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# 物理代写|固体力学代写Solid Mechanics代考|PHYS440 Elastic moduli

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## 物理代写|固体力学代写Solid Mechanics代考|Elastic moduli

We have so far introduced four different elastic parameters, namely the Young modulus, the Poisson ratio and the two Lamé coefficients. We previously introduced also the bulk modulus $B$ when we investigated thermal expansion in section 4.2.1. Since $B$ was defined as the inverse of the isothermal compressibility (see appendix $C$ ), that is it deals with volume variations, it is expected to be related to some elastic parameter. In order to elucidate this issue, let us consider a hydrostatic stress $T_{i j}=P \delta_{i j}$ (where $P$ is the macroscopic hydrostatic pressure) and insert it into the constitutive equation (5.39) so as to get
$$\mathbb{S}=\frac{1}{3} \frac{1}{\lambda+\frac{2}{3} \mu} P \mathbb{.} .$$
The connection with equation (C.11) is established by defining
$$B=\lambda+\frac{2}{3} \mu,$$

so that
$$\mathbb{S}=\frac{1}{3 B} P \rrbracket \quad \rightarrow \quad \operatorname{Tr}(\mathbb{S})=\sum_i \epsilon_{i i}=\frac{\Delta V}{V}=\frac{P}{B},$$
which leads to the following definition
$$\frac{1}{B}=\frac{1}{V} \frac{\Delta V}{P},$$
representing the finite difference counterpart of equation (C.11). This result reconciles the thermodynamical and elastic treatment of deformations affecting the system volume and it allows us to recast the stress-strain relation of a homogeneous and isotropic linear elastic medium in the form
\begin{aligned} \mathbb{U} &\left.=2 \mu \mathbb{S}+\left(B-\frac{2}{3} \mu\right) \operatorname{Tr}(\mathbb{S})\right] \ &\left.=3 B\left[\frac{1}{3} \operatorname{Tr}(\mathbb{S})\right]\right]+2 \mu\left[\mathbb{S}-\frac{1}{3} \operatorname{Tr}(\mathbb{S}) \mathbb{]},\right. \end{aligned}
where the first and second term on the right-hand side are, respectively, named spherical part and deviatoric part of the stress tensor: they describe the hydrostatic volume variation and the change in shape of the solid body subject to $\mathbb{I}$.

## 物理代写|固体力学代写Solid Mechanics代考|Thermoelasticity

We have so far implicitly assumed that the deformations are imposed to the system at zero temperature. While this assumption was useful to define a clean purely-elastic problem, we must duly generalise our theory to include stress actions applied at $T>0 \mathrm{~K}$ as well [6].

The starting point is of course the energy balance stated by the first law of thermodynamics (see equation (C.5)) which for a system with volume $V$ in equilibrium at temperature $T$ under some elastic action is written as
$$d \mathcal{U}=V \sum_{i j} T_{i j} d \epsilon_{i j}+T d S,$$
where the mechanical work contributing to the internal energy $\mathcal{U}$ has been written in terms of the stress tensor since we know that this latter describes any possible kind of volume and shape variation of the system. It is easy to reconcile equation (5.47) with the standard thermodynamical formulation by simply considering the case of a hydrostatic stress $T_{i j}=-P \delta_{i j}$, where $P$ is the applied pressure whose negative sign indicates that the mechanical action is compressive. We assume it to operate quasistatically, like anywhere else in the remaining of this chapter. By inserting the hydrostatic stress into equation (5.47) we get
\begin{aligned} d \mathcal{U} &=V \sum_{i j}\left(-P \delta_{i j}\right) d \epsilon_{i j}+T d S \ &=-P V \sum_i d \epsilon_{i i}+T d S \ &=-P V \frac{d V}{V}+T d S \ &=-P d V+T d S \end{aligned}

consistently with the first law of thermodynamics. Equation (5.47) is valid for any arbitrary deformation and, therefore, it allows for a thermodynamical definition of the stress tensor
$$T_{i j}=\left.\frac{1}{V} \frac{\partial \mathcal{U}}{\partial \epsilon_{i j}}\right|S=\left.\frac{1}{V} \frac{\partial \mathcal{F}}{\partial \epsilon{i j}}\right|T,$$ where $\mathcal{F}=\mathcal{U}-T S$ is the Helmholtz free energy, corresponding to the work exchanged quasi-statically during the constant-temperature deformation (see appendix C). This result defines the next task to accomplish, namely: deriving the explicit dependence $\mathcal{F}=\mathcal{F}\left(\epsilon{i j}\right)$.

## 物理代写|固体力学代写Solid Mechanics代考|弹性模量

$$\mathbb{S}=\frac{1}{3} \frac{1}{\lambda+\frac{2}{3} \mu} P \mathbb{.} .$$
。通过定义
$$B=\lambda+\frac{2}{3} \mu,$$ ，建立了与方程(C.11)的联系

，使
$$\mathbb{S}=\frac{1}{3 B} P \rrbracket \quad \rightarrow \quad \operatorname{Tr}(\mathbb{S})=\sum_i \epsilon_{i i}=\frac{\Delta V}{V}=\frac{P}{B},$$
，从而得到以下定义
$$\frac{1}{B}=\frac{1}{V} \frac{\Delta V}{P},$$

\begin{aligned} \mathbb{U} &\left.=2 \mu \mathbb{S}+\left(B-\frac{2}{3} \mu\right) \operatorname{Tr}(\mathbb{S})\right] \ &\left.=3 B\left[\frac{1}{3} \operatorname{Tr}(\mathbb{S})\right]\right]+2 \mu\left[\mathbb{S}-\frac{1}{3} \operatorname{Tr}(\mathbb{S}) \mathbb{]},\right. \end{aligned}
，其中右侧的第一项和第二项分别称为应力张量的球形部分和偏离部分:它们描述了静压体积变化和固体形状的变化$\mathbb{I}$。

## 物理代写|固体力学代写Solid Mechanics代考|热弹性

.

$$d \mathcal{U}=V \sum_{i j} T_{i j} d \epsilon_{i j}+T d S,$$
，其中贡献于内能$\mathcal{U}$的机械功被写成应力张量的形式，因为我们知道后者描述了任何可能的体积和形状变化系统。通过简单地考虑静水应力$T_{i j}=-P \delta_{i j}$的情况，很容易使式(5.47)与标准热力学公式相一致，其中$P$是施加的压力，其负号表示机械作用为压缩。我们假设它是准静态的，就像本章其余部分的其他内容一样。将静水应力代入(5.47)式，得到
\begin{aligned} d \mathcal{U} &=V \sum_{i j}\left(-P \delta_{i j}\right) d \epsilon_{i j}+T d S \ &=-P V \sum_i d \epsilon_{i i}+T d S \ &=-P V \frac{d V}{V}+T d S \ &=-P d V+T d S \end{aligned}

$$T_{i j}=\left.\frac{1}{V} \frac{\partial \mathcal{U}}{\partial \epsilon_{i j}}\right|S=\left.\frac{1}{V} \frac{\partial \mathcal{F}}{\partial \epsilon{i j}}\right|T,$$的热力学定义，其中$\mathcal{F}=\mathcal{U}-T S$是亥姆霍尔兹自由能，对应于在恒温变形过程中准静态交换的功(见附录C)。这个结果定义了下一个要完成的任务，即:导出显式依赖$\mathcal{F}=\mathcal{F}\left(\epsilon{i j}\right)$ .

.

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