Advanced Dynamics and Vibrations: Lagrange’s equations applied to dynamic systems
Analytical Mechanics – Lagrange’s Equations
Up to the present we have formulated problems using newton’s laws in which the main disadvantage of this approach is that we must consider individual rigid body components and as a result, we must deal with interaction forces that we really have no interest in. These forces come from constraints which one part of the system puts on another.
An alternate approach referred to as analytical mechanics, considers the system as a whole and formulates the equations of motion starting from scalar quantities – the kinetic and potential energies and an expression of the virtual work associated with non-conservative forces. It relies heavily on the concept of virtual displacements.
Any set of numbers that serve to specify the configuration of the system are examples of generalized coordinates. To these we can add constraints. The number of DOF is then the number of generalized coordinates minus the number of independent equations of constraint.
e.g a point moving on the surface of a sphere of radius R with center must obey:
There are 3 coordinates and 1 equation of constraint. Therefore it has 2 DOF. If we used two angles, say & then we would not need the constraint.
These constraints are classified as:
If the constraint can be written in finite form (as above) they are called holonomic if not (then in differential form) then they are non-holonomic.
If they are equalities then they are bilateral and unilateral if inequalities
If there is no explicit dependence on time then they are scleronomic and if there is explicit time dependence they are rheonomic.
Therefore, 2 DOF holonomic, scleronomic bilateral system.
however if we allow the point of attachment to move (in the x-y plane)
Vertical displacement & Virtual Work
The work done during the motion is
where & are vectors and is a scalar. For a particle where may be the sum of several forces. Therefore,
where this is a true differential of the kinetic energy.
If the particle moves from positive to then
if depends only on ,
where is a scalar function that depends only on the position and not on or time. is term the potential energy.
The total energy is constant.
When we have a conservative system we can use Rayleigh’s to estimate the natural frequencies.
Consider a simple problem and use visual work to solve.
Two bars of length . What is the equilibrium position
Consider a rigid beam (uniform) of mass loaded by a uniform time dependent load and supported by a spring & dashpot as shown
Using determine the equation of motion.
Virtual work of forces applied Virtual work of inertia forces
We can use virtual work to calculate the energy in a multi-degree of freedom system.
For a displacement of only
and similarly for , alone
So for the combination (assuming superposition)
The work done by force in is energy stored due to that force:
If there are n forces and the displacements are (i = 1,2, … n) then the total energy is:
We can also show that the kinetic energy is
& are examples of so-called quadratic forms. Quadratic forms are said to be positive definite if they are never negative and only zero when the variables are zero ()
They are said to be positively semi-definite if they are never negative but can be zero for non-negative variables.
This is useful to know because if we have a stiffness matrix that is positive semi-definite it means there is a rigid body mode with zero natural frequency [m] is positive definite
For example a submarine:
has 2 rigid body modes
Principle of Virtual Work
We will consider a system of particles moving in a 3D space and define the virtual displacements as for particle 1 for particle 2 etc.
These are not true displacements but small variations in the so-called generalized coordinates. These are imagined displacements (slight variations) which are at the same time and which satisfy the constraints given by
Then for virtual changes in these coordinates:
Expanding this in a Taylor series gives:
So that there are arbitrary displacements
On each of these particles the forces acting are
Where the resultant force is composed of the external forces and the constraint forces.
When all the particles are in static equilibrium
where this represents the virtual work performed by the resultant force on the particle.
If we rule out work by the constraint forces ( as the constraint force is perpendicular to the motion), (therefore cannot handle friction).
Consider two particles held by a rigid massless rod.
But because the rod is rigid
Therefore for the constraint forces
and the Principal of Virtual Work says
that the work performed by the applied forces through infinitesimal virtual displacements compatible with the constraints is zero.
To make the virtual work equations more useful we write it in terms of the generalized coordinates
the generalized forces. These are not necessarily forces as they can be moments. It is only necessary that the quantity has the units of work.
For the situation in which we have a conservative system the work can be expressed differently.
as we can select the virtual displacement arbitrarily (i.e. only one non-zero at a time)
This states that has a stationary value at the equilibrium position (can use this for stability consideration). To extend this to dynamical situations we use D’Alembert’s principle.
Where the are the inertia forces. The virtual work for a system of particles is:
and since we showed the virtual work of the interned forces is zero.
This is called the generalized D ‘Alembert’s principle and is really not useful as it is still in vector form. To get it in a scalar form we rewrite the second term in terms of any generalized coordinates and this leads to Lagrange’s equations of motion.
Consider a coordinate transformation where radius vector is written in terms of instead .
Note: that since do not depend explicitly on time that.
now consider a virtual change in
and apply this to the second term of D ‘Alembert’s principle
Consider a typical term in the bracket:
Now interchange the order of differentiation in the last term.
As a result, the total second term becomes:
As is the kinetic energy of the entire system. To complete the conversion, we must rewrite the forces.
And we call:
Some of the forces will be conservative (derivable from a potential and some non-conservative).
So that the rewritten D’Alembert’s principle becomes:
And since the displacements are arbitrary and independent.
However, since the potential energy does not depend on the velocities, this can be written as:
- Does not have to be only for “small” oscillations.
- Does not matter whether the conservative forces are included in the .
- Sometimes the are able to be put into a “potential like” function.
- Can include reaction if wanted using Lagrange multipliers.
- Determine the using virtual work.
Find the equations of motion:
NOTE: Sometimes write the damping forces as a dissipation function.
Find the equations of motion:
The potential energy is:
And the damping force is the same. Note that:
The equation of motion comes from:
For small deflection and :
2 DOF Example
The uniform rod of length and mass carries a slider of mass which is attached by a spring of stiffness . The spring is unstretched at .
Also, note that:
The equilibrium positions are:
We could now study the stability of these configurations.
Example with External Forces
Using , we can say:
Where is the datum. Now we must find the generalized forces. For a virtual displacement only:
For a virtual displacement only:
Then, note that:
Therefore, (1) becomes:
And (2) becomes:
For small angles these become:
If we have support motion instead of forcing,
then we cannot neccesairly use a potential, so now give the system a and . For a :
as no work is done for alone. Thus, is the same, and:
For small motions:
Consider the double pendulum:
Consider only small angles. Thus:
Now try a different set of generalized coordinates for the same problem.
If we consider small angles:
For mode shapes:
These are different values of ratios than in the first case. However, when plotted:
Now use the newtonain approach to the same problem:
Note both mass and stiffness matrices are non-symmetric!
Calculating the eigenvalues gives the same result and the mode shapes are identical with the other 2 formulations.
NOTE: These sets (first and third formulation) are linearly dependent.
Lagrange’s Equation for Small Oscillations
Our main interest is in the motion of MDOF systems in the neighborhood of equilibrium positions. WLOG we assume the equilibrium position is and that the displacements are sufficiently small that the linear force- displacement and force-velocity relations hold. This means that the generalized coordinates and their time derivatives appear in the differential equations to only the first power.
And are constants.
Set as the generalized masses.
Similarly we can write the potential energy as and consider a Taylor expansion of V about the equilibrium configuration.
Where the partial derivatives are all evaluated at .
As the equilibrium point is a condition in which:
Where are the stiffness coefficients. If we now use Lagrange’s equation the equations of motion become:
While this is very straightforward, we can use this result to obtain approximate formulations of the equations of motion as continuous systems it replaces them with a degree of freedom system. It can also extend the technique to find approximate response to a specific forcing function. This is called the assumed modes method (AMM).
For discrete system it replaces them with a smaller number of DOF, while for continuous system it replaces them with an degree of freedom system. We can also easily extend it to a forced system.
Then the method assumes:
Where are trial functions and are the generalized coordinates.
needs only satisfy the geometric boundary conditions.
For continuous systems:
While for discrete systems
The potential energy can be determined is a similar manner.
Consider a generalized beam:
This will allow calculation of a consistent mass and stiffness matrices.
Example: Discrete System
We want to estimate the lowest natural frequency using assumed modes method.
Now calculate the consistent mass and stiffness using this assumed shape:
What if we choose .
Then and are the same.
So we have replace the system by:
Now calculate the ‘s.
The answer is dependent on the assumption of mode shape
EXAMPLE: Use static deflection as mode.
Add masses to the diagram
It is sometimes useful to consider these approximate solutions to finding the eigenvalues and eigenvectors. The UAMM gives upper bounds as do most techniques. We can get lower bounds as well.
Consider again the equations and motion.
To get to an eigenvalue problem we can do it in two ways.
- Premultiply by
- Premultiply by
either gives the same eigenvalues and equivalent eigenvectors.
One can use these to find the eigenvalues/vectors through matrix iteration.
One clever idea that gives a lower bound is Dunkerleys formula. Consider the Dynamic Matrix formulation
and expand the determinant of the coefficients
Now consider that we have the roots of this polynomial . This equation is equivalent to (factored)
and expand this to get
Now equate the coefficients
However the largest term is the first one and often dominates
A special case occurs when the mass matrix is diagonal. Then tr D is
and we need calculate only to
In this case there is a physical interpretation of the expressions
namely that is the natural frequency of the system with only as the mass (all other masses set to zero).
This is also sometimes shown as Dunkerley’s formula.
Consider previous example
- Apply unit load to
Apply unit load to
Apply unit load to
Consider an example of a continuous system
Therefore the eigenvalue problem is
The mode shape is
What if we apply some dynamic loading to the structure? Then we must calculate the generalized forces for our model. Consider a virtual displacement
Consider the previous example
Whirling of Rotating Shafts
Rotating shafts tend to bow out at certain speeds and whirl in a complicated manner. Whirling is defined as the rotation of the bent shaft and the line of centers of the bearing. This phenomenon results from various causes as mass unbalance, hysteresis damping, gyroscopic effects, fluid friction, etc. This whirling can take place in the same or opposite direction as that of the rotation of the shaft. The whirling speed may or may not be equal to the rotation speed.
Let us assume a simple system
Note: is the center of the disc, is the center of gravity, and is an eccentricity.
is the stiffness of the system to lateral deflection.
We will assume the shaft to be rotating at an instant angular velocity and OS to be whirling at angular speed
These are nonlinear equations and there are more than one solution.
Consider the simplest case of synchronous whirl i.e. try
This looks exactly the same as rotating inbalance excitation
So that synchronous whirl occurs when
Self Excited Vibrations
Almost all of the vibrations considered so far were either free or forced vibrations. A fundamentally different class are the so-called self-excited vibrations.
“In a self-excited vibration the alternating force that sustains the motion is created or controlled by the motion itself ; when the motion stops the alternating force disappears”
In a forced vibration the sustaining alternating force exists independently of the motion and persists even when the vibratory motion is stopped.
- Use vibration phenomena for the display of classic ideas that ask the question “Why does that occur”
- Search kinetic art
- Naum Gabo Stationary Wave (1919)
Consider the stability of the SHM of the simple pendulum that is rotating with angular velocity . The total velocity of mass is:
The frequency of oscillation of the pendulum is:
Therefore the pendulum will oscillate () until at which point the pendulum will simply fly horizontally and never cross the .
To see the frequencies try a length of pendulum of inches.
Friction Testing (Kinetic)
Find the natural frequency in terms of
Therefore, frequency of oscillation is
“Friction keeps the world in harmonious motion”
Let , and :
The equilibrium configurations are:
Therefore always stable
therefore, always unstable.
This is easy to do for a SDOF (MDOF), however for general systems it is not as straightforward. For the general case, there are mathematical techniques that are used and referred to by the work of Routh, E.J. done in ‘Rigid Dynamics’ 1897. We will consider it for the case of the ‘1 1/2” and two DOF systems only.
7.2. Mathematical Criterion of Stability
(J.P. Den Hartog, Mechanical Vibrations, 4th Ed. 1956)
For single-degree-of-freedom systems, the criterion of dynamic stability can be derived by physical rather than by mathematical means. With systems of two or three degrees of freedom, a physical conception is always very helpful but usually does not give a complete interpretation of what happens. A mathematical approach is necessary, and this involves at first the setting up of the differential equations of the problem. As long as we deal with small vibrations (and thus disregard any non-linearities that may exist), the equations are all linear and of the second order. Their solution, as usual, is found by assuming the following:
where is a complex number the real part of which determines the damping and the imaginary part of which is the natural frequency. Substituting into the differential equations of the free vibration transforms these equations into a set of n homogeneous, linear algebraic equations in the (complex) unknowns . A process of algebraic elimination is then performed with the result that one equation is obtained which does not contain any of these variables. This equation, known as the “frequency equation,” is generally of the degree 2n in s. Thus, for a two-degree-of-freedom system we obtain a quartic; for a three-degree-of freedom system we obtain a sixth-degree equation, etc.
An algebraic equation of degree 2n in the variable s has 2n roots or 2n values of s. Real roots of s would lead to terms in the solution, which rarely occur in ordinary vibrating systems. The roots of s are usually complex and then they always occur in conjugate pairs:
The solution of the first differential equation is
We know that these terms can be combined by pairs as follows:
so that the imaginary part of s is the frequency, and the real part of determines the rate of damping. If the real parts of all the values of s are negative, the system is dynamically stable; but if the real part of any one of the values of s is positive, the system is dynamically unstable.
Therefore the stability can be determined by an examination of the signs of the real parts of the solutions of the frequency equation. It is not necessary to solve the equation, because certain rules exist by which from an inspection of the coefficients of the equation a conclusion regarding the stability or instability can be drawn. These rules, which were given by Routh in 1877, are rather complicated for frequency equations of higher degree, but for the most practical cases (third and fourth degree) they are sufficiently simple.
Let us consider first the cubic equation
(7.4) which occurs in the case of two degrees of freedom where one mass or spring is zero (in a sense one and one-half degrees of freedom)If its roots are and , can be written
or, worked out,
A comparison shows the following:
One of the three roots of a cubic equation must always be real, and the other two are either real or conjugate complex. Separating the roots into their real and imaginary parts, we may write
Substituted into this leads to the following:
The criterion of stability is that both and be negative. It is seen in the first place that all coefficients , and must be positive, because, if any one of them were negative, the above requires that either or , or both and , must be positive. This requirement can be proved to hold for higher degree equations as well. Hence a frequency equation of any degree with one or more negative coefficients determines an unstable motion.
Granted that the coefficients, and are all positive, the third equation requires that be negative. No information about is available as yet. However, on the boundary between stability and instability, must pass from a positive to a negative value through zero. Make = 0 and we can derive the following:
These relations must be satisfied on the boundary of stability. By eliminating and we find
We do not know yet on which side of this relation stability exists. That can be found in the simplest manner by trying out one particular case. For example let and , which obviously is a stable solution. Substitution gives
The complete criterion for stability of the cubic is that all coefficients A are positive and that
Next consider the quartic
for which the procedure is similar. Since a quartic can be resolved into two quadratic factors, we may write for the roots:
and substitite in, which leads to:
The requirement for stability is that both and be negative. Substitution of negative values of and makes all four A‘s positive, so that the first requirement for stability is that all coefficients A be positive. Granted that this is so, the first equation requires that at least one of the quantities or be negative. Let be negative. We still need another requirement to make also negative. On the boundary between stability and instability, = 0, which substituted in gives:
being four equations in the three variables
Elimination of these variables leads to a relations between the ‘s:
To find out on which side of this equality stability exists, we try out a simple stable case, for example:
Which, on substitution, gives:
So that . Thus, the complete criterion for stability of the quartic (7.10) is that all coefficients are positive and that:
Systems with three degrees of freedom generally have a sextic for their frequency equation and in degenerated cases, a quintic. In such cases, there are three real parts of the roots , and besides the requirements of positive signs for all coefficients A, there are two other requirements, each of which is rather lengthy. For further information, the reader is referred to the original work of Routh.
Investigate the stability of motion of a pendulum of length and mass rotating with constant angular velocity about its vertical axis.
Assume during the steady state rotation about the vertical axis as small angle is given to the pendulum and is the corresponding variation in the angular velocity. Use the Lagrangian formulation to determine the equations of motion.
If there was viscous damping , the equation would be:
Therefore, assuming with :
So that if is negative, one of the roots is positive and the motion is unstable.
Look at the stability of the steady rotation of a horizontal flywheel carrying two masses attached to hub by springs and free to slide along the spokes. If is the moment of inertia of the flywheel, is the angular velocity, and is the upstretched lengths of the springs. Let be the small change in and the small change in .
Using Lagrange’s equations and omitting terms of higher order in the small quantities leaves:
But the equilibrium configuration when . Therefore:
If we put with , we obtain:
For a non-trivial solution:
This equation will have a positive root if: