数学代写|CE595 Finite Element Method

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这是普渡大學有限元方法的代写成功案例。

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CE595 Finite Element Method HELP（EXAM HELP， ONLINE TUTOR）

Solution: The strain energy for the Bernoulli-Euler beam is given by Eq. (2.3.10). First we compute $d u / d x$ and $d_2 w / d x_2$ in terms of the generalized displacements $\left(u_1, u_2, w_1, w_2, \theta_1, \theta_2\right)$ :
$$
\begin{aligned}
\frac{d u}{d x} & =\frac{u_2-u_1}{L}, \
\frac{d^2 w}{d x^2} & =-\frac{6}{L^2}\left(1-2 \frac{x}{L}\right) w_1-\frac{2}{L}\left(3 \frac{x}{L}-2\right) \theta_1+\frac{6}{L^2}\left(1-2 \frac{x}{L}\right) w_2-\frac{2}{L}\left(3 \frac{x}{L}-1\right) \theta_2
\end{aligned}
$$
Then the strain energy due to stretching is
$$
U_s=\frac{E A}{2} \int_0^L\left(\frac{d u}{d x}\right)^2 d x=\frac{E A}{2 L}\left(u_2-u_1\right)^2=\frac{E A}{2 L}\left{\begin{array}{l}
u_1 \
u_2
\end{array}\right}^{\mathrm{T}}\left[\begin{array}{rr}
1 & -1 \
-1 & 1
\end{array}\right]\left{\begin{array}{l}
u_1 \
u_2
\end{array}\right}
$$
Next, we evaluate the integrals $(\xi=x / L$ and $d x=L d \xi)$ :
$$
\begin{aligned}
& \int_0^1(1-2 \xi)^2 L d \xi=\frac{L}{3}, \quad \int_0^1(1-2 \xi)(3 \xi-2) L d \xi=-\frac{L}{2} \
& \int_0^1(3 \xi-2)^2 L d \xi=L, \quad \int_0^1(1-2 \xi)(3 \xi-1) L d \xi=-\frac{L}{2} \
& \int_0^1(3 \xi-1)^2 L d \xi=L, \quad \int_0^1(3 \xi-2)(3 \xi-1) L d \xi=\frac{L}{2}
\end{aligned}
$$
Then the strain energy due to bending is
$$
\begin{aligned}
U_b= & \frac{E I}{2} \int_0^L\left(\frac{d^2 w}{d x^2}\right)^2 d x \
= & \frac{E I}{L^3}\left(6 w_1^2-6 L w_1 \theta_1-12 w_1 w_2-6 L w_1 \theta_2+2 L^2 \theta_1^2+6 L \theta_1 w_2\right. \
& \left.+2 L^2 \theta_1 \theta_2+6 w_2^2+6 L w_2 \theta_2+2 L^2 \theta_2^2\right) \
= & \frac{E I}{L^3}\left{\begin{array}{c}
w_1 \
\theta_1 \
w_2 \
\theta_2
\end{array}\right}\left[\begin{array}{rrrr}
-3 L & -6 & -3 L \
-3 L & 2 L^2 & 3 L & L^2 \
-6 & 3 L & 6 & 3 L \
-3 L & L^2 & 3 L & 2 L^2
\end{array}\right]\left{\begin{array}{c}
w_1 \
\theta_1 \
w_2 \
\theta_2
\end{array}\right}
\end{aligned}
$$
Thus, the strain energy due to stretching and bending of the beam $\left(U=U_s+\right.$ $U_b$ ) is
$$
\begin{aligned}
& U=\left{\begin{array}{l}
u_1 \
u_2
\end{array}\right}^{\mathrm{T}} \frac{E A}{2 L}\left[\begin{array}{rr}
1 & -1 \
-1 & 1
\end{array}\right]\left{\begin{array}{l}
u_1 \
u_2
\end{array}\right} \
&+\left{\begin{array}{c}
w_1 \
\theta_1 \
w_2 \
\theta_2
\end{array}\right} \frac{E I}{L^3}\left[\begin{array}{rrrr}
6 & -3 L & -6 & -3 L \
-3 L & 2 L^2 & 3 L & L^2 \
-6 & 3 L & 6 & 3 L \
-3 L & L^2 & 3 L & 2 L^2
\end{array}\right]\left{\begin{array}{c}
w_1 \
\theta_1 \
w_2 \
\theta_2
\end{array}\right}
\end{aligned}
$$

Solution: The displacement field of a beam under the Bernoulli-Euler kinematic hypothesis is given by [see Eq. (2.3.7)]
$$
u_1(x, y, z)=u(x)+z \theta_x, u_2=0, u_3(x, y, z)=w(x) ; \quad \theta_x \equiv-\frac{d w}{d x}
$$
If we assume that the strains are small, the only nonzero strain is given by
$$
\varepsilon_{x x}=\frac{d u}{d x}+z \frac{d \theta_x}{d x}=\frac{d u}{d x}-z \frac{d^2 w}{d x^2}
$$
Let the virtual displacements be $\delta u$ and $\delta w$, which are completely arbitrary because there are no specified geometric boundary conditions for the problem at hand. Then the virtual strain $\delta \varepsilon_{x x}$ is given by
$$
\delta \varepsilon_{x x}=\delta\left(\frac{d u}{d x}+z \frac{d \theta_x}{d x}\right)=\frac{d \delta u}{d x}+z \frac{d \delta \theta_x}{d x}
$$
Then the internal and external virtual works due to the virtual displacements $\delta u$ and $\delta w$ are given by
$$
\begin{aligned}
\delta W_E= & -\left{\int_0^L\left[f(x) \delta u+q(x) \delta w\left(x, h_t\right)\right] d x+\int_0^L\left(-F_s\right) \delta w\left(x, h_b\right) d x\right. \
& +P \delta u(L)+F \delta w(L)} \
= & -\left{\int_0^L[f(x) \delta u+q(x) \delta w(x)] d x-\int_0^L k w(x) \delta w(x) d x\right. \
& +P \delta u(L)+F \delta w(L)}
\end{aligned}
$$
$$
\begin{aligned}
\delta W_I & =\int_0^L \int_A \sigma_{x x} \delta \varepsilon_{x x} d x d A \
& =\int_0^L \int_A \sigma_{x x}\left(\frac{d \delta u}{d x}-z \frac{d^2 \delta w}{d x^2}\right) d x d A
\end{aligned}
$$
where $L$ is the length, $h_t$ is the distance from the $x$-axis to the top of the beam, $h_b$ is the distance from the $x$-axis to the bottom of the beam, and $A$ is the cross-sectional area of the beam. The foundation reaction force $F_s$ (acting downward) is replaced with $F_s=k w(x)$ using the linear elastic constitutive equation for the foundation.
The principle of virtual displacements requires that $\delta W=\delta W_I+\delta W_E=0$, which gives
$$
\begin{aligned}
0= & \int_0^L \int_A \sigma_{x x}\left(\frac{d \delta u}{d x}-z \frac{d^2 \delta w}{d x^2}\right) d A d x \
& -\int_0^L[f \delta u+(q-k w) \delta w] d x-P \delta u(L)-F \delta w(L) \
= & \int_0^L\left(N \frac{d \delta u}{d x}-M \frac{d^2 \delta w}{d x^2}\right) d x-\int_0^L[f \delta u+(q-k w) \delta w] d x \
& -P \delta u(L)-F \delta w(L) \
= & \int_0^L\left(-\frac{d N}{d x} \delta u-\frac{d^2 M}{d x^2} \delta w\right) d x-\int_0^L[f \delta u+(q-k w) \delta w] d x \
& -P \delta u(L)-F \delta w(L)+\left[N \delta u+\frac{d M}{d x} \delta w-M \frac{d \delta w}{d x}\right]0^L \end{aligned} $$ where $N$ and $M$ are the stress resultants defined by $$ N=\int_A \sigma{x x} d A, M=\int_A \sigma_{x x} z d A
$$
The Euler equations are obtained by setting the coefficients of $\delta u$ and $\delta w$ under the integral separately to zero:
$$
\delta u: \quad-\frac{d N}{d x}=f
$$
$$
\delta w: \quad-\frac{d^2 M}{d x^2}+k w=q
$$
in $0<x<L$.
We note that $\delta u, \delta w$, and $d \delta w / d x=\delta(d w / d x)$ appear in the boundary terms. Hence, $u, w$, and $(d w / d x)$ are the primary variables of the theory, and their specification constitutes the displacement, geometric, or essential boundary conditions. Since nothing is said about the primary variables being specified, $\delta u, \delta w$, and $d \delta w / d x$ are arbitrary at
$$
\begin{aligned}
& N=0, \quad V \equiv \frac{d M}{d x}=0, \quad M=0 \text { at } x=0 \
& N=P, \quad \frac{d M}{d x}=F, \quad M=0 \text { at } x=L
\end{aligned}
$$

Solution: The strain energy of a straight beam in pure bending according to the Euler-Bernoulli beam theory is given in terms of the generalized end displacements $w_1, \theta_1, w_2$, and $\theta_2$ in Eq. (6) of Example 2.3.3 as
$$
\begin{aligned}
U= & \frac{E I}{L^3}\left(6 w_1^2-6 L w_1 \theta_1-12 w_1 w_2-6 L w_1 \theta_2+2 L^2 \theta_1^2+6 L \theta_1 w_2\right. \
& \left.+2 L^2 \theta_1 \theta_2+6 w_2^2+6 L w_2 \theta_2+2 L^2 \theta_2^2\right)
\end{aligned}
$$
Then by Castigliano’s theorem I, we have
$$
\begin{aligned}
& F_1=\frac{\partial U}{\partial w_1}=\frac{E I}{L^3}\left(12 w_1-6 L \theta_1-12 w_2-6 L \theta_2\right) \
& M_1=\frac{\partial U}{\partial \theta_1}=\frac{E I}{L^3}\left(-6 L w_1+4 L^2 \theta_1+6 L w_2+2 L^2 \theta_2\right) \
& F_2=\frac{\partial U}{\partial w_2}=\frac{E I}{L^3}\left(-12 w_1+6 L \theta_1+12 w_2+6 L \theta_2\right) \
& M_2=\frac{\partial U}{\partial \theta_2}=\frac{E I}{L^3}\left(-6 L w_1+2 L^2 \theta_1+6 L w_2+4 L^2 \theta_2\right)
\end{aligned}
$$

which can be expressed in matrix form as
$$
\left{\begin{array}{c}
F_1 \
M_1 \
F_2 \
M_2
\end{array}\right}=\frac{2 E I}{L^3}\left[\begin{array}{rrrr}
6 & -3 L & -6 & -3 L \
-3 L & 2 L^2 & 3 L & L^2 \
-6 & 3 L & 6 & 3 L \
-3 L & L^2 & 3 L & 2 L^2
\end{array}\right]\left{\begin{array}{c}
w_1 \
\theta_1 \
w_2 \
\theta_2
\end{array}\right}
$$

MY-ASSIGNMENTEXPERT™可以为您提供ENGINEERING MATH4604 FINITE ELEMENT METHOD有限元方法的代写代考和辅导服务！