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# 数学代写|有限元方法代写Finite Differences Method代考|MEM234019

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## 数学代写|有限元方法作业代写finite differences method代考|Generalized Hooke’s law for two-dimensional problems

In a $2 \mathrm{D}$ problem, the elastic material property matrix given in Eq. (2.60) becomes,
$$[E]=\left[\begin{array}{lll} c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33} \end{array}\right]$$
where $c_{i j}(i, j=1,2,3)$ are material properties obtained experimentally. For isotropic materials, where the material properties are independent of the material direction, $c_{i j}$ are simplified.

If the thickness of the domain is small in comparison with the other two dimensions, and if the top and bottom surfaces are traction free (Fig. 2.8A), the stress components acting on the z-planes can be neglected,
$$\sigma_{z z}=0, \tau_{z x}=0, \tau_{z y}=0$$
for all $x, y, z$-locations. By using the generalized Hooke’s law the above equation implies that the transverse shear strain components are zero,
$$\gamma_{z x}=0, \gamma_{z y}=0$$
Moreover, by using Eq. (2.61a) with the condition $\sigma_{z z}=0$ we find,
$$\varepsilon_{z z}=-\frac{v}{E}\left(\sigma_{x x}+\sigma_{y y}\right)$$
Thus by using Eqs. (2.61a), it can be shown that the plane stress problem is defined by the following constitutive relationship,
\begin{aligned} \varepsilon_{x x} & =\frac{1}{E}\left(\sigma_{x x}-v \sigma_{y y}\right) \ \varepsilon_{y y} & =\frac{1}{E}\left(\sigma_{y y}-v \sigma_{x x}\right) \ \gamma_{x y} & =\frac{\tau_{x y}}{G} \end{aligned}

The elasticity matrix for the plane stress problem can be given as follows:
$$[E]=\frac{E}{1-v^2}\left[\begin{array}{ccc} 1 & v & 0 \ v & 1 & 0 \ 0 & 0 & \frac{1}{2}(1-v) \end{array}\right]$$

## 数学代写|有限元方法作业代写finite differences method代考|Plane strain problem

Consider a long cylindrical body whose axis is parallel to the $z$ axis (Fig. 2.8B). If the external loading does not vary along the $z$-axis, such a geometry renders itself to be analyzed by using the assumption that strain variation is confined to $(x, y)$ plane. The plane-strain conditions are described as follows:
i. $u_x=u_x(x, y)$ and $u_y=u_y(x, y)$
ii. $u_z=0$ (no extension in the $z$-direction).
Considering the strain-displacement relations in Eq. (2.47) in light of these assumptions one finds,
$$\varepsilon_{z z}=0, \gamma_{z x}=0, \gamma_{z y}=0 .$$
An equivalent statement can be made in terms of stresses by using the constitutive relations given in Eq. (2.62a), as follows:
$$\tau_{z y}=0, \tau_{z x}=0$$
Moreover, by using Eqs. (2.81) $\left(\varepsilon_{z z}=0\right)$ and (2.61) we get,
$$\sigma_{z z}=v\left(\sigma_{x x}+\sigma_{y y}\right)$$
Using this relationship, it can then be shown that the constitutive relations for a plane-strain problem are,
\begin{aligned} \varepsilon_{x x} & =\frac{1-v^2}{E} \sigma_{x x}-\frac{v(1+v)}{E} \sigma_{y y} \ \varepsilon_{y y} & =\frac{1-v^2}{E} \sigma_{y y}-\frac{v(1+v)}{E} \sigma_{x x} \ \gamma_{x y} & =\frac{\tau_{x y}}{G} \end{aligned}
The elasticity matrix can then be expressed by using the Lamé constants as follows:
$$[E]=\left[\begin{array}{ccc} (\lambda+2 G) & \lambda & 0 \ \lambda & (\lambda+2 G) & 0 \ 0 & 0 & G \end{array}\right]$$

## 数学代写|有限元方法作业代写finite differences method代考|Generalized Hooke’s law for two-dimensional problems

$$[E]=\left[\begin{array}{lll} c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33} \end{array}\right]$$

$$\sigma_{z z}=0, \tau_{z x}=0, \tau_{z y}=0$$

$$\gamma_{z x}=0, \gamma_{z y}=0$$

$$\varepsilon_{z z}=-\frac{v}{E}\left(\sigma_{x x}+\sigma_{y y}\right)$$

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

$$[E]=\frac{E}{1-v^2}\left[\begin{array}{ccc} 1 & v & 0 \ v & 1 & 0 \ 0 & 0 & \frac{1}{2}(1-v) \end{array}\right]$$

## 数学代写|有限元方法作业代写finite differences method代考|Plane strain problem

2$u_z=0$(没有$z$ -方向的扩展)。

$$\varepsilon_{z z}=0, \gamma_{z x}=0, \gamma_{z y}=0 .$$

$$\tau_{z y}=0, \tau_{z x}=0$$

$$\sigma_{z z}=v\left(\sigma_{x x}+\sigma_{y y}\right)$$

\begin{aligned} \varepsilon_{x x} & =\frac{1-v^2}{E} \sigma_{x x}-\frac{v(1+v)}{E} \sigma_{y y} \ \varepsilon_{y y} & =\frac{1-v^2}{E} \sigma_{y y}-\frac{v(1+v)}{E} \sigma_{x x} \ \gamma_{x y} & =\frac{\tau_{x y}}{G} \end{aligned}

$$[E]=\left[\begin{array}{ccc} (\lambda+2 G) & \lambda & 0 \ \lambda & (\lambda+2 G) & 0 \ 0 & 0 & G \end{array}\right]$$

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