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# 数学代写|有限元方法代写finite differences method代考|Total Potential Energy

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## 数学代写|有限元方法作业代写finite differences method代考|Total Potential Energy

When $W_E$ involves only the work done by external forces that are independent of the deformation, we call it potential due to external loads and denote it by $V_E=W_E$ [see Eq. (2.3.3)]. The sum of the strain energy $U$ and the potential due to external forces $V_E$ is termed the total potential energy, and it is denoted by $\Pi=U+V_E\left(\sigma: \varepsilon=\sigma_{i j} \varepsilon_{i j}\right)$ :
$$\Pi(\mathbf{u})=U+V_E=\frac{1}{2} \int_{\Omega} \sigma(\varepsilon): \varepsilon d \Omega-\left[\int_{\Omega} \mathbf{f} \cdot \mathbf{u} d \Omega+\oint_{\Gamma} \mathbf{t} \cdot \mathbf{u} d \Gamma\right]$$
where $\Omega$ denotes the volume occupied by the body and $\Gamma$ is its total boundary.
For example, the total potential energy expression for the Bernoulli-Euler beams (which include bars, i.e., members subjected to uniaxial forces as a special case) is

\begin{aligned} \Pi(u, w)= & \frac{1}{2} \int_0^L\left[E A\left(\frac{d u}{d x}\right)^2+E I\left(\frac{d^2 w}{d x^2}\right)^2\right] d x \ & -\left{\int_0^L[f(x) u(x)+q(x) w(x)] d x\right} \end{aligned}
where $f(x)$ is the distributed axial force along the line $z=0$ and $q(x)$ is the transverse distributed load at $z=h / 2$, both measured per unit length. The expression in Eq. (2.3.13) must be appended with any terms corresponding to the work done by external point forces and moments.

## 数学代写|有限元方法作业代写finite differences method代考|Strain Energy and Strain Energy Density

For deformable elastic bodies under isothermal conditions and infinitesimal deformations, the internal energy per unit volume, denoted by $U_0$ and termed strain energy density, consists of only stored elastic strain energy (for rigid bodies, $U_0=0$ ). An expression for strain energy density can be derived as follows.
First, we consider axial deformation of a bar of area of cross section $A$. The free-body diagram of an element of length $d x_1$ of the bar is shown in Fig. 2.3.2(a). Note that the element is in static equilibrium, and we wish to determine the work done by the internal force associated with stress $\sigma_{11}^f$, where the superscript $f$ indicates that it is the final value of the quantity. Suppose that the element is deformed slowly so that axial strain varies from 0 to its final value $\varepsilon_{11}^f$. At any instant during the strain variation from $\varepsilon 11$ to $\varepsilon_{11}$ $+d \varepsilon_{11}$, we assume that $\sigma_{11}$ (due to $\varepsilon_{11}$ ) is kept constant so that equilibrium is maintained. Then the work done by the force $A \sigma_{11}$ in moving through the displacement $d \varepsilon_{11} d x_1$ is
$$A \sigma_{11} d \varepsilon_{11} d x_1=\sigma_{11} d \varepsilon_{11}\left(A d x_1\right) \equiv d U_0\left(A d x_1\right)$$
where $d U_0$ denotes the work done per unit volume in the element.
Referring to the stress-strain diagram in Fig. 2.3.2(b), $d U_0$ represents the elemental area under the stress-strain curve. The elemental area in the complement (in the rectangle formed by $\varepsilon_{11}$ and $\sigma_{11}$ ) is given by $d U_0^*=\varepsilon_{11} d \sigma_{11}$ and it is called the complementary strain energy density of the bar of length $d x_1$. In the present study, we will not be dealing with the complementary strain energy. The total area under the curve, $U_0$, is obtained
by integrating from zero to the final value of the strain (during which the stress changes according to its relation to the strain):
$$U_0=\int_0^{\varepsilon_{11}} \sigma_{11} d \varepsilon_{11}$$
where the superscript $f$ is omitted as the expression holds for any value of $\varepsilon$.
The internal work done (or strain energy stored) by $A \sigma_{11}$ over the whole element of length $d x$ during the entire deformation is
$$d U=\int_0^{\varepsilon_{11}} \sigma_{11} d \varepsilon_{11}\left(A d x_1\right)=U_0\left(A d x_1\right)$$
The total energy stored in the entire bar is obtained by integrating over the length of the bar:
$$U=\int_0^L A U_0 d x_1$$
This is the internal energy stored in the body due to deformation of the bar and it is called the strain energy. We note that no stress-strain relation is used, except that the stress-strain diagrams shown in Fig. 2.3.2(b) implies that it can be nonlinear. When the stress-strain relation is linear, say $\sigma_{11}=$ $E \varepsilon_{11}$, we have $U_0=(1 / 2) E \varepsilon_{11}^2=(1 / 2) \sigma_{11} \varepsilon_{11}$.

## 数学代写|有限元方法作业代写finite differences method代考|Work and Energy

$$W=\int_{\Omega} \mathbf{F}(\mathbf{x}) \cdot \mathbf{u}(\mathbf{x}) d \Omega$$

$$W=\int_{\Omega}\left(\int_0^{\mathbf{u}} \mathbf{F}(\mathbf{x}) \cdot d \mathbf{u}(\mathbf{x})\right) d \Omega$$

$$W_E=\int_0^{e_0} F_0 d e=F_0 e_0$$

$$W_I=\int_0^{e_0} F_s(e) d e$$