lll} n_x & n_y & n_z \end{array}\right}^T$$

$$n_x^2+n_y^2+n_z^2=1$$

$$n_i=\frac{\Delta A_i}{\Delta A_n} \text { for } i=x, y, z$$

$$\vec{T}n \Delta A_n=\vec{T}x \Delta A_x+\vec{T}_y \Delta A_y+\vec{T}_z \Delta A_z$$其中$\vec{T}_x, \vec{T}_y$和$\vec{T}_z$是作用在四面体其他面上的牵引力$\left(\Delta A_x, \Delta A_y\right.$和$\Delta A_z$)。通过使用Eq.(2.10)，这个表达式变成:$$\vec{T}_n=\vec{T}_x n_x+\vec{T}_y n_y+\vec{T}_z n_z$$在四面体的每个侧面，我们可以用方程来表示拉力的分量。(2.1)及(2.2)条款如下: \begin{aligned} & \vec{T}_x=\sigma{x x} \hat{i}+\tau{x y} \hat{j}+\tau_{x z} \hat{k} \ & \vec{T}y=\tau{y x} \hat{i}+\sigma_{y y} \hat{j}+\tau_{y z} \hat{k} \ & \vec{T}z=\tau{z x} \hat{i}+\tau_{z y} \hat{j}+\sigma_{z z} \hat{k} \end{aligned}

## 数学代写|有限元方法代写Finite Element Method代考|Principal stresses and directions

\begin{aligned} & \vec{T}{n_p}=\lambda \vec{n}p \ & \vec{T}{n_p}=[\sigma]^T\left{n_p\right} \end{aligned} 结合这两种关系，我们发现， \begin{aligned} \lambda\left(n{p_x} \hat{i}+n_{p_y} \hat{j}+n_{p_z} \hat{k}\right)= & {\left[\left(\sigma_{x x} n_{p_x}+\tau_{y x} n_{p_y}+\tau_{z x} n_{p_z}\right) \hat{i}+\left(\tau_{x y} n_{p_x}+\sigma_{y y} n_{p_y}+\tau_{z y} n_{p_z}\right) \hat{j}\right.} \ & \left.+\left(\tau_{x z} n_{p_x}+\tau_{y z} n_{p_y}+\sigma_{z z} n_{p_z}\right) \hat{k}\right] \end{aligned}

\begin{aligned} & \left(\sigma_{x x} n_{p_x}+\tau_{y x} n_{p_y}+\tau_{z x} n_{p_z}\right)=\lambda n_{p_x} \ & \left(\tau_{x y} n_{p_x}+\sigma_{y y} n_{p_y}+\tau_{z y} n_{p_z}\right)=\lambda n_{p_y} \ & \left(\tau_{x z} n_{p_x}+\tau_{y z} n_{p_y}+\sigma_{z z} n_{p_z}\right)=\lambda n_{p_z} \end{aligned}

$$\left[\begin{array}{ccc} \left(\sigma_{x x}-\lambda\right) & \tau_{y x} & \tau_{z x} \ \tau_{x y} & \left(\sigma_{y y}-\lambda\right) & \tau_{z y} \ \tau_{x z} & \tau_{y z} & \left(\sigma_{z z}-\lambda\right) \end{array}\right]\left{\begin{array}{l} n_{p_x} \ n_{p_y} \ n_{p_z} \end{array}\right}=0$$

$$([\sigma]-\lambda[I])\left{n_p\right}=0$$

## 有限元方法代写

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