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Lecture 2. A Day of Principles The principle of virtual work d’Alembert’s principle Hamilton’s principle 1 (with an example that applies ‘em all at the.

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Presentation on theme: "Lecture 2. A Day of Principles The principle of virtual work d’Alembert’s principle Hamilton’s principle 1 (with an example that applies ‘em all at the."— Presentation transcript:

1 Lecture 2. A Day of Principles The principle of virtual work d’Alembert’s principle Hamilton’s principle 1 (with an example that applies ‘em all at the end)

2 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 thatcan be treated as an inertial force 2

3 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 3

4 4 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

5 5 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.

6 6 m1m1 m2m2 y1y1 y2y2 k1k1 k2k2 k3k3 Simple example of possible qs

7 7 Each r can depend on all the qs. repeated indices summation convention does not apply to i in this formula

8 8 Let’s do some rearranging Define the generalized force This force goes with the k th generalized coordinate

9 9 The left hand side looks a lot like a momentum change dotted into something. If you like manipulations you’ll LOVE what comes next First a bit of what I call “integration by parts” This is a useful trick in deriving things

10 10 is not a function of

11 11 so turns into We still aren’t done. The second term on the right is cool but we need to play with the first term

12 12 so sort of a trick

13 13 turns into Remember that there is a sum on k here! I can split Q k into two pieces: the potential and nonpotential parts so

14 14 For our purposes

15 15 The Lagrangian If the q k are independent, then we have the Euler-Lagrange equations From here on in Q k will be understood to be the nonpotential generalized forces We’ll have an algorithm for the calculation later.

16 16 What did we do and what did we assume? Principle of virtual work d’Alembert’s principle (inertial forces) Independence of the generalized coordinates We will spend a lot of time dealing with cases where the qs are not independent..... but not today... All the rest was clever manipulation

17 17 ??

18 18 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  qs). 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

19 19 We write the action integral The action integral depends on the path between the two end points.

20 20 Hamilton’s principle states The actual path will be the path that minimizes the action integral. Suppose thatwhere the zero denotes the desired path The first piece is fine; we need to play with the second.

21 21 I can integrate the first part. All the paths hit the end points, and the integral is zero. I do my integration by parts trick again

22 22 so I have and this gives us the Euler-Lagrange equations No matter how we look at it, we have a governing system

23 23 from a governing integral We get the Euler-Lagrange equations if the  q k 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.

24 24 ??

25 25 A special kind of friction: “viscous” friction, damper/dashpot y damping constant

26 26 We can get the force by differentiating something called the Rayleigh dissipation function The (unconstrained) Euler-Lagrange equations become where Q k no longer includes the friction forces note double summation the coefficients are the most general they will usually be much simpler

27 27 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...

28 28 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

29 29 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 to Q 1 Repeat until you have done all the coordinates

30 30 M y 1, f 1 OVERHEAD CRANE (y 2, z 2 ) m  Steps 1-4 lead us to (start without the generalized forces)

31 31 For q 1

32 32 For q 2

33 33 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

34 34 M y 1, f 1 OVERHEAD CRANE (y 2, z 2 ) m  If y 1 changes, f 1 does work Q 1 = f 1 If  changes,  1 does work Q 2 =  1 add a torque,  1 If y 1 changes,  1 does no work Q 1 does not change If  changes, f 1 does no work Q 2 = 0 (with forces)

35 35 The governing equations were We added the generalized forces Now we need the Rayleigh dissipation function

36 36 The damper works when the angle changes, but not when the cart moves So, the Rayleigh dissipation function for this problem is

37 37 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

38 38 Then we’ll have If you supply a force, a torque and initial conditions you can solve this set numerically.

39 39 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

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