\begin{aligned} a \frac{\partial^4 W(x,t)}{\partial x^4}+b\frac{\partial^2 W(x,t)}{\partial t^2}&=c\frac{\partial^2 U(y,t)}{\partial t^2}W(x,t)\delta(x-\zeta)\\ d\frac{\partial^2 U(y,t)}{\partial y^2}&=c\frac{\partial^2 U(y,t)}{\partial t^2} \end{aligned}
How to solve the first partial differential equation which is coupled as shown above. The first PDE is the governing equation of motion of the beam and the second PDE is the governing equation of motion of the bar. If we assume the harmonic dependency that is \begin{aligned} W(x,t)&=W(x)e^{i\omega t}\\ U(y,t)&=U(y)e^{i\omega t}\\ \end{aligned} The above PDEs now converted into ODE. And the associated boundary conditions are \begin{align} W\big|_{x=0} = W\big|_{x=1}&=0\\ \frac{\mathrm{d}W}{\mathrm{d}x}\bigg|_{x=0} = \frac{\mathrm{d}W}{\mathrm{d}x}\bigg|_{x=1} &= 0\\ \end{align} and the boundary conditions for 2nd PDE are \begin{align} U\big|_{y=0} &= 0\\ U\big|_{y=1} &= W\big|_{x=\zeta} \end{align}
How to solve these PDEs? Below figure shows the Line diagram of bar and beam
A similar case with a spring instead of a bar is tackled in Ref. [A].
[A] J.-S. Wu, H.-M. Chou: "Free vibration analysis of a cantilever beam carrying any number of elastically mounted point masses with the analytical-and-numerical-combined method", J. Sound Vib. 213-2 (1998), 317-332. doi:10.1006/jsvi.1997.1501
Let us assume that both beam and bar are homogeneous. They have the same geometry and the same material properties. Perfect vertical coupling between both sub-systems is achieved (perfect mechanical linkage with negligible size, cf. diagram), and gravity is neglected.
In the interior of the beam ($0<x<1$), the vertical deflection $w$ satisfies the dynamic beam equation $$ \rho S\frac{\partial^2 w}{\partial t^2} = -EI \frac{\partial^4 w}{\partial x^4} + f^v_{\text{bar/beam}} \, , $$ where $\rho$ is the mass density, $E$ is Young's modulus, $I$ is the moment of inertia, $S$ is the cross-section area, and $f^v_{\text{bar/beam}}$ is the vertical volume force due to the bar. In the interior of the bar ($0<y<1$), the vertical displacement $u$ satisfies D'Alembert's equation $$ \rho \frac{\partial^2 u}{\partial t^2} = E\frac{\partial^2 u}{\partial y^2} \, . $$ At $y=1$, the vertical displacement is $u|_{y=1} = w|_{x=\zeta}$. The vertical force in the bar is $\left.ES\frac{\partial u}{\partial y}\right|_{y=1}$. According to Newton's second law of motion, the force by unit volume reads $$ f^v_{\text{bar/beam}} = -\left.E\frac{\partial u}{\partial y}\right|_{y=1} \delta(x-\zeta)\, , $$ where $\delta$ is the Dirac distribution. At the other boundaries, we have $w|_{x=0} = 0 = w|_{x=1}$, and $u|_{y=0} = 0$.
One can apply separation of variables, to seek time-harmonic solutions. Setting $u(y,t) = U(y)e^{\text i \omega t}$ and $w(x,t) = W(x)e^{\text i \omega t}$, we are left with \begin{aligned} W^{(4)}(x) - \frac{\rho S \omega^2}{E I} W(x) &= - \frac{U'(1)}{I} \delta(x-\zeta) \\ U''(y) + \frac{\rho \omega^2}{E} U(y) &= 0 \\ W(0) = W(1) = W'(0) = W'(1) &= 0 \\ U(0) &= 0 \\ U(1) &= W(\zeta) \, .\\ \end{aligned} Thus, we deduce that $$U(y) = c_1\cos(k y) + c_2\sin(k y)$$ where $k = \omega\sqrt{\frac{\rho}{E}}$. We obtain $W(x)$ from the sum of the homogeneous solution and a particular solution (or from Fourier transforms like here): \begin{aligned} W(x) &= \frac{-U'(1)}{2 I \kappa^3}\left[\sinh(\kappa (x-\zeta))-\sin(\kappa (x-\zeta))\right] \theta(x-\zeta) \\ &+ d_1 \cos(\kappa x) + d_2 \sin(\kappa x) + d_3 \cosh(\kappa x) + d_4 \sinh(\kappa x)\, , \end{aligned} where $\kappa = \sqrt[4]{\frac{\rho S \omega^2}{E I}}$ and $\theta$ is the Heaviside function. The boundary conditions give the trivial solution if the linear system of equations satisfied by $c_1$, $c_2$, $d_1$, $d_2$, $d_3$, $d_4$ has an invertible matrix. Otherwise, this linear system gives the eigenfrequencies/eigenmodes.
Notes:
In the case of a spring instead of the bar, we have only the PDE for $w$ to solve. The volume force reads $$ f^v_{\text{bar/beam}} = -\frac{K}{S}w|_{x=\zeta} \delta(x-\zeta)\, , $$ where $K$ is the spring constant.
In the case of two beams, deflection and longitudinal displacements must be considered for each beam. Since only transverse forces are taken into account, longitudinal forces do not come into play. One could consider resetting the problem as the interaction of three beams (picture below). The interaction is purely written as a displacement boundary condition with the black beam, while the blue and red beams satisfy also a force boundary condition.