数学代写|有限元方法代写Finite Differences Method代考|AERO9301

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数学代写|有限元方法作业代写finite differences method代考|Balance law for the one-dimensional problem

The equation of motion for a one-dimensional problem depends on the geometry that is under consideration. If the geometry is a one-dimensional slab, extending to infinity in two directions (e.g., $y$ and $z$ ) there is no possibility of a surface force influence. In this case, equilibrium of an infinitesimally small section of the body will result in the following relationship,
$$\sigma_{x x} A-\left(\sigma_{x x}+\frac{\partial \sigma_{x x}}{\partial x} d x\right) A+F_{B x} A d x=\rho A \frac{\partial^2 u_x}{\partial t^2} d x$$
where $F_{B x}$ is body force vector, $\rho$ is the mass density per unit volume, $A$ is the area of the slab in the $y z$-plane and $\partial^2 u_x / \partial t^2$ is the local acceleration of the material points. The equation of motion of an infinitely wide one-dimensional medium is then found as follows:
$$\frac{\partial \sigma_{x x}}{\partial x}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2}$$
More often, the one-dimensional geometry is a slender member with crosssectional area $A$. In this case, we need to consider the possibility of surface traction acting along the peripheral surface (Fig. 2.9). Applying Newton’s second law on an infinitesimally small section of the body results in the following relationship,
$$\sigma_{x x} A-\left(\sigma_{x x} A+\frac{\partial \sigma_{x x} A}{\partial x} d x\right)+q_x d x+F_{B x} A d x=\rho A \frac{\partial^2 u_x}{\partial t^2} d x$$
where $q_x$ is the force per unit length acting on the outer periphery of the body along the $x$-direction. The equation of motion of slender, one-dimensional body is then found as follows:
$$\frac{\partial \sigma_{x x} A}{\partial x}+q_x+F_{B x} A=\rho A \frac{\partial^2 u_x}{\partial t^2}$$

数学代写|有限元方法作业代写finite differences method代考|Balance laws for two-dimensional problems

body. An infinitesimally small rectangular segment of this body of size ( $d x . d y)$ with unit depth, is shown in Fig. 2.10A. By considering the Newton’s second law and adding up the forces acting along the $x$-axis we find,
$$\sigma_{x x} d y-\left(\sigma_{x x}+\frac{\partial \sigma_{x x}}{\partial x} d x\right) d y-\tau_{y x} d x+\left(\tau_{x y}+\frac{\partial \tau_{x y}}{\partial y} d y\right) d x+F_{B x} d x d y=\rho d x d y \frac{\partial^2 u_x}{\partial t^2}$$
where $F_{B x}$ is the projection of the body force vector onto the $x$-axis. Note that the body force is defined as force per unit volume. Dividing this equation by $(d x . d y)$ the equation of motion in the $x$-direction is found as follows:
$$\frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \tau_{y z}}{\partial y}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2}$$
Similarly, the equation of motion in the $y$-direction can be stated as follows:
$$\frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y y}}{\partial y}+F_{B y}=\rho \frac{\partial^2 u_y}{\partial t^2}$$
where $F_{B y}$ is the projection of the body force vector onto the $y$-axis.

Fig. 2.10B shows the stresses acting on the sides of an infinitesimally small segment $(d x d y d z)$ of a three-dimensional deformable body under. The details of the derivation for the three-dimensional equations of motion are similar to the two-dimensional derivation just presented. Therefore, we directly give the equations of motion for a three-dimensional solid as follows:
\begin{aligned} & \frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \tau_{y x}}{\partial y}+\frac{\partial \tau_{z x}}{\partial z}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2} \ & \frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y y}}{\partial y}+\frac{\partial \tau_{z y}}{\partial z}+F_{B y}=\rho \frac{\partial^2 u_y}{\partial t^2} \ & \frac{\partial \tau_{x z}}{\partial x}+\frac{\partial \tau_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+F_{B z}=\rho \frac{\partial^2 u_z}{\partial t^2} \end{aligned}

数学代写|有限元方法作业代写finite differences method代考|Balance law for the one-dimensional problem

$$\sigma_{x x} A-\left(\sigma_{x x}+\frac{\partial \sigma_{x x}}{\partial x} d x\right) A+F_{B x} A d x=\rho A \frac{\partial^2 u_x}{\partial t^2} d x$$

$$\frac{\partial \sigma_{x x}}{\partial x}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2}$$

$$\sigma_{x x} A-\left(\sigma_{x x} A+\frac{\partial \sigma_{x x} A}{\partial x} d x\right)+q_x d x+F_{B x} A d x=\rho A \frac{\partial^2 u_x}{\partial t^2} d x$$

$$\frac{\partial \sigma_{x x} A}{\partial x}+q_x+F_{B x} A=\rho A \frac{\partial^2 u_x}{\partial t^2}$$

数学代写|有限元方法作业代写finite differences method代考|Balance laws for two-dimensional problems

$$\sigma_{x x} d y-\left(\sigma_{x x}+\frac{\partial \sigma_{x x}}{\partial x} d x\right) d y-\tau_{y x} d x+\left(\tau_{x y}+\frac{\partial \tau_{x y}}{\partial y} d y\right) d x+F_{B x} d x d y=\rho d x d y \frac{\partial^2 u_x}{\partial t^2}$$

$$\frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \tau_{y z}}{\partial y}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2}$$

$$\frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y y}}{\partial y}+F_{B y}=\rho \frac{\partial^2 u_y}{\partial t^2}$$

\begin{aligned} & \frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \tau_{y x}}{\partial y}+\frac{\partial \tau_{z x}}{\partial z}+F_{B x}=\rho \frac{\partial^2 u_x}{\partial t^2} \ & \frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y y}}{\partial y}+\frac{\partial \tau_{z y}}{\partial z}+F_{B y}=\rho \frac{\partial^2 u_y}{\partial t^2} \ & \frac{\partial \tau_{x z}}{\partial x}+\frac{\partial \tau_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+F_{B z}=\rho \frac{\partial^2 u_z}{\partial t^2} \end{aligned}

MATLAB代写

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