Chapter I: Lagrange’s Equations Section I.1 Lagrange’s ...Chapter I: Lagrange’s Equations...

Chapter I: Lagrange’s Equations

I.1-1 ME 564 - Spring 2000 cmk

Section I.1 Lagrange’s Equations for Discrete Systems

The first step in vibrational analysis is the development of an appropriate mathematical model. Many decisions need to be made along the way: whether to use a discrete or continuous model; whether to use a 1-D, 2-D or 3-D model; if a discrete model is to be used, the choice of number of degrees of freedom; what method to use in developing the EOM’s. In this section, we will focus on the background of the Lagrangian formulation. This formulation will be used throughout most of this course in developing EOM’s for discrete systems due to its ease of use and the form of the resulting differential EOM’s. We will start with a review of the Newton-Euler formulation (on which we will base our derivation), a review of the concept of work and an introduction of the concepts of generalized displacements and forces. Using these concepts, the Lagrangian formulation will be developed. A discussion and examples of the use of this method follow at the end. It is possible that many of you have already been introduced to the use of Lagrange’s equations in other course work. If so, then this material will provide you with a review of the method and with an opportunity to better understand the concepts involved. If you have not used Lagrange’s equations before, hopefully these notes are written at a level that you can grasp the important concepts. In either case, do not hesitate to ASK questions when they arise. If you can get your questions answered right away, you will be able to come away with a much better understanding of these ideas.



Newton-Euler Equations The most direct way that we have learned in the past to derive the governing equations of motion (EOM’s) for a system of particles/rigid bodies is through use of the Newton-Euler equations. For planar motion of a system of n rigid bodies,

Gi

A

!i

aGi

ith rigid body

these vector equations are for i = 1, 2, …, n:

!

Fi

i

" = miaGi

= mi

drGi

dt (1)

!

MA i

i

" = IAi#

i= I

Ai

d$i

dt (2)

where mi is the mass of the ith body,

!

rGi

is the position of the center of mass of the ith body

!

IAi

is the mass moment of inertia about point A,

!

"i is the angular acceleration for the ith body and point A is

EITHER: • the center of mass of the body, OR • a fixed point for the rigid body.

For particles, equations (1) and (2) are completely equivalent; only (1) is generally used in this case. Remarks

• With the Newton-Euler formulation one must deal with all forces, which are acting on or within the system. This is particularly cumbersome when dealing with a system of interconnected bodies and many of these forces are forces of reaction and constraint, forces with which one is generally not concerned when wanting only to describe motion.

• For a system of n rigid bodies, the Newton-Euler formulation produces a set of 6n scalar equations, a number that generally exceeds the number of degrees of freedom (DOF’s) by a considerable amount. So, in order to solve for the motion of the system with this formulation, one must solve many more equations than the number of motion coordinates sought.

• Although we will still use the Newton-Euler formulation in this course, we will rely on a energy-based formulation (Lagrange’s equations) for deriving the equations of motion for the systems of interest. With this approach, we can simplify the derivation by ignoring forces that do no virtual work on the system eliminating the need to deal with many forces of constraint.



The Concept of Work Recall the definition of the concept of “work” done by a force F along the path of the force from position 1 to position 2:

!

W1"2 = dW

1

2

# = F • dr1

2

#

where dr is a differential vector that is tangent to the path of the point at which F acts. With this definition, it can be observed that any force component that is perpendicular to the path of the point at which it acts will do NO work on the system. Furthermore, it can be shown with the aid of Newton’s Third Law that many forces of reaction will do no work on the system as a whole (although the force may do work on an individual body in the system). With this in mind, it is productive to consider a formulation of the problem based on work in that many forces about which we are not concerned will not appear in the EOM’s. Say we consider for now a system of n particles that are governed by Newton’s second law (equation (1) above):

!

Ri= F

i

i

" = mi

dri

dt= m

i˙ r

i (1a)

where Ri is the resultant force acting on the ith particle. Finding the total (differential) work done by all forces acting on the system, and using (1a) gives:

!

dW = dWi

i

" = Ri• dr

i

i

" = mi˙ r

i• dr

i

i

" (3)

Note that at this point we have eliminated the presence of all forces that do no net work on our system. It is important to note that we have also reduced a set of n vector equations (3n scalar equations) down to a single scalar equation. That is, for a system with N DOF’s we have only one governing EOM. This is NOT progress!! But let us continue…

F

dr

path of P

P



Generalized Coordinates For a holonomic system with N DOF’s, we can chose a set of N “generalized coordinates” qj(t) ; j = 1, 2, …, N which are independent and, along with time, completely describe the configuration of the system. Using the “chain rule of differentials” we can write:

!

dri ="ri"q j

dq j

j=1

N

# +"ri"t

dt (4)

or, alternately, as the chain rule of differentiation:

!

˙ r i =dri

dt=

"ri

"q j

˙ q jj=1

N

# +"ri

"t (4a)

Note: Recall that a partial derivative only sees an explicit appearance of the variable with

which the differentiation is done. Specifically, the partial derivatives with respect to time in (4) and (4a) appear only if the position vector is explicitly dependent of time. An example of this is when there are constraints on the motion of the ith particle that are prescribed as a function of time. Otherwise, the last term in (4) and (4a) will not appear. Systems for which the position vectors are NOT an explicit function of time are called “scleronomic”. Those that do have an explicit appearance of time in the position vectors are known as “rheonomic” systems. We will see examples of both types of systems in examples contained in this section.

Substitution of (4) into (3) gives:

!

Ri •"ri

"q j

dq j

j=1

N

# +"ri

"tdt

$

% & &

'

( ) ) i=1

n

# = mi˙ r i •

"ri

"q j

dq j

j=1

N

# +"ri

"tdt

$

% & &

'

( ) ) i=1

n

# (5)

From (5), we can see that some of the work done on the system (the left-hand-side of the equation) comes from a possible explicit time dependence of

!

ri (note that

!

ri has an explicit time dependence

when

!

"ri/"t # 0 ).



Virtual Displacements, Virtual Work and Generalized Forces Here is where Lagrange showed great insight on simplifying the motion equations of a system. He chose to define “work” in a different fashion. In his definition, he defined a vector quantity called a “virtual displacement”, dri , as the differential displacement of (4) ignoring any explicit dependence of ri on time. That is, he ignored the ∂rj /∂t term in equation (4). With this, we see that the virtual displacement of the ith particle is:

!ri = !ri

!qj"j = 1

N

!qj (6)

Along with this definition comes a definition of “virtual work”:

!W = Ri • !ri!i = 1

n

= m i ri • !ri!i = 1

n

(7)

Substituting in (6) gives:

Ri • !ri

!qj"i = 1

n

!qj"j = 1

N

= m i ri • !ri

!qj"i = 1

n

!qj"j = 1

N

or

m i ri • !ri

!qj"i = 1

n

- Qj !qj"j = 1

N

= 0 (8)

where

Qj = Ri • !ri

!qj"i = 1

n

= "generalized force" for the jth gen. coord. (9)

Since we have supposed that the generalized coordinates qj are independent1, equation (8) says that each coefficient of dqj in (8) must be zero; that is,

m i ri • !ri

!qj"i = 1

n

= Qj ; j = 1, 2, …, N (10)

Equation (10) is often referred to as d’Alembert’s Principle.

1 Recall that if functions

f j t( ) are independent, then a linear combination of these functions being

equal to zero:

aj

f j t( )j=1

N! = 0

implies that EACH of the coefficients a

j individually must be zero.



Remarks: i) In words, d’Alembert’s Principle says that for each generalized coordinate qj , we have a

dynamics equation for which the corresponding generalized force Qj is balanced by an effective

“inertia-type” force of

mi ˙ r i •!ri

!qji=1

n

" .

ii) Note that, although this set of equations has been derived directly from vector form of Newton’s Second Law, d’Alembert’s principle is a scalar representation of the dynamics. Importantly:

• for each particle, we have a vector statement of Newton’s Second Law. • for each generalized coordinate, we have a scalar statement of d’Alembert’s Principle.

iii) Forces that do NO virtual work do not appear in d’Alembert’s Principle. In contrast, ALL forces

must be included in Newton’s Second Law, regardless of whether they do work or not. iv) For scleronomic systems (those systems for which no time-dependent constraints exist) there is no

distinction between the differential displacements dri , and virtual displacements dri . For these systems, no distinction exists between differential work and virtual work.

v) In this course, we will deal with a number of systems that are rheonomic (those having time-dependent constraints). For example, spring-mass systems with prescribed motion for supports and rotating systems with prescribed rotation rates are rheonomic. For these systems the time-dependent constraints will introduce “inertial” and “gyroscopic” forces into our equations of motion, as we will see in a later section of the notes.

vi) Generally forces that do virtual work on a system are a subset of those forces that do differential work on the system. By using virtual work as the basis for d’Alembert’s Principle, we potentially have eliminated from the system equations of motion some forces that do work.



Lagrange’s Equations for Particles We will now convert the LHS of (10) to a form for which the scalar nature of the terms is more apparent. To do so, observe that

•

!

" ˙ r i

" ˙ q j="ri

"q j

(11)

To see this, take a partial derivative of equation (4a) above with respect to qj:

!

" ˙ r i

" ˙ q j=

"ri

"qk

" ˙ q k

" ˙ q jk=1

N

# +"

" ˙ q j

"ri

"t="ri

"q j

+ 0 $" ˙ r i

" ˙ q j="ri

"q j

• d dt (

∂ri∂qj

) = ∂ri∂qj

(order of partial and full derivatives are interchangeable) (12)

• r • ∂ri∂qj

= d dt (ri •

∂ri∂qj

) - ri • d dt (

∂ri∂qj

) ; using product rule

= d dt (ri •

∂ri

∂qj) - ri •

∂ri∂qj

; using (11) and (12)

= d dt {

∂ ∂qj

[ 12 (ri • ri) ] } - ∂ ∂qj [ 12 (ri • ri) ] ; using product rule (13)

Substituting (13) into (10):

d dt

!T

!qj -

!T

!qj = Qj (14)

where

T = 12

m i ri • ri!i = 1

n

= 12

m i vi2!

i = 1

n

= kinetic energy for system

The set of equations in (14) are referred to as Lagrange’s equations for a system of n particles whose motion is described by a set of N generalized coordinates. Note there is only a difference in “form” between the d’Alembert and Lagrange equations; all remarks previously made about the d’Alembert equations (10) also apply to Lagrange’s equations (14).



Lagrange’s Equations for Rigid Bodies - Planar Motion We can easily extend our Lagrangian development for a system of particles for the inclusion of rigid bodies. Here, for expediency sake, we will limit our development to include rigid bodies undergoing planar motion, motion for which the following Newton-Euler equations are valid:

Gi

Gi

Ri

MGi

!

Fi

i

" = miaGi

= mi

drGi

dt (1)

!

MA i

i

" = IAi

d#i

dt (2)

where Ai is either a fixed point or center of mass for the rigid body. In the following derivation, we will use Ai as the center of mass Gi . We will dot (1) with the virtual displacement δrGi ,

R

i•!r

Gi= m i

˙ r Gi•!r

Gi (3)

dot (2) with the virtual rotation δθι,

M

Gi•!"i = IGi

˙ " i •!" i (4) sum over the number of n particles and add the results:

Ri

•!rGi

+ MGi

•!"i[ ]i =1

n

# = mi˙ r

Gi•!r

Gi+ IGi

˙ " i •!"i[ ]i=1

n

# (5)

Substituting the following expressions for virtual displacements and rotations:

! rGi

="r

Gi

"qj

!qjj=1

N

#

!$i=

"$i

"qj

!qjj=1

N

#

into (5) gives:



Qj !qjj=1

N

" = mi ˙ r Gi

•#r

Gi

#qj

+ IGi˙ $ i •

#$i

#qj

%

& '

(

) *

i= 1

n

"+ , -

. / 0 j=1

N

" !qj (6)

where

Q j = Ri•!r

Gi

!qj

+ MGi

•!"

i

!qj

#

$ %

&

' (

i=1

n

) (7)

Using the same procedure as was used for the formulation for particles, equation (6) can be rewritten as:

d

dt

!T

! ˙ q j

"

# $

%

& ' (

!T

!qj

( Qj

) * +

, - . j=1

N

/ 0qj = 0 (8)

where

T =1

2mi

˙ r Gi

• ˙ r Gi

+1

2IGi

˙ ! i • ˙ ! i" #

$ %

i=1

n

& (9)

Again, if a set of independent generalized coordinates qj (j = 1, 2, …, N) are used to describe the motion of the system, equation (8) becomes:

d

dt

!T

! ˙ q j

"

# $

%

& ' (

!T

!qj

= Q j ; j = 1, 2, …, N (10)

Remarks: i) Note that the form of Lagrange’s equations for rigid bodies is the same as for particles;

however, the expression for the kinetic energy for a rigid body includes both translation and rotation, as seen in (9) above.

ii) Although we assumed that Ai was the center of mass Gi in arriving at (9), we can actually use Ai to be either the center of mass or a fixed point on the rigid body in writing down the kinetic energy expression for a rigid body.


I.1-10 ME 564 - Spring 2000 cmk

Alternate Form of Lagrange’s Equations Some forces contained within the generalized forces Qj may be conservative. Conservative forces are those for which the work done by these forces is independent of path. A consequence of this is that the work done by these forces can be expressed in terms of a potential, V. That is, we can write the virtual work done on the system as:

δW = δW(nc) + δW(c) = δW(nc) - δV If this potential V is a function of the q’s (and not the q ’s), we can write (recall the previous definition of the virtual differential operation in equation (6):

!V ="V

"qjj=1

N

# !qj

In addition we can define a so-called Rayleigh dissipation function Ri for each viscous dashpot in the system:

Ri =1

2ci

˙ ! i2

where ci and ∆i are the damping constant and extension/compression, respectively, of the dashpot. If “r” dashpots are included in the system, the total Rayleigh dissipation function for the system is:

R = Ri

i =1

r

! =1

2ci

˙ " i2

i= 1

r

!

It can be shown that the ith generalized force that originates from the dashpots in the system can be found from:

Q jnc( )

damping= !

"R

" ˙ q j

With the above, we can alternately write (14) as:

d

dt

!T

! ˙ q j

"

# $

%

& ' (

!T

!qj

+!U

!qj

+!R

! ˙ q j= Q j

nc( ) (14a)


I.1-11 ME 564 - Spring 2000 cmk

Summary and Discussion of Lagrange’s Equations The equations of motion (EOM’s) derived in (14) are completely equivalent to a set of EOM’s which would be derived through the use of the Newton-Euler equations when using the same generalized coordinates. The “appearance” of the EOM’s may be different, particularly when the system includes forces of constraint that do no work. However, the elimination of these constraint forces from the Newton-Euler formulation will produce the same equations as Lagrange. One may ask a question as to why introduce a new method when it gives the same EOM’s as the more fundamental approach. In answering this, let’s consider some of the advantages of each approach. Some advantages of the Lagrangian formulation:

(a) As can be seen from equation (5), forces doing no work on the system as a whole will NOT remain in the Lagrangian formulation. In addition, those forces doing work but not virtual work will also be absent from the equations, as seen in equation (7). (This distinction is particular to rheonomic systems, that is, systems involving time-dependent constraints.)

(b) The Lagrangian formulation produces the minimum possible EOM’s needed to describe the motion of the system. Since the Newton-Euler formulation will involve all forces (including reaction forces) acting on the system, more equations are needed in the dynamic formulation.

(c) The kinematics involved in the Lagrangian formulation are generally simpler than that of the Newton-Euler approach since more effort is spent on deriving acceleration relations than on velocity relations.

(d) Sign conventions on “spring forces” and “spring torques” are less cumbersome in forming potential energy than in setting up force balances. Along a similar line of thought, sign conventions are less a source of error in setting up kinetic energy expressions than in setting up accelerations. However, one still does need to be concerning with sign conventions at various steps in Lagrange, so one should not get too sloppy!

(e) As we will see in the next section, the Lagrangian formulation of the EOM’s has a more readily observed structure than the Newton-Euler formulation. Such a structure is helpful in being able to detect errors in the derivation of the EOM’s.


I.1-12 ME 564 - Spring 2000 cmk

Some advantages of the Newton-Euler formulation:

(a) With the Lagrangian formulation, one needs to have a priori knowledge of the number of degrees of freedom for the system of interest. That is, one needs to know exactly the number of generalized coordinates to be used before applying Lagrange’s equations. Such knowledge is not needed before using the Newton-Euler formulation.

(b) One must have a set of independent generalized coordinates before applying Lagrange’s equations, as seen in going from equation (8) to (10) above (this is really the same issue as in (a) above). For some systems (known as “non-holonomic” systems), this is NOT possible. There is a Lagrangian formulation available for use on non-holonomic systems (one which involves so-called Lagrange multipliers), but it tends to be considerably more tedious than that for holonomic systems. Such a distinction between non-holonomic and holonomic is not necessary with the Newton-Euler formulation.

(c) The inclusion of friction involves considerable complications with the Lagrangian formulation. This is true since quantifying friction forces relies on being able to quantify “normal” reactions forces. Since the normal forces themselves do not do virtual work, they will not appear in the EOM’s. Here one needs to use the non-holonomic formulation of Lagrange’s equations (mentioned above in (b)) to have the normal forces appear. Inclusion of friction forces in any formulation can get tricky, so this is not necessarily an advantage for Newton-Euler.

(d) Physical insight can be lost in using the Lagrangian formulation, whereas it may be more apparent in the Newton-Euler formulation. For example, terms arising from the centripetal and Coriolis components of acceleration are not as readily apparent in the Lagrangian formulation as with Newton-Euler.


I.1-13 ME 564 - Spring 2000 cmk

Example I.1 Use Lagrange’s equations to derive the equations of motion (EOM’s) for the two-DOF system shown below

a) using the absolute coordinates x1 and x2 shown. b) using the relative coordinates z1 = x1 and z2 = x2 - x1.

k

x1

c

m

f(t)

k m

x2


I.1-14 ME 564 - Spring 2000 cmk

Example I.2 A horizontal force F acts at point B of a double pendulum hanging in a vertical plane. The top link is known to be rotating with a constant rate of Ω. Find the

(a) total differential work done on the system by the force F. (b) total virtual work done on the system by the force F. (c) generalized forces using θ1 and θ2 as generalized coordinates. (d) generalized forces using φ1 and φ2 as generalized coordinates where

!

"1

= #1 and

!

"2

= #2$#

1.

L

L

m

mF

g

!

"1

"2


I.1-15 ME 564 - Spring 2000 cmk

Example I.3 Find the equations of motion of the problem in Example I.2 using

(a) Lagrange’s equations and the absolute angles θ1 and θ2 as generalized coordinates. (b) the Newton-Euler formulation.

L

L

m

mF

g

!

"1

"2


I.1-16 ME 564 - Spring 2000 cmk

Example I.4

Consider the three-DOF spring-mass-dashpot system shown below. Using Lagrange’s equations, find the EOM’s of the system using the absolute coordinates x1, x2 and x3.

x1

F

c

k

k k r

m

m

m

1 2 3

x2 x3

no slip


I.1-17 ME 564 - Spring 2000 cmk

Example I.5

A two-particle system is acted upon by a force F which always acts along the line of the massless link. All springs are unstretched when x = θ = 0. Find the EOM’s in terms of the generalized coordinates x and θ. Assume that the system moves in a horizontal plane (i.e., ignore gravity).

k

c

F

x

!M

m

L

K

Date post:	05-Feb-2020
Category:	Documents
Upload:	others
View:	5 times
Download:	0 times

Chapter I: Lagrange’s Equations Section I.1 Lagrange’s ...Chapter I: Lagrange’s Equations...

Documents