# 物理代写|固体力学代写Solid Mechanics代考|ME885

## 物理代写|固体力学代写Solid Mechanics代考|Tresca Yields CRITERION

For Tresca yield criterion, yielding is initiated when the maximum shear stress, i.e. the largest of the three developed maximum shear stresses, exceeds the threshold value of material.

Let define maximum shear stress as it is in Eq. (2.44), where $\sigma_1$ and $\sigma_3$ are the maximum and minimum principal stresses respectively. Also, let $k$ be the threshold value that defines whether the developed shear stress is enough to make the body initiate yielding. Thus, at yielding point:
$$\frac{\left(\sigma_1-\sigma_3\right)}{2}=k$$
Consider the case where only tensile stress instead of shear stress developed in the body. Express Eq. (6.1) in the following form:
$$\sigma_1-\sigma_3=2 k$$
The yield stress of material, $\sigma_y$ is determined through uniaxial load test. Therefore, $\sigma_y$ is the threshold value that defines whether the developed tensile stress is enough to make the body initiate yielding. For this case, $\sigma_1$ should equal to the yield strength of material, $\sigma_y$, while principal stress along other directions, $\sigma_2$ and $\sigma_3$ are zero at yielding point. By applying the condition to equation above yields the follows:
$$\sigma_y=2 k$$
Express $k$ in term of $\sigma_y$ produces the expression below:
$$k=\frac{\sigma_y}{2}$$

## 物理代写|固体力学代写Solid Mechanics代考|VON MISES YIELDS CRITERION

Von Mises yield criterion is devised based on distortion energy theory. According to this criterion, yielding is initiated when the second deviatoric invariant exceeds the threshold value of material.

The equation below is obtained by substituting the expressions of $I_I$ and $I_2$ (Eqs. (2.35) and (2.36)) into $J_2$ (Eq. (2.47)):
$$J_2=\frac{\left(\sigma_x+\sigma_y+\sigma_z\right)^2}{3}-\left(\sigma_x \sigma_y+\sigma_y \sigma_z+\sigma_x \sigma_z-\tau_{x y}{ }^2-\tau_{y z}^2-\tau_{x z}{ }^2\right)$$
After expanding the equation above it becomes:
\begin{aligned} J_2=& \frac{\sigma_x{ }^2+\sigma_x \sigma_y+\sigma_x \sigma_z+\sigma_y \sigma_x+\sigma_y{ }^2+\sigma_y \sigma_z+\sigma_z \sigma_x+\sigma_z \sigma_y+\sigma_z^2}{3} \ &-\sigma_x \sigma_y-\sigma_y \sigma_z-\sigma_x \sigma_z+\tau_{x y}{ }^2+\tau_{y z}{ }^2+\tau_{x z}{ }^2 \end{aligned}
Simplify the equation above yields the follows:
$$J_2=\frac{\sigma_x^2+\sigma_y^2+\sigma_z^2}{3}-\frac{1}{3} \sigma_x \sigma_y-\frac{1}{3} \sigma_y \sigma_z-\frac{1}{3} \sigma_x \sigma_z+\tau_{x y}{ }^2+\tau_{y z}{ }^2+\tau_{x z}^2$$
Express the equation above in terms of principal stresses, where $\sigma_x=\sigma_1, \sigma_y=\sigma_2$, $\sigma_z=\sigma_3$ and $\tau_{x y}=\tau_{y z}=\tau_{x z}=0$ and we get:
$$J_2=\frac{\sigma_1^2+\sigma_2^2+\sigma_3^2}{3}-\frac{1}{3} \sigma_1 \sigma_2-\frac{1}{3} \sigma_2 \sigma_3-\frac{1}{3} \sigma_1 \sigma_3$$
Factorization of the equation above with $\frac{1}{6}$ gives us:
$$J_2=\frac{1}{6}\left(2 \sigma_1^2+2 \sigma_2^2+2 \sigma_3^2-2 \sigma_1 \sigma_2-2 \sigma_2 \sigma_3-2 \sigma_1 \sigma_3\right)$$

## 物理代写|固体力学代写Solid Mechanics代考|Tresca屈服准则

$$\frac{\left(\sigma_1-\sigma_3\right)}{2}=k$$

$$\sigma_1-\sigma_3=2 k$$

$$\sigma_y=2 k$$

$$k=\frac{\sigma_y}{2}$$

## 物理代写|固体力学代写Solid Mechanics代考|冯·米塞斯屈服准则

Von Mises屈服准则是基于变形能理论设计的。根据该准则，当第二个偏差不变量超过材料的阈值时开始屈服

$$J_2=\frac{\left(\sigma_x+\sigma_y+\sigma_z\right)^2}{3}-\left(\sigma_x \sigma_y+\sigma_y \sigma_z+\sigma_x \sigma_z-\tau_{x y}{ }^2-\tau_{y z}^2-\tau_{x z}{ }^2\right)$$

\begin{aligned} J_2=& \frac{\sigma_x{ }^2+\sigma_x \sigma_y+\sigma_x \sigma_z+\sigma_y \sigma_x+\sigma_y{ }^2+\sigma_y \sigma_z+\sigma_z \sigma_x+\sigma_z \sigma_y+\sigma_z^2}{3} \ &-\sigma_x \sigma_y-\sigma_y \sigma_z-\sigma_x \sigma_z+\tau_{x y}{ }^2+\tau_{y z}{ }^2+\tau_{x z}{ }^2 \end{aligned}
.
.
$$J_2=\frac{\sigma_x^2+\sigma_y^2+\sigma_z^2}{3}-\frac{1}{3} \sigma_x \sigma_y-\frac{1}{3} \sigma_y \sigma_z-\frac{1}{3} \sigma_x \sigma_z+\tau_{x y}{ }^2+\tau_{y z}{ }^2+\tau_{x z}^2$$将上式用主应力表示，其中 $\sigma_x=\sigma_1, \sigma_y=\sigma_2$， $\sigma_z=\sigma_3$ 和 $\tau_{x y}=\tau_{y z}=\tau_{x z}=0$ 我们得到:
$$J_2=\frac{\sigma_1^2+\sigma_2^2+\sigma_3^2}{3}-\frac{1}{3} \sigma_1 \sigma_2-\frac{1}{3} \sigma_2 \sigma_3-\frac{1}{3} \sigma_1 \sigma_3$$

$$J_2=\frac{1}{6}\left(2 \sigma_1^2+2 \sigma_2^2+2 \sigma_3^2-2 \sigma_1 \sigma_2-2 \sigma_2 \sigma_3-2 \sigma_1 \sigma_3\right)$$

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