Using the Ritz Method we can express vibration in a beam as follows: $$w(x,t) = \displaystyle\sum_{j=1}^{n}\psi_{j}(x)u_{j}(t),$$
where $\psi_{j}(x)$ is an $\textit{admissible}$ function, i.e. a continuous function that satisfies the boundary conditions of the system.
Using the following form of Euler-Lagrange,
$$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}+\frac{\partial V}{\partial q_{i}}=0 \text{ } \forall \text{ }i\in \mathbb{N}.$$
Deriving expressions for kinetic energy ($T$) and potential energy ($V$) and substituting,
$$ 0=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{x}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}\frac{1}{2}\rho A_{c} \int_{0}^{a} \psi_{j}\psi_{k}\text{ } dx\right)\right)-\frac{\partial}{\partial x}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}\frac{1}{2}\rho A_{c} \int_{0}^{a} \psi_{j}\psi_{k}\text{ } dx\right) \\ +\frac{\partial}{\partial x}\left(\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{k=1}^{n}u_{j}u_{k}\frac{EI}{2}\int_{0}^{a}\frac{\partial^{2}\psi_{j}}{\partial x^{2}}\frac{\partial^{2}\psi_{k}}{\partial x^{2}}\text{ }dx\right).$$
Frankly, I don't really know where to go from here. I'm stuck primarily because of the $\dot{x}$ term. I think the next step is to find some new way to express $\dot{x}$, but I'm not sure.
Short Answer:
$$\frac{\partial}{\partial \dot{x}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}\frac{1}{2}A_{c}\int_{0}^{a}\psi_{j}\psi_{k}dx\right)=0$$
Long Answer:
I was taking Euler-Lagrange with respect to the wrong variables. I should have set $q_{i}=u_{i}$ $\forall$ $i\in\mathbb{N}$. Below I will correct the mistake and go through with the derivation for completeness and to attempt to eliminate all potential confusion.
For a beam of length $a$,
$$ T=\frac{1}{2}\rho A_{c} \int_{0}^{a}\dot{w}^{2}\text{ }dx $$
Substituting our definition for $w$,
$$ T=\frac{1}{2}\rho A_{c} \int_{0}^{a}\left(\displaystyle\sum_{j=1}^{N}\psi_{j}\dot{u}_{j}\right)^{2}\text{ }dx \\ T=\frac{1}{2}\rho A_{c} \int_{0}^{a}\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\psi_{j}\psi_{k}\dot{u}_{j}\dot{u}_{k}\text{ }dx. $$
I believe that this should play nicely with Fubini's Theorem, but I can't exactly show why this is true since I don't know the details well enough, but I believe it should have something to do with the fact that the integral is finite over $[0,a]$. Blindly applying Fubini, and recognizing that $u_{j}$ is not a function of $x$,
$$ T=\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}\frac{1}{2}\rho A_{c} \int_{0}^{a} \psi_{j}\psi_{k}\text{ } dx. $$
Defining,
$$ (m_{jk})_{b} \equiv \rho A_{c} \int_{0}^{a} \psi_{j}\psi_{k}\text{ } dx. $$
Substituting,
$$ T=\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\frac{1}{2}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}. $$
For a beam we can express the potential energy as $$ V=\int_{0}^{a}\frac{EI}{2}\left(\frac{\partial^{2}w}{\partial x^{2}}\right)^{2} dx. $$
Substituting in our definition for $w$, $$ V=\frac{EI}{2}\int_{0}^{a}\left(\frac{\partial^{2}}{\partial x^{2}}\displaystyle\sum_{j=1}^{n}\psi_{j}u_{j}\right)^{2}dx \\ V=\frac{EI}{2}\int_{0}^{a}\left(\displaystyle\sum_{j=1}^{n}\frac{\partial^{2}\psi_{j}}{\partial x^{2}}u_{j}\right)^{2}dx. $$
Taking similar steps to before (to shorten the answer length), $$ V=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{k=1}^{n}\frac{EI}{2}\int_{0}^{a}\frac{\partial^{2}\psi_{j}}{\partial x^{2}}\frac{\partial^{2}\psi_{k}}{\partial x^{2}}u_{j}u_{k}\text{ }dx. $$
Defining, $$ (k_{jk})_{b} \equiv EI\int_{0}^{a}\frac{\partial^{2}\psi_{j}}{\partial x^{2}}\frac{\partial^{2}\psi_{k}}{\partial x^{2}}\text{ }dx. $$
Substituting, $$ V=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{k=1}^{n}\frac{1}{2}u_{j}u_{k}(k_{jk})_{b}. $$
The general formulation for Euler-Lagrange can be expressed as $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}+\frac{\partial V}{\partial q_{i}}=0 \text{ } \forall \text{ }i\in \mathbb{N}. $$
Choosing $q_{i}=u_{i}$ $\forall$ $i\in \mathbb{N}$, $$ 0=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{u}_{i}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right)\right)-\frac{\partial}{\partial u_{i}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right)+\frac{\partial}{\partial u_{i}}\left(\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{k=1}^{n}u_{j}u_{k}(k_{ij})_{b}\right)\\ 0=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{u}_{i}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right)\right)+\frac{\partial}{\partial u_{i}}\left(\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{k=1}^{n}u_{j}u_{k}(k_{ij})_{b}\right). $$
Examining the case where i=1 for the first term, $$ \frac{\partial}{\partial \dot{u}_{1}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\frac{1}{2}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right) = \begin{matrix} \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{1}\dot{u}_{1}(m_{11})_{b}\right)\text{ }+ & \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{1}\dot{u}_{2}(m_{12})_{b}\right)\text{ }+ & \cdots & +\displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{1}\dot{u}_{N}(m_{1N})_{b}\right)+ \\ \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{2}\dot{u}_{1}(m_{21})_{b}\right)\text{ }+ & \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{2}\dot{u}_{2}(m_{22})_{b}\right)\text{ }+ & \cdots & +\displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{2}\dot{u}_{N}(m_{2N})_{b}\right)+ \\ \vdots & \vdots & \ddots & \vdots \\ \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{N}\dot{u}_{1}(m_{N1})_{b}\right)\text{ }+ & \displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{N}\dot{u}_{2}(m_{N2})_{b}\right)\text{ }+ & \cdots & +\displaystyle\frac{\partial}{\partial \dot{u}_{1}}\left(\frac{1}{2}\dot{u}_{N}\dot{u}_{N}(m_{NN})_{b}\right) \end{matrix} \\ \\ \\ \\ =\begin{matrix} \dot{u}_{1}(m_{11})_{b}\text{ }+ & \frac{1}{2}\dot{u}_{2}(m_{12})_{b}\text{ }+ & \cdots & +\frac{1}{2}\dot{u}_{N}(m_{1N})_{b}+ \\ \frac{1}{2}\dot{u}_{2}(m_{21})_{b}\text{ }+ & 0\text{ }+ & \cdots & +0+ \\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{2}\dot{u}_{N}(m_{N1})_{b}\text{ }+ & 0\text{ }+ & \cdots & +0 \end{matrix}. \nonumber $$
Recognizing that $(m_{jk})_{b}=(m_{kj})_{b}$, $$ \frac{\partial}{\partial \dot{u}_{1}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right) = \begin{matrix} \dot{u}_{1}(m_{11})_{b}\text{ }+ & \frac{1}{2}\dot{u}_{2}(m_{12})_{b}\text{ }+ & \cdots & +\frac{1}{2}\dot{u}_{N}(m_{1N})_{b}+ \\ \frac{1}{2}\dot{u}_{2}(m_{12})_{b}\text{ }+ & 0\text{ }+ & \cdots & +0+ \\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{2}\dot{u}_{N}(m_{1N})_{b}\text{ }+ & 0\text{ }+ & \cdots & +0 \end{matrix} \nonumber \\ \frac{\partial}{\partial \dot{u}_{1}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right) = \displaystyle\sum_{j=1}^{N} \dot{u}_{j}(m_{1j})_{b}. $$
Generalizing for all $i$, $$ \frac{\partial}{\partial \dot{u}_{i}}\left(\displaystyle\sum_{j=1}^{N}\displaystyle\sum_{k=1}^{N}\dot{u}_{j}\dot{u}_{k}(m_{jk})_{b}\right) = \displaystyle\sum_{j=1}^{N} \dot{u}_{j}(m_{ij})_{b}. $$
Similarly for the second term, $$ \frac{\partial}{\partial u_{i}}\left(\displaystyle\sum_{j=1}^{N} \displaystyle\sum_{k=1}^{N}u_{j}u_{k}(k_{jk})_{b}\right)=\displaystyle\sum_{j=1}^{N}u_{j}(k_{ij})_{b}. $$
Substituting, $$ \frac{d}{dt}\left(\displaystyle\sum_{j=1}^{N} \dot{u}_{j}(m_{ij})_{b}\right)+\displaystyle\sum_{j=1}^{N}u_{j}(k_{ij})_{b}=0 \\ \displaystyle\sum_{j=1}^{N} \ddot{u}_{j}(m_{ij})_{b}+\displaystyle\sum_{j=1}^{N}u_{j}(k_{ij})_{b}=0. $$
Rewriting as a matrix equation, $$ [m]\ddot{u}+[k]u=\boldsymbol{0}. $$
From this point we can solve this system as an N degree of freedom system in $u$ using classical modal analysis. To get back to physical coordinates there will have to be 2 coordinate transformations.
Please let me know if there is anything I can do to improve this answer.