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.
