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# 数学代写|有限元代写Finite Element Method代考|CE507 EQUATIONS FOR TRUSS MEMBERS

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## 数学代写|有限元代写Finite Element Method代考|EQUATIONS FOR TRUSS MEMBERS

A typical truss structure is shown in Figure 2.7. Each truss member in a truss structure is a solid whose dimension in one direction is much larger than in the other two directions as shown in Figure 2.8. The force is applied only in the $x$ direction. Therefore a truss member is actually a one-dimensional (1D) solid. The equations for 1D solids can be obtained by further omitting the stress related to the $y$ direction, $\sigma_{y y}, \sigma_{x y}$, from the $2 \mathrm{D}$ case.

Stress and Strain
Omitting the stress terms in the $y$ direction, the stress in a truss member is only $\sigma_{x x}$, which is often simplified as $\sigma_x$. The corresponding strain in a truss member is $\varepsilon_{x x}$, which is simplified as $\varepsilon_x$. The strain-displacement relationship is simply given by
$$\varepsilon_x=\frac{\partial u}{\partial x}$$

## 数学代写|有限元代写Finite Element Method代考|Constitutive Equations

Hooke’s law for $1 \mathrm{D}$ solids has the following simple form, with the exclusion of the $y$ dimension and hence the Poisson effect:
$$\sigma=E \varepsilon$$
This is actually the original Hooke’s law in one dimension. The Young’s module $E$ can be obtained using a simple tensile test.
Dynamic Equilibrium Equations
By eliminating the $y$ dimension term from Eq. (2.33), for example, the dynamic equilibrium equation for $1 \mathrm{D}$ solids is
$$\frac{\partial \sigma_x}{\partial x}+f_x=\rho \ddot{u}$$
Substituting Eqs. (2.38) and (2.39) into Eq. (2.40), we obtain the governing equation for elastic and homogenous ( $E$ is independent of $x)$ trusses as follows:
$$E \frac{\partial^2 u}{\partial x^2}+f_x=\rho \ddot{u}$$
The static equilibrium equation for trusses is obtained by eliminating the inertia term in Eq. (2.40):
$$\frac{\partial \sigma_x}{\partial x}+f_x=0$$
The static equilibrium equation in terms of displacement for elastic and homogenous trusses is obtained by eliminating the inertia term in Eq. (2.41):
$$E \frac{\partial^2 u}{\partial x^2}+f_x=0$$
For bars of constant cross-sectional area $A$, the above equation can be written as
$$E A \frac{\partial^2 u}{\partial x^2}+F_x=0$$
where $F_X=f_x A$ is the external force applied in the axial direction of the bar.

## 数学代写|有限元代写有限元法代考|桁架构件的方程

$$\varepsilon_x=\frac{\partial u}{\partial x}$$ 给出

## 数学代写|有限元代写有限元法代考|本构方程

$$\sigma=E \varepsilon$$

$$\frac{\partial \sigma_x}{\partial x}+f_x=\rho \ddot{u}$$

$$E \frac{\partial^2 u}{\partial x^2}+f_x=\rho \ddot{u}$$

$$\frac{\partial \sigma_x}{\partial x}+f_x=0$$

$$E \frac{\partial^2 u}{\partial x^2}+f_x=0$$

$$E A \frac{\partial^2 u}{\partial x^2}+F_x=0$$
，其中$F_X=f_x A$为施加在杆轴向的外力

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