How to formulate the Lagrangian for a cantileverbeam

[Jompa]

Hello smarties!
Anyone who can help me out here?

The problem statement is in the attachment, problem.jpg

My attempt:
Lagrangian: L=T-U where T kin. energy and U elastic potential.

Stating T: T=$$\int$$0.5*$$\rho$$A(w(x,t)^2)dx from 0 to L

Stating U: U=L/6EI *[(-$$\int$$q(x,t)xdx+M(t))^2 +M(t)(-$$\int$$q(x,t)xdx+M(t)) +M(t)^2]- what?
(the integrals from 0 to L, sorry couldn't get the hang of how to get the limits to look nice with latex)

The get the expression for W (the expression before "what") comes from W=$$\overline{W}$$=L/6EI*[M$$_{A}$$M$$_{B}$$+M$$_{A}$$^2+M$$_{B}$$^] where M$$_{A}$$ and M$$_{B}$$ are the moments at the ends of the beam. They are determined with help of moment equilibrium.

How should the "what" term look like? Should it look something like this $$^{}_{S_t}$$$$\int$$F*w(L,t)dS
Does my elastic energy come out right?

Thanks for helping!
Most grateful for all comments.
/J

 

[Jhero]

Hello J,

Thank you for reaching out for help with this problem. Based on the information you provided, it seems like you are trying to find the expression for the elastic potential energy in a beam subjected to a distributed load and two end moments. The "what" term in your equation should represent the contribution of the distributed load to the elastic potential energy. It should look something like this:

U = L/6EI * [(-∫q(x,t)xdx+M(t))^2 + M(t)(-∫q(x,t)xdx+M(t)) + M(t)^2 + ∫q(x,t)w(x,t)dx]

Where ∫q(x,t)w(x,t)dx represents the work done by the distributed load on the beam, and q(x,t) is the distributed load function. This term is necessary because the distributed load also contributes to the deformation of the beam, and therefore, to the elastic potential energy.

As for your expression for T (kinetic energy), it seems correct. However, it might be helpful to specify the units of the variables and constants in your equation, as well as the limits of integration. This will make it easier to understand and follow your calculations.

I hope this helps. If you have any further questions, please don't hesitate to ask. Good luck with your research!