### 数学代写|有限元方法代写Finite Element Method代考|GENG5514

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## 数学代写|有限元方法代写Finite Element Method代考|Generalized Hooke’s law for isotropic materials with symmetric stress and strain tensors

In case the material is elastically isotropic and the stress and strain tensors are symmetric the material behavior can be characterized with two material constants,
E: Elastic modulus or Young’s modulus
v: Poisson’s ratio
For a three-dimensional problem, it can be shown that the following relationships exist between the stresses and strains,
\begin{aligned} \varepsilon_{x x} & =\frac{1}{E}\left[\sigma_{x x}-v\left(\sigma_{y y}+\sigma_{z z}\right)\right] \ \varepsilon_{y y} & =\frac{1}{E}\left[\sigma_{y y}-v\left(\sigma_{z z}+\sigma_{x x}\right)\right] \ \varepsilon_{z z} & =\frac{1}{E}\left[\sigma_{z z}-v\left(\sigma_{x x}+\sigma_{y y}\right)\right] \ \tau_{x y} & =G \gamma_{x y} \ \tau_{y z} & =G \gamma_{y z} \ \tau_{z x} & =G \gamma_{z x} \end{aligned}
where shear modulus $G=E / 2(1+v)$.

Note that Eq. (2.61a) can be inverted and expressed as follows:
\begin{aligned} \sigma_{x x} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{x x} \ \sigma_{y y} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{y y} \ \sigma_{z z} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{z z} \ \tau_{x y} & =\mu \gamma_{x y} \ \tau_{y z} & =\mu \gamma_{y z} \ \tau_{z x} & =\mu \gamma_{z x} \end{aligned}
where, the Lamé constants are defined as follows:
\begin{aligned} & \lambda=\frac{v E}{(1+v)(1-2 v)} \ & \mu=G \end{aligned}

## 数学代写|有限元方法代写Finite Element Method代考|Effects of initial stress/strain and thermal strain

Thermal stress in a one-dimensional problem: Consider a long and slender bar of length $L$ and initial temperature $T^{(0)}$. If the temperature of the bar is changed by $\Delta T$, material points in the bar would experience thermal strain proportional to the temperature change,
$$\varepsilon^{(t h)}=\alpha \Delta T$$
the proportionality constant $\alpha$ is a material property known as the coefficient of thermal expansion with units of $\mathrm{K}^{-1}$ or $\left({ }^{\circ} \mathrm{C}\right)^{-1}$. If the bar is not constrained on its ends, its length will change by an amount,
$$\Delta L=\int_0^L \alpha \Delta T d x$$
but no internal stress will develop.

On the other hand if both ends of the bar are constrained, internal forces and hence stress will develop in the bar. If such constraint conditions exist, the thermal stress in the bar can be found from Hooke’s law as follows:
$$\sigma^{(t h)}=E \alpha \Delta T$$
Next, consider a constrained bar subjected to external forces and change of temperature. The total strain in this bar can be found by using the superposition of the mechanical component of the strain and the thermal strain,
$$\varepsilon=\frac{\sigma}{E}+\varepsilon^{(t h)}=\frac{\sigma}{E}+\alpha \Delta T$$
The inverse of this relation gives the corresponding total stress,
$$\sigma=E(\varepsilon-\alpha \Delta T)$$
Generalized stress-strain relations with thermal effects: For materials with isotropic material properties temperature change only causes normal strain in the material. The stress-strain relations for a three-dimensional isotropic material subjected to a temperature change $\Delta T$ are expressed as follows [8]:
\begin{aligned} \varepsilon_{x x}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{x x}-v\left(\sigma_{y y}+\sigma_{z z}\right)\right] \ \varepsilon_{y y}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{y y}-v\left(\sigma_{z z}+\sigma_{x x}\right)\right] \ \varepsilon_{z z}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{z z}-v\left(\sigma_{x x}+\sigma_{y y}\right)\right] \ \gamma_{x y} & =\frac{\tau_{x y}}{G} \ \gamma_{y z} & =\frac{\tau_{y z}}{G} \ \gamma_{z x} & =\frac{\tau_{z x}}{G} \end{aligned}

## 数学代写|有限元方法代写Finite Element Method代考|Generalized Hooke’s law for isotropic materials with symmetric stress and strain tensors

E:弹性模量或杨氏模量
v:泊松比

\begin{aligned} \varepsilon_{x x} & =\frac{1}{E}\left[\sigma_{x x}-v\left(\sigma_{y y}+\sigma_{z z}\right)\right] \ \varepsilon_{y y} & =\frac{1}{E}\left[\sigma_{y y}-v\left(\sigma_{z z}+\sigma_{x x}\right)\right] \ \varepsilon_{z z} & =\frac{1}{E}\left[\sigma_{z z}-v\left(\sigma_{x x}+\sigma_{y y}\right)\right] \ \tau_{x y} & =G \gamma_{x y} \ \tau_{y z} & =G \gamma_{y z} \ \tau_{z x} & =G \gamma_{z x} \end{aligned}

\begin{aligned} \sigma_{x x} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{x x} \ \sigma_{y y} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{y y} \ \sigma_{z z} & =\lambda\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right)+2 \mu \varepsilon_{z z} \ \tau_{x y} & =\mu \gamma_{x y} \ \tau_{y z} & =\mu \gamma_{y z} \ \tau_{z x} & =\mu \gamma_{z x} \end{aligned}

\begin{aligned} & \lambda=\frac{v E}{(1+v)(1-2 v)} \ & \mu=G \end{aligned}

## 数学代写|有限元方法代写Finite Element Method代考|Effects of initial stress/strain and thermal strain

$$\varepsilon^{(t h)}=\alpha \Delta T$$

$$\Delta L=\int_0^L \alpha \Delta T d x$$

$$\sigma^{(t h)}=E \alpha \Delta T$$

$$\varepsilon=\frac{\sigma}{E}+\varepsilon^{(t h)}=\frac{\sigma}{E}+\alpha \Delta T$$

$$\sigma=E(\varepsilon-\alpha \Delta T)$$

\begin{aligned} \varepsilon_{x x}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{x x}-v\left(\sigma_{y y}+\sigma_{z z}\right)\right] \ \varepsilon_{y y}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{y y}-v\left(\sigma_{z z}+\sigma_{x x}\right)\right] \ \varepsilon_{z z}-\alpha \Delta T & =\frac{1}{E}\left[\sigma_{z z}-v\left(\sigma_{x x}+\sigma_{y y}\right)\right] \ \gamma_{x y} & =\frac{\tau_{x y}}{G} \ \gamma_{y z} & =\frac{\tau_{y z}}{G} \ \gamma_{z x} & =\frac{\tau_{z x}}{G} \end{aligned}

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