物理代写|固体力学代写Solid Mechanics代考|BELTAMI–MICHELL STRESS COMPATIBILITY EQUATIONS

Under plane stress condition, $\sigma_z=\tau_{y z}=\tau_{\mathrm{xz}}=0$. Eq. (4.80) will be written as follows:
$$\varepsilon_{\mathrm{x}}=\frac{1}{E}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)$$
Similarly, Eq. (4.81) will be written as below:
$$\varepsilon_y=\frac{1}{E}\left(\sigma_y-\nu \sigma_x\right)$$
From Eq. (4.94), $\tau_{x y}$ can be expressed as follows:
$$\tau_{x y}=\frac{E}{2(1+v)} \gamma_{x y}$$
By expressing $\gamma_{x y}$ in term of $\tau_{x y}$ yields the following equation:
$$\gamma_{x y}=\frac{2(1+\nu)}{E} \tau_{x y}$$
Equation below is obtained by substituting the relationships in Eqs. (5.14), (5.15) and (5.16) into Eq. (3.20):
$$\frac{\partial^2}{\partial y^2}\left[\frac{1}{E}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)\right]+\frac{\partial^2}{\partial x^2}\left[\frac{1}{E}\left(\sigma_y-\nu \sigma_x\right)\right]=\frac{\partial^2}{\partial x \partial y}\left[\frac{2(1+\nu)}{E} \tau_{x y}\right]$$
Remove the common term $\frac{1}{E}$ from both sides gives us the follows:
$$\frac{\partial^2}{\partial y^2}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)+\frac{\partial^2}{\partial x^2}\left(\sigma_y-v \sigma_x\right)=\frac{\partial^2}{\partial x \partial y}\left[2(1+\nu) \tau_{x y}\right]$$

物理代写|固体力学代写Solid Mechanics代考|AIRY STRESS FUNCTION

When at rest, a body possesses potential energy, say $\psi$. Body force can thus be expressed in term of such potential energy:
$$f_x=-\frac{\partial \psi}{\partial x} f_y=-\frac{\partial \psi}{\partial y} f_z=-\frac{\partial \psi}{\partial z}$$
Also, the stress component can be expressed in term of stress function, $\phi$. For a 2-D scênario,
$$\sigma_x=\frac{\partial^2 \phi}{\partial y^2}+\psi \sigma_y=\frac{\partial^2 \phi}{\partial x^2}+\psi \tau_{x y}=-\frac{\partial^2 \phi}{\partial x \partial y}$$
Under plane stress condition, the equilibrium equation can be written as follows by substituting relationships in Eq. (5.28) into Eq. (5.18):
$$\frac{\partial \sigma_x}{\partial x}+\frac{\partial \tau_{y x}}{\partial y}-\frac{\partial \psi}{\partial x}=0$$
Simplify the equation above yields the following equation:
$$\frac{\partial}{\partial x}\left(\sigma_x-\psi\right)+\frac{\partial \tau_{y x}}{\partial y}=0$$
Similarly, the following equation can be derived from Eq. (5.19):
$$\frac{\partial}{\partial y}\left(\sigma_y-\psi\right)+\frac{\partial \tau_{x y}}{\partial x}=0$$
We get the equation below by substituting relationships in Eqs. (5.28) and (5.29) into Eq. (5.23):
$$\nabla^2\left(\frac{\partial^2 \phi}{\partial y^2}+\psi+\frac{\partial^2 \phi}{\partial x^2}+\psi\right)=-(1+\nu)\left[\frac{\partial}{\partial x}\left(-\frac{\partial \psi}{\partial x}\right)+\frac{\partial}{\partial y}\left(-\frac{\partial \psi}{\partial y}\right)\right]$$

物理代写|固体力学代写Solid Mechanics代考|beltami – michelle应力相容方程

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$$\varepsilon_{\mathrm{x}}=\frac{1}{E}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)$$类似地，Eq.(4.81)将被写成如下:
$$\varepsilon_y=\frac{1}{E}\left(\sigma_y-\nu \sigma_x\right)$$

$$\tau_{x y}=\frac{E}{2(1+v)} \gamma_{x y}$$

$$\gamma_{x y}=\frac{2(1+\nu)}{E} \tau_{x y}$$

$$\frac{\partial^2}{\partial y^2}\left[\frac{1}{E}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)\right]+\frac{\partial^2}{\partial x^2}\left[\frac{1}{E}\left(\sigma_y-\nu \sigma_x\right)\right]=\frac{\partial^2}{\partial x \partial y}\left[\frac{2(1+\nu)}{E} \tau_{x y}\right]$$

$$\frac{\partial^2}{\partial y^2}\left(\sigma_{\mathrm{x}}-\nu \sigma_{\mathrm{y}}\right)+\frac{\partial^2}{\partial x^2}\left(\sigma_y-v \sigma_x\right)=\frac{\partial^2}{\partial x \partial y}\left[2(1+\nu) \tau_{x y}\right]$$

物理代写|固体力学代写Solid Mechanics代考|艾里应力函数

$$f_x=-\frac{\partial \psi}{\partial x} f_y=-\frac{\partial \psi}{\partial y} f_z=-\frac{\partial \psi}{\partial z}$$此外，应力分量可以用应力函数表示， $\phi$。对于二维scênario，
$$\sigma_x=\frac{\partial^2 \phi}{\partial y^2}+\psi \sigma_y=\frac{\partial^2 \phi}{\partial x^2}+\psi \tau_{x y}=-\frac{\partial^2 \phi}{\partial x \partial y}$$在平面应力条件下，将式(5.28)中的关系式代入式(5.18):
，平衡方程为$$\frac{\partial \sigma_x}{\partial x}+\frac{\partial \tau_{y x}}{\partial y}-\frac{\partial \psi}{\partial x}=0$$

$$\frac{\partial}{\partial x}\left(\sigma_x-\psi\right)+\frac{\partial \tau_{y x}}{\partial y}=0$$同样地，从式(5.19)可以推导出以下等式:
$$\frac{\partial}{\partial y}\left(\sigma_y-\psi\right)+\frac{\partial \tau_{x y}}{\partial x}=0$$我们通过代入等式中的关系得到下面的等式。(5.28)和(5.29)放入式(5.23):
$$\nabla^2\left(\frac{\partial^2 \phi}{\partial y^2}+\psi+\frac{\partial^2 \phi}{\partial x^2}+\psi\right)=-(1+\nu)\left[\frac{\partial}{\partial x}\left(-\frac{\partial \psi}{\partial x}\right)+\frac{\partial}{\partial y}\left(-\frac{\partial \psi}{\partial y}\right)\right]$$

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