slide1 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Lecture 2. A Day of Principles PowerPoint Presentation
Download Presentation
Lecture 2. A Day of Principles

Loading in 2 Seconds...

play fullscreen
1 / 39

Lecture 2. A Day of Principles - PowerPoint PPT Presentation

  • Uploaded on

Lecture 2. A Day of Principles. (with an example that applies ‘ em all at the end). The principle of virtual work. d’Alembert’s principle. Hamilton’s principle. Principle of virtual work says the work done by a virtual displacement from an equilibrium position must be zero .

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'Lecture 2. A Day of Principles' - morgan

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Lecture 2. A Day of Principles

(with an example that applies ‘em all at the end)

The principle of virtual work

d’Alembert’s principle

Hamilton’s principle


Principle of virtual work says

the work done by a virtual displacement from an equilibrium position must be zero.

This can be used in statics to find forces

We need to move this up to dynamics

d’Alembert’s principle says that

can be treated as an

inertial force


Combine these

We eliminated internal forces last time, so the fs are external

We suppose we know them, and since the displacement is tiny

they don’t change during the virtual displacement


The virtual displacements, if unconstrained, can be written

What I’d like to do is gather all of this up into what I’ll call configuration space

The dimension of q is N = 3K


That’s just an example of a configuration space.

We’ll see more as we go along.

The general rule, which we’ll investigate as we go along,

there are (at least) as many qs as there are degrees of freedom.

We’ll take up the matter of constraints — relations among the qs —later

Bottom line: we can write the rs in terms of the qs.


Each r can depend on all the qs.



summation convention does not apply to i in this formula


Let’s do some rearranging

Define the generalized force

This force goes with the kth generalized coordinate


The left hand side looks a lot like a momentum change

dotted into something. If you like manipulations

you’ll LOVEwhat comes next

First a bit of what I call “integration by parts”

This is a useful trick in deriving things



turns into

We still aren’t done. The second term on the right is cool

but we need to play with the first term



turns into

Remember that there is a sum on k here!

I can split Qk into two pieces: the potential and nonpotential parts


The Lagrangian

From here on in Qk will be understood to be the nonpotential generalized forces

We’ll have an algorithm for the calculation later.

If the qk are independent, then we have the Euler-Lagrange equations


What did we do and what did we assume?

Principle of virtual work

d’Alembert’s principle (inertial forces)

Independence of the generalized coordinates

All the rest was clever manipulation

We will spend a lot of time dealing with cases where the qs are not independent. . . . . but not today . . .


Hamilton’s Principle

This is a formalism that leads to the Euler-Lagrange equations

and it will help us when we need to consider constraints

(relations between the dqs).

This is generally posed in terms of the Lagrangian

but that eliminates the generalized forces

which I’d like to include

Let me define

where W denotes the work, potential and nonpotential


We write the action integral

The action integral depends on the path between the two end points.


Hamilton’s principle states

The actual path will be the path that minimizes the action integral.

Suppose that

where the zero denotes the desired path

The first piece is fine; we need to play with the second.


I do my integration by parts trick again

I can integrate the first part. All the paths hit the end points, and the integral is zero.


so I have

and this gives us the Euler-Lagrange equations

No matter how we look at it, we have a governing system


from a governing integral

We get the Euler-Lagrange equations if the dqk are independent

We will have issues regarding independence

and solutions for them.

We will have issues regarding the generalized forces

and we’ll develop techniques for finding them.


We can get the force by differentiating something called

the Rayleigh dissipation function

note double summation

the coefficients are the most general

they will usually be much simpler

The (unconstrained) Euler-Lagrange equations become

where Qk no longer includes the friction forces


I’d like to put all this together in some sort of procedure.

We can look at some mechanical systems that can be viewed as

collections of point masses


OK. Away we go . . .


The Euler-Lagrange process

1. Find T and V as easily as you can

2. Apply geometric constraints to get to N coordinates

3. Assign generalized coordinates

4. Define the Lagrangian


5. Differentiate the Lagrangian with respect to the

derivative of the first generalized coordinate

6. Differentiate that result with respect to time

7. Differentiate the Lagrangian

with respect to the same generalized coordinate

8. Subtract that and set the result equal toQ1

Repeat until you have done all the coordinates



(start without the generalized forces)

y1, f1


Steps 1-4 lead us to



(y2, z2)


The governing equations are then

Put the physical variables back so it looks more familiar

This is all without forcing or damping — let’s add those



add a torque, t1

(with forces)

y1, f1

If y1 changes, f1 does work

Q1 = f1


If q changes, f1 does no work

Q2 = 0

If q changes, t1 does work

Q2 = t1


If y1 changes, t1 does no work

Q1 does not change


(y2, z2)


The governing equations were

We added the generalized forces

Now we need the Rayleigh dissipation function


The damper works when the angle changes, but not when the cart moves

So, the Rayleigh dissipation function for this problem is


We aren’t really up to discussing solving these problems,

but let me say a few things that we will revisit.

a pair of coupled second order ordinary differential equations

It would be nice to have first order equations

There are lots of ways to do this, and we’ll look at many of them

but the simplest is to let


Then we’ll have

If you supply a force, a torque and initial conditions

you can solve this set numerically.


You can solve for the variables, which I won’t do because it is pretty messy,

and you’ll wind up with

Later on we’ll learn to call this a state vector