Notebook

Universidade Federal do Rio Grande do Sul (UFRGS)
Programa de Pós-Graduação em Engenharia Civil (PPGEC)

PEC00025: Introduction to Vibration Theory¶

Class 18 - Lagrange's equation¶

1. The Lagrange's equation
2. Example: generic s.d.o.f. system
3. Example: generic m.d.o.f. system
4. Example: simply supported beam
4.1. Discrete flexibility and distributed mass
4.2. Discrete mass and distributed flexibility
4.3. Distributed mass and distributed flexibility
5. Example: nonlinear simple pendulum
6. Example: tower with flexible foundation
7. Example: nonlinear double pendulum
8. Assignments

Prof. Marcelo M. Rocha, Dr.techn. (ORCID)
Porto Alegre, RS, Brazil

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# Importing Python modules required for this notebook
# (this cell must be executed with "shift+enter" before any other Python cell)

import numpy as np
import matplotlib.pyplot as plt
from   scipy.integrate import odeint

1. The Lagrange's equation ¶

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2. Example: generic s.d.o.f. system ¶

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3. Example: generic m.d.o.f. system ¶

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4. Example: simply supported beam ¶

Now we shall use the Lagrange's equation for solving a simply supported beam, considering three different models:

Distributed mass and discrete flexibility,
Distributed flexibility and discrete mass, and
Distributed mass and distributed flexibility.

The three beam models are useful for illustrating the use of Lagrange's equation with different energy evaluation approaches. At the end, the three solutions may be compared for evaluating the best estimates of discrete parameters $k_\theta$ and $M$ such that the three models give the same beam natural vibration frequency.

4.1. Distributed mass and discrete flexibility ¶

For the sake of compatibility with the other two models, we define the generalized coordinate $u$ as being the vertical displacement at beam center. The potential elastic energy depends on the relative rotation, $\theta$ , between the two segments:

$\theta = \arcsin \left( \frac{2u}{L}\right) \approx \frac{2u}{L}$

where the small displacements assumption ( $\sin \beta \approx \beta$ ) has been used. Hence:

$V = \frac{1}{2} k_\theta \theta^2 \approx \frac{2 k_\theta}{L^2} u^2$

The kinetic energy is obtained by integrating the velocities along the two bars:

$T = 2 \cdot \frac{1}{2} \int_0^{L/2} {\left( \dot{r}_x^2 + \dot{r}_y^2 \right) \; \mu d\xi}$

with:

$\begin{align*} r_x &= \xi \cos\left( \frac{2u}{L}\right) \\ r_y &= \xi \sin\left( \frac{2u}{L}\right) \end{align*}$

Observe that, without the small displacements assumption, the dummy variable $\xi$ is a line differential, not the projection in $x$ direction. Corresponding time derivatives are:

$\begin{align*} \dot{r}_x &= \frac{dr_x}{du} \; \frac{du}{dt} = -\frac{2\xi}{L} \sin\left( \frac{2u}{L}\right) \, \dot{u} \approx -\frac{4 u \dot{u}}{L^2} \xi \approx 0 \\ \dot{r}_y &= \frac{dr_y}{du} \; \frac{du}{dt} = \frac{2\xi}{L} \cos\left( \frac{2u}{L}\right) \, \dot{u} \approx \frac{2\dot{u}}{L} \xi \\ \end{align*}$

where the small displacements assumption is made ( $\sin \beta \approx \beta$ and $\cos \beta \approx 1$ ) and the velocity in direction $x$ is neglected. Replacing these results in the kinetic energy yields:

$T \approx \frac{4\dot{u}^2}{L^2} \int_0^{L/2} {\xi^2 \; \mu d\xi} = \frac{\mu L}{6} \, \dot{u}^2$

Now the two terms of Lagrange's equation can be calculated. Firstly the kinectic energy:

$\frac{\partial T}{\partial \dot{u}} = \frac{\mu L}{3} \, \dot{u}$

and then for the potential elastic energy:

$\frac{\partial V}{\partial u} = \frac{4 k_\theta }{L^2} \, u$

Applying Lagrange's equation gives finally the equilibrium equation:

$\frac{\mu L}{3} \, \ddot{u} + \frac{4 k_\theta }{L^2} \, u = 0$

which means that the natural vibration frequency is:

$\omega_{\rm n} = \sqrt{\frac{12 k_\theta}{\mu L^3}}$

4.2. Distributed flexibility and discrete mass ¶

In this case we must assume a deflected shape. A good option is the elastic line resulting from a point load at beam center:

$w(\xi) = \frac{P L^3}{12 EI} \left( \frac{3}{4} \xi - \xi^3 \right), \hspace{1cm} \xi \leq \frac{1}{2}$

where the variable $\xi$ is a non-dimensional coordinate, $\xi = x/L$ . This function is valid up to beam center and its symmetry must be accounted for. The assumed shape must be rescaled, such that the displacement at beam center is our generalized coordinate, $u$ . The maximumum central deflection from the function above is:

$w_{\rm max} = \frac{P L^3}{48 EI}$

Re-scaling gives:

$w(\xi) = 4 u \left( \frac{3}{4} \xi - \xi^3 \right), \hspace{1cm} \xi \leq \frac{1}{2}$

The rotation and the curvature functions are, respectively:

$\begin{align*} w^{\prime}(\xi) &= \frac{ 4}{L } u \left( \frac{3}{4} - 3\xi^2 \right) \\ w^{\prime\prime}(\xi) &= -\frac{24}{L^2} u \xi \end{align*}$

where one must be aware that:

$\frac{d}{dx} = \frac{d}{d\xi}\,\frac{d\xi}{dx} = \frac{1}{L} \, \frac{d}{d\xi}$

The potential elastic energy is then given by:

$V = 2 \cdot \frac{1}{2} \int_{0}^{1/2} {EI \left[ w^{\prime\prime}(\xi) \right]^2 \, L d\xi} = \frac{24 EI}{L^3} \, u^2$

On the other hand, the kinetic energy evaluation is straighforward:

$T = \frac{1}{2} M\dot{u}^2$

Now we calculate the two terms of Lagrange's equation, starting with the kinectic energy:

$\frac{\partial T}{\partial \dot{u}} = M \, \dot{u}$

and then for the potential elastic energy:

$\frac{\partial V}{\partial u} = \frac{48 EI}{L^3} \, u$

Applying Lagrange's equation gives finally the equilibrium equation:

$M \, \ddot{u} + \frac{48 EI}{L^3} \, u = 0$

which means that the natural vibration frequency is:

$\omega_{\rm n} = \sqrt{\frac{48 EI}{ML^3}}$

4.3. Distributed mass and distributed flexibility ¶

Here we assume a half-sine as the deflected shape (that we know to be the right solution):

$w(\xi) = u \sin\left( \pi \xi \right)$

The rotation and the curvature functions are, respectively:

$\begin{align*} w^{\prime}(\xi) &= \frac{\pi u}{L } \, \cos \left(\pi \xi \right) \\ w^{\prime\prime}(\xi) &= -\frac{\pi^2 u}{L^2} \, \sin \left(\pi \xi \right) \end{align*}$

The velocity is assumed (small displacements) to have only the vertical component, which is then calculated as:

$\dot{w}(\xi) = \frac{dw}{du} \, \frac{du}{dt} = \dot{u} \sin\left( \pi \xi \right)$

The potential elastic energy is then given by:

$V = \frac{1}{2} \int_{0}^{1} {EI \left[ w^{\prime\prime}(\xi) \right]^2 \; L d\xi} = \frac{\pi^4 EI}{4 L^3} \, u^2$

A comparison with the previous model leads to $\pi^4/4 \approx 24$ , such that the potential energy for both models are quite close (error is $\approx 1.5$ %) even by using different functions for the deflected shape. Comparison with the discrete flexibility model demands that:

$\frac{2 k_\theta}{L^2} = \frac{\pi^4 EI}{4 L^3}$

and consequently the hinge flexibility to match the continuous beam flexibility must be:

$k_\theta = \frac{\pi^4 EI}{8 L}$

On the other hand, the kinetic energy is given by:

$T = \frac{1}{2} \int_{0}^{1} {\mu \left[ \dot{w}(\xi) \right]^2 \; L d\xi} = \frac{\mu L}{4} \, \dot{u}^2$

A comparison with the discrete mass model demands that:

$\frac{M}{2} = \frac{\mu L}{4}$

and consequently the discrete mass must be $M = \mu L/2$ , which is half the total beam mass.

Finally, taking derivatives and applying Lagrange's equation gives the equilibrium equation:

$\frac{\mu L}{2} \, \ddot{u} + \frac{\pi^4 EI}{2 L^3} \, u = 0$

and the natural vibration frequency results as :

$\omega_{\rm n} = \left( \frac{\pi}{L} \right)^2 \sqrt{\frac{EI}{\mu}}$

which is the exact solution for the first natural frequency of a simply supported beam.

5. Example: simple pendulum ¶

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6. Example: tower with flexible foundation ¶

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7. Example: nonlinear double pendulum ¶

Define $d\vec{y}/dt$ function:

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def double_pendulum(y, t, m1, h1, m2, h2, g):
    
    th1, w1, th2, w2 = y
    
    mt  = m1  + m2
    dth = th1 - th2
    
    cd  = np.cos(dth)
    sd  = np.sin(dth)
    s1  = np.sin(th1)
    s2  = np.sin(th2)
    
    A   = np.array([[mt*h1,    m2*h2*cd], 
                    [m2*h1*cd, m2*h2  ]])

    B   = np.array([-m2*h2*w2*w2*sd - g*mt*s1,
                     m2*h1*w1*w1*sd - g*m2*s2])
    
    a   = np.linalg.solve(A, B)
    
    return [w1, a[0], w2, a[1]]

Call scipy for differential equation solution:

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t  =  np.linspace(0, 4, 16384)

m1 =  1.0;   h1 = 1.0
m2 =  1.5;   h2 = 0.5

g  =  9.81

Y0 = [np.pi+0.001, 0., np.pi/2, 0.]

plt.figure(1, figsize=(12, 5.5), clear=True)

for k in range(2):
    
    Y  =  odeint(double_pendulum, Y0, t, args=(m1, h1, m2, h2, g))

    x1 =  h1*np.sin(Y[:,0])
    y1 = -h1*np.cos(Y[:,0])

    x2 =  x1 + h2*np.sin(Y[:,1])
    y2 =  y1 - h2*np.cos(Y[:,1])

    plt.subplot(1,2,k+1)
    plt.plot(x1,y1, 'r:')
    plt.plot(x2,y2, 'b', linewidth=0.5)

    plt.xlim(-2, 2);   plt.xlabel('X2 (m)') 
    plt.ylim(-2, 2);   plt.ylabel('Y2 (m)') 
    plt.title('Simulation {0}'.format(k+1))

    plt.grid(True) 
    
    Y0 = [np.pi+0.002, 0, np.pi/2, 0.]   # repeat with butterfly wing flap

8. Assignments ¶

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