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

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

Work done by a force (or moment) is defined to be the product of the force (or moment) and the displacement (or rotation) in the direction of the force (moment). Thus, if $\mathbf{F}$ is the force vector and $\mathbf{u}$ is the displacement vector, each having its own magnitude and direction, then the scalar product (or dot product) $\mathbf{F} \cdot \mathbf{u}$ gives the work done. The work done is a scalar with units of $\mathrm{N}-\mathrm{m}$ (Newton meters). When the force and displacement are functions of position $\mathbf{x}$ in a domain $\Omega$, then the integral of the product over the domain gives the work done:
$$W=\int_{\Omega} \mathbf{F}(\mathbf{x}) \cdot \mathbf{u}(\mathbf{x}) d \Omega$$
where $d \Omega$ denotes a volume element of the body. If $\mathbf{F}$ is also a function of $\mathbf{u}$, then
$$W=\int_{\Omega}\left(\int_0^{\mathbf{u}} \mathbf{F}(\mathbf{x}) \cdot d \mathbf{u}(\mathbf{x})\right) d \Omega$$
To further understand the difference between work done by a force that is independent of the displacement and one that is a function of the displacement, consider a spring-mass system in static equilibrium [see Fig. 2.3.1(a)]. Suppose that the mass $m$ is placed slowly (to eliminate dynamic effects) at the end of a linear elastic spring (i.e., the force in the spring is linearly proportional to the displacement in the spring). Under the action of the externally applied force of fixed magnitude $F_0$, the spring will elongate by an amount $e_0$, measured from its undeformed state. In this case, the force $F_0$ is due to gravity and it is equal to $F_0=m g$. Clearly, $F_0$ is independent of the extension $e_0$ in the spring, and $F_0$ does not change during the course of the extension $e$ going from 0 to its final value $e_0$ [see Fig. 2.3.1(b)]. The work done by $F_0$ in moving through $d e$ is $F_0 d e$. Then the total (external) work done by $F_0$ is
$$W_E=\int_0^{e_0} F_0 d e=F_0 e_0$$
Next consider the force in the spring $F_s$. The force in the spring is proportional to the displacement in the spring. Therefore, the value of $F_s$ goes from 0 when $e=0$ to its final value $F_s^f$ when $e=e_0$ [see Fig. 2.3.1(c)]. The work done by $F_s$ in moving through de is $F_s d e$. The total (internal) work done by $F_s$ is
$$W_I=\int_0^{e_0} F_s(e) d e$$

## 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

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

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

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

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