Lagrangian Involving an Integral

Question

This is my first post here, so do let me know if I am asking anything or in any manner improperly!

I am trying to construct a Lagrangian for a dynamic Euler-Bernoulli beam (with one end cramped) with a large deflection. The linearize case is well recorded by Wikipedia:

However, with large deflection, I need to use the arclength together with the local deflection angle to represent the status of the beam, so the Cartesian coordinates of the beam located $s$ from the cramped end can be represented as:

$$x(s,t)=\int_0^s\cos(\theta(s',t))\,\mathrm ds,\space y(s,t)=\int_0^s\sin(\theta(s',t))\,\mathrm ds \tag 1$$

Thus, at a specific location $s$, with linear density $\mu$ and external load $q$ (which is perpendicular to the beam locally), the Lagrangian can be written as:

$$\mathcal{L}=\frac{1}{2}\mu \dot{x}^2+\frac{1}{2}\mu \dot{y}^2-\frac{1}{2}EI\left(\frac{\partial\theta(s,t)}{\partial s}\right)^2\\[3ex]+q(s,t)x(s,t)\sin(\theta(s,t)) +q(s,t)y(s,t)\cos(\theta(s,t)) \tag 2\\$$

Here the term $\frac{1}{2}EI\left(\frac{\partial\theta(s,t)}{\partial s}\right)^2$ is associated with the energy stored in the bending moment. From here if we plug in $(1)$ to $(2)$, we obtain a Lagrangian containing an integral:

$$\,\\\mathcal{L}=\frac{1}{2}\mu\left(\int_0^s \sin(\theta(s',t))\dot{\theta}(s',t)\,\mathrm ds\right)^2+\frac{1}{2}\mu\left(\int_0^s \cos(\theta(s',t))\dot{\theta}(s',t)\,\mathrm ds\right)^2\\[3ex]-\frac{1}{2}EI\left(\left(\frac{\partial\theta(s,t)}{\partial s}\right)^2 +q(s,t)\sin(\theta(s,t)\right) \int_0^s\cos(\theta(s',t))\,\mathrm ds+q(s,t)\cos(\theta(s,t))\\[3ex]\cdot\int_0^s \sin(\theta(s',t))\,\mathrm ds$$

Obviously, this Lagrangian contains some integrals that we don't see in those simpler cases we find in the textbook. Does anyone have an idea on how to deal with this sort of Lagrangian? A reference to a similar problem would definitely help as well.

Many thanks!

Answer 1

The boundary conditions do not affect the Lagrangian of the problem. You may use von-Karman strains to take the large rotations into account, which is simplified form of Green's strains.

First, define displacement fields:

$$ u_1(x,t) = u(x,t) - z \frac{\partial v(x,t)}{\partial x}$$ $$ u_2(x,t) = v(x,t) $$

I consider this problem as two dimensional problem, where $u_1$ is the axial displacement, $u_2$ is the transverse displacement. Since this is a beam, we only one spatial variable $x$ and temporal variable $t$.

There will be only axial strain due to beam assumption, which is given by:

$$ \varepsilon_x = \frac{\partial u_1}{\partial x} + \frac{1}{2} \left( \frac{\partial u_2}{\partial x} \right)^2 $$

Expanding this term by substituting the displacement relations:

$$ \varepsilon_x = \frac{\partial u}{\partial x} - z \frac{\partial^2 v}{\partial x^2} + \frac{1}{2} \left( \frac{\partial v}{\partial x} \right)^2 $$

The Lagrangian is difference between kinetic and strain energy. Note that you have included external work in Lagrangian, which is not correct.

$$ \mathcal{L} = T - U$$

Kinetic energy:

$$ T = \int_\Omega \frac{1}{2}\rho \left[ \left( \frac{\partial u}{\partial t} \right)^2 + \left( \frac{\partial v}{\partial t} \right)^2 \right] d\Omega $$

$$ T = \int_x \frac{1}{2} \mu \left( \ddot{u}^2 + \ddot{v}^2 \right) dx $$

Strain energy (assuming linear isotropic material):

$$ U = \int_\Omega \frac{1}{2} E \varepsilon_x ^2 d\Omega $$

$$ U = \int_\Omega \frac{1}{2} E \left[ \left(\frac{\partial u}{\partial x}\right)^2 + z^2 \left(\frac{\partial^2 v}{\partial x^2}\right)^2 + \frac{1}{4} \left(\frac{\partial v}{\partial x}\right)^4 + 2\left(\frac{\partial u}{\partial x}\right)\left(-z \frac{\partial^2 v}{\partial x^2}\right) + 2\left(\frac{\partial u}{\partial x}\right)\frac{1}{2} \left(\frac{\partial v}{\partial x}\right)^2 + 2\left(-z \frac{\partial^2 v}{\partial x^2}\right)\frac{1}{2} \left(\frac{\partial v}{\partial x}\right)^2 \right] d\Omega$$

Assuming that coordinate system is attached to the principal centroidal axes, the equation reduces by recalling that $ \int z dA = 0 $ and $ I = \int z^2 dA $

$$ U = \int_x \frac{1}{2} \left[ EA \left(\frac{\partial u}{\partial x}\right)^2 + EI \left(\frac{\partial^2 v}{\partial x^2}\right)^2 + EA \frac{1}{4} \left(\frac{\partial v}{\partial x}\right)^4 + EA \left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial v}{\partial x}\right)^2 \right] dx$$

Finally, your Lagrangian is given as:

$$ \mathcal{L} = \int_x \frac{1}{2} \mu \left( \ddot{u}^2 + \ddot{v}^2 \right) dx - \int_x \frac{1}{2} \left[ EA \left(\frac{\partial u}{\partial x}\right)^2 + EI \left(\frac{\partial^2 v}{\partial x^2}\right)^2 + EA \frac{1}{4} \left(\frac{\partial v}{\partial x}\right)^4 + EA \left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial v}{\partial x}\right)^2 \right] dx $$

I hope this answers your question.