Finite elements/Axial bar approximate solution

To find the finite element solution, we can either start with the strong form and derive the weak form, or we can start with a weak form derived from a variational principle.

Let us assume that the approximate solution is ${\displaystyle \mathbf {u} _{h}(\mathbf {x} )}$ and plug it into the ODE. We get

${\displaystyle AE{\cfrac {d^{2}\mathbf {u} _{h}}{dx^{2}}}+a\mathbf {x} =R_{h}(\mathbf {x} )}$

where ${\displaystyle R_{h}}$ is the residual. We now try to minimize the residual in a weighted average sense

${\displaystyle \int _{0}^{L}R_{h}(\mathbf {x} )\mathbf {w} (\mathbf {x} )~dx=0}$

where ${\displaystyle \mathbf {w} (\mathbf {x} )}$ is a weighting function. Notice that this equation is similar to equation (5) (see 'Weak form: integral equation') with ${\displaystyle \mathbf {w} }$ in place of the variation ${\displaystyle \delta \mathbf {u} }$. For the two equations to be equivalent, the weighting function must also be such that ${\displaystyle \mathbf {w} (0)=0}$.

Therefore the approximate weak form can be written as

${\displaystyle {\int _{0}^{L}AE{\cfrac {d\mathbf {u} _{h}}{dx}}{\cfrac {d\mathbf {w} }{dx}}~dx=\int _{0}^{L}\mathbf {q} \mathbf {w} ~dx+\left.{\boldsymbol {R}}~\mathbf {w} \right|_{x=L}~.}}$

In Galerkin's method we assume that the approximate solution can be expressed as

${\displaystyle \mathbf {u} _{h}(\mathbf {x} )=a_{1}\varphi _{1}(\mathbf {x} )+a_{2}\varphi _{2}(\mathbf {x} )+\dots +a_{n}\varphi _{n}(\mathbf {x} )=\sum _{i=1}^{n}a_{i}\varphi _{i}(\mathbf {x} )~.}$

In the Bubnov-Galerkin method, the weighting function is chosen to be of the same form as the approximate solution (but with arbitrary coefficients),

${\displaystyle \mathbf {w} (\mathbf {x} )=b_{1}\varphi _{1}(\mathbf {x} )+b_{2}\varphi _{2}(\mathbf {x} )+\dots +b_{n}\varphi _{n}(\mathbf {x} )=\sum _{j=1}^{n}b_{j}\varphi _{j}(\mathbf {x} )~.}$

If we plug the approximate solution and the weighting functions into the approximate weak form, we get

${\displaystyle \int _{0}^{L}AE\left(\sum _{i=1}^{n}a_{i}{\cfrac {d\varphi _{i}}{dx}}\right)\left(\sum _{j=1}^{n}b_{j}{\cfrac {d\varphi _{j}}{dx}}\right)~dx=\int _{0}^{L}\mathbf {q} \left(\sum _{j=1}^{n}b_{j}\varphi _{j}\right)~dx+\left.{\boldsymbol {R}}~\left(\sum _{j=1}^{n}b_{j}\varphi _{j}\right)\right|_{x=L}~.}$

This equation can be rewritten as

${\displaystyle \sum _{j=1}^{n}b_{j}\left[\int _{0}^{L}AE\left(\sum _{i=1}^{n}a_{i}{\cfrac {d\varphi _{i}}{dx}}{\cfrac {d\varphi _{j}}{dx}}\right)~dx\right]=\sum _{j=1}^{n}b_{j}\left[\int _{0}^{L}\mathbf {q} \varphi _{j}~dx+\left.\left({\boldsymbol {R}}~\varphi _{j}\right)\right|_{x=L}\right]~.}$

From the above, since ${\displaystyle b_{j}}$ is arbitrary, we have

${\displaystyle \int _{0}^{L}AE\left(\sum _{i=1}^{n}a_{i}{\cfrac {d\varphi _{i}}{dx}}{\cfrac {d\varphi _{j}}{dx}}\right)~dx=\int _{0}^{L}\mathbf {q} \varphi _{j}~dx+\left.{\boldsymbol {R}}~\varphi _{j}\right|_{x=L}~,~j=1\dots n.}$

After reorganizing, we get

${\displaystyle \sum _{i=1}^{n}\left[\int _{0}^{L}{\cfrac {d\varphi _{j}}{dx}}AE{\cfrac {d\varphi _{i}}{dx}}~dx\right]a_{i}=\int _{0}^{L}\varphi _{j}\mathbf {q} ~dx+\left.\varphi _{j}{\boldsymbol {R}}\right|_{x=L}~,~j=1\dots n}$

which is a system of ${\displaystyle n}$ equations that can be solved for the unknown coefficients ${\displaystyle a_{i}}$. Once we know the ${\displaystyle a_{i}}$s, we can use them to compute approximate solution. The above equation can be written in matrix form as

${\displaystyle \mathbf {K} \mathbf {a} =\mathbf {f} \qquad \leftrightarrow \qquad K_{ji}a_{i}=f_{j}}$

where

${\displaystyle \mathbf {K} =\int _{0}^{L}\mathbf {B} ^{T}\mathbf {D} \mathbf {B} ~dx\qquad \leftrightarrow \qquad K_{ji}=\int _{0}^{L}{\cfrac {d\varphi _{j}}{dx}}AE{\cfrac {d\varphi _{i}}{dx}}~dx}$

and

${\displaystyle f_{j}=\int _{0}^{L}\varphi _{j}\mathbf {q} ~dx+\left.\varphi _{j}{\boldsymbol {R}}\right|_{x=L}~.}$

The problem with the general form of the Galerkin method is that the functions ${\displaystyle \varphi _{i}}$ are difficult to determine for complex domains.