Basic Course on finite element analysis (Zeng) Note Two-beam element finite element equation derivation

I have to say, Mathematica is a really good thing, before learning finite element, for the equation deduction in the book, see the past, never thought to deduce it yourself again, because the manual derivation is too complex. With the MM, the original very complex things suddenly become simple.

1. Unit Geometry Description

is pure curved beam unit, length L, modulus e, area A, moment of inertia I. The displacement arrays of two nodes 1 and 2 are

\[
Q^{e}=[v_{1},\theta_{1},v_{2},\theta_{2}]^{t}
\]

$v $ is deflection (defection), or displacement; $\theta$ is the corner (slope). Note that the direction of $v$ and $\theta$, one is up, the other is counterclockwise.

The node force matrix of two nodes is

\[
P^{e}=[p_{v1},m_{1},p_{v2},m_{2}]^{t}
\]

Of course, the actual situation is often in the length of the beam in the direction of the role of the load, rather than only at the node, then the load equivalent, will be explained later. Note that both matrices are column matrices.

It is important to note that the Nodal force matrix represents all the forces on the node, not only the equivalent nodal forces caused by the load, but also the counter forces of the joints, the counter torque, etc.

2. Cell displacement field expression

Given the known conditions of 4 displacement nodes, it is assumed that the displacement deflection function of a pure curved beam element has four undetermined coefficients, as follows

\begin{equation}
V (x) =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}
\end{equation}
For both ends of the node, the displacement and the rotation angle are $v_{1},\theta_{1},v_{2},\theta_{2}$ respectively, note that the deflection curve equation at a point out of the guide value is the Pity Dorado Corner, so four boundary conditions for

$$
\begin{cases}
V (0) =v_{1} & V ' (0) =\theta_{1}\\
V (l) =v_{2} & V ' (l) =\theta_{2}
\end{cases}
$$

Solving equations with mm

The undetermined coefficients obtained are brought into the original equation, which can be

By merging four displacements, the similar terms can be

That is, the final flex-Curve equation Vfea
$$
Vfea=\text{$\theta $ \left (\frac{x^3}{l^2}-\frac{2 x^2}{l}+x\right) +\text{$\theta $ \left (\frac{x^3}{L^2}-\frac {x^2} {l}\right) +\text{v1} \left (\frac{2 x^3}{l^3}-\frac{3 x^2}{l^2}+1\right) +\text{v2} \left (\frac{3 x^2}{L^2}-\frac{2 x^ 3}{l^3}\right)
$$

If the $\zeta=\frac{x}{l}$, the matrix of the coefficients before the displacement of the upper formula is called the Matrix, which is often called the form function.

That
$$
V (x) =n (x) Q^{e}
$$

3. Cell strain field, stress field expression

The expression of the strain is
$$
\varepsilon=-yv "(x) =b (x) Q^{e}
$$

where $b (x) =-yn "(x) $, $B (x) $ is called the geometric matrix of the element, representing the relationship between strain and displacement.

The expression of stress is

$$
\sigma (x) =e\cdot B (x) q^{e}=s (x) Q^{e}
$$

where $s (x) =e\cdot B (x) $, called the element's stress matrix, represents the relationship between the stress and displacement of the unit.

4. Cell Potential Energy expression

The potential energy of the Unit is

\begin{equation} \label{potential Equation}
\varpi^{e}=u^{e}-w^{e}
\end{equation}

The expression of strain energy $u^{e}$ is
$$
\begin{split}u^{e} & =\dfrac{1}{2}\int_{0}^{l}\int_{a}\sigma\cdot\varepsilon Da\cdot dx\\
& =\dfrac{1}{2}\int_{0}^{l}\int_{a}e\cdot B (x) Q^{e}\cdot B (x) Q^{e}da\cdot dx\\
& =\dfrac{1}{2}\int_{0}^{l}\int_{a}e\cdot (Q^{e}) ^{t}\cdot B (x) ^{t}\cdot B (x) Q^{e}da\cdot dx\\
& =\dfrac{1}{2}q^{et}[\int_{0}^{l}\int_{a}e\cdot B (x) ^{t}\cdot B (x) Da\cdot dx]q^{e}\\
& =\dfrac{1}{2}q^{et}k^{e}q^{e}
\end{split}
$$

Note that because $b (x) $ and $q^{e}$ are single-line single-column matrices, the multiplication result is the same as after the transpose point is multiplied.

For a large lump in the middle, replace with $k^{e}$, $K ^{e}$ is the element stiffness matrix. Using mm to find the specific results, as follows

$$
B (x) =n "(x) =[\frac{12x}{l^{3}}-\frac{6}{l^{2}},l\left (\frac{6x}{l^{3}}-\frac{4}{l^{2}}\right), \frac{6}{L^{2}}- \frac{12x}{l^{3}},l\left (\frac{6x}{l^{3}}-\frac{2}{l^{2}}\right)]
$$

$$
K^{e}=ei\cdot\int_{0}^{l}b (x) ^{t}\cdot B (x) dx=ei\cdot\left (\BEGIN{ARRAY}{CCCC}
\FRAC{12}{L^{3}} & \frac{6}{l^{2}} &-\frac{12}{l^{3}} & \frac{6}{l^{2}}\\
\FRAC{6}{L^{2}} & \frac{4}{l} &-\frac{6}{l^{2}} & \frac{2}{l}\\
-\FRAC{12}{L^{3}} &-\frac{6}{l^{2}} & \frac{12}{l^{3}} &-\frac{6}{l^{2}}\\
\FRAC{6}{L^{2}} & \frac{2}{l} &-\frac{6}{l^{2}} & \frac{4}{l}
\end{array}\right) =\dfrac{ei}{l^{3}}\left (\BEGIN{ARRAY}{CCCC}
& 6L & -12 & 6l\\
6L & 4l^{2} & -6l & 2l^{2}\\
-12 & -6l & & -6l\\
6L & 2l^{2} & -6l & 4l^{2}
\end{array}\right)
$$

Since all forces are simplified into nodal forces, the force is
$$
W^{e}=p^{et}\cdot Q^{e}
$$

5. Stiffness equation of the unit

The stiffness equation of the element can be obtained by minimizing the $\varpi^{e}$ in the \ref{potential energy equation of the $q^{e}$ by the principle of the minimum potential energy.

$$
K^{e}\cdot Q^{e}=p^{e}
$$

Because the above equation is based on the principle of energy, it is universal. Note that the $p^{e}$ is the force matrix of all nodes, including the equivalent nodal forces of the external loads and the counter forces of the joints.

In fact, how to put $\varpi^{e}$ to $q^{e}$ to take the minimum value I now also can not its solution, this should be used to the variational principle of things, a bit similar to the derivation, this problem to learn the principle of variational principles later.

6. Equivalent Nodal load

Since the finite element method can only be loaded on the node of the element, if there is load on the unit, it is necessary to convert it to equivalent nodal load, the principle of conversion is equal work. For example, for a beam element subjected to uniform loading

It can be equivalent to the force and bending moment of the node

At this point, all equivalent before the external load on the unit to do the work, should be equal to the equivalent post-nodal force on the node displacement of the work done.

The deflection displacement field of the unit $v (x) =n (x) q^{e}$.

Equivalent to the work done before the external load

$$
W=\int_{0}^{l}p (x) v (x) dx=[\int_{0}^{l} (-p_{0}) N (x) Dx]\cdot Q^{e}
$$

Equivalent after the external load of the work done

$$
W=p^{et}\cdot Q^{e}
$$

Because they are equal,

$$
P^{et}=\int_{0}^{l} (-p_{0}) N (x) dx==[-\dfrac{p_{0}l}{2},-\dfrac{p_{0}l^{2}}{12},-\dfrac{p_{0}l}{2},\dfrac{p_{0} L^{2}}{12}]
$$

Note that the above is the equivalent nodal load of the $p_{0}$ load on the beam element, the process of establishing the equation is irrelevant to the constraints at both ends, so the equivalent nodal force is independent of the constraints on both ends. It is not possible to confuse "equivalent nodal loads" with "nodal forces caused by loads", for example, for simply supported Beams, there is no bending moment on the hinge supports at both ends, but the equivalent nodal moment is present on the node, which is equivalent to the equivalent nodal moment causing the angle of the end point.

Basic Course on finite element analysis (Zeng) Note Two-beam element finite element equation derivation