Euler–Bernoulli Beam Theory - Dynamic Beam Equation

Dynamic Beam Equation

The dynamic beam equation is the Euler-Lagrange equation for the following action

$S = \int_0^L \left dx$

The first term represents the kinetic energy where is the mass per unit length; the second one represents the potential energy due to internal forces (when considered with a negative sign) and the third term represents the potential energy due to the external load . The Euler-Lagrange equation is used to determine the function that minimizes the functional . For a dynamic Euler-Bernoulli beam, the Euler-Lagrange equation is

$\cfrac{\partial^2 }{\partial x^2}\left(EI\cfrac{\partial^2 w}{\partial x^2}\right) = - \mu\cfrac{\partial^2 w}{\partial t^2} + q(x)$

Derivation of Euler–Lagrange equation for beams
Since the Lagrangian is $\mathcal{L} := \tfrac{1}{2} \mu \left( \frac{\partial w}{\partial t} \right)^2 - \tfrac{1}{2} EI \left( \frac{ \partial^2 w}{\partial x^2} \right)^2 + q(x) w(x,t) = \tfrac{\mu}{2}\dot{w}^2 - \tfrac{EI}{2}w_{xx}^2 + qw \equiv \mathcal{L}(x, t, w, \dot{w}, w_{xx})$ the corresponding Euler-Lagrange equation is $\cfrac{\partial\mathcal{L}}{\partial w} - \frac{\partial}{\partial t}\left(\frac{\partial \mathcal{L}}{\partial \dot{w}}\right) + \frac{\partial^2}{\partial x^2}\left(\frac{\partial \mathcal{L}}{\partial w_{xx}}\right) = 0$ Now, $\cfrac{\partial\mathcal{L}}{\partial w} = q ~;~~ \frac{\partial \mathcal{L}}{\partial \dot{w}} = \mu\dot{w} ~;~~ \frac{\partial \mathcal{L}}{\partial w_{xx}} = -EI w_{xx} ~.$ Plugging into the Euler-Lagrange equation gives $q - \mu\ddot{w} - (EI w_{xx})_{xx} = 0$ or, $\cfrac{\partial^2 }{\partial x^2}\left(EI\cfrac{\partial^2 w}{\partial x^2}\right) = - \mu\cfrac{\partial^2 w}{\partial t^2} + q$ which is the governing equation for the dynamics of an Euler-Bernoulli beam.

Derivation of Euler–Lagrange equation for beams

Since the Lagrangian is

$\mathcal{L} := \tfrac{1}{2} \mu \left( \frac{\partial w}{\partial t} \right)^2 - \tfrac{1}{2} EI \left( \frac{ \partial^2 w}{\partial x^2} \right)^2 + q(x) w(x,t) = \tfrac{\mu}{2}\dot{w}^2 - \tfrac{EI}{2}w_{xx}^2 + qw \equiv \mathcal{L}(x, t, w, \dot{w}, w_{xx})$

the corresponding Euler-Lagrange equation is

$\cfrac{\partial\mathcal{L}}{\partial w} - \frac{\partial}{\partial t}\left(\frac{\partial \mathcal{L}}{\partial \dot{w}}\right) + \frac{\partial^2}{\partial x^2}\left(\frac{\partial \mathcal{L}}{\partial w_{xx}}\right) = 0$

Now,

$\cfrac{\partial\mathcal{L}}{\partial w} = q ~;~~ \frac{\partial \mathcal{L}}{\partial \dot{w}} = \mu\dot{w} ~;~~ \frac{\partial \mathcal{L}}{\partial w_{xx}} = -EI w_{xx} ~.$

Plugging into the Euler-Lagrange equation gives

$q - \mu\ddot{w} - (EI w_{xx})_{xx} = 0$

or,

$\cfrac{\partial^2 }{\partial x^2}\left(EI\cfrac{\partial^2 w}{\partial x^2}\right) = - \mu\cfrac{\partial^2 w}{\partial t^2} + q$

which is the governing equation for the dynamics of an Euler-Bernoulli beam.

Read more about this topic: Euler–Bernoulli Beam Theory

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