1. ME451 Kinematics and Dynamics of Machine Systems
Dynamics of Planar Systems
Thursday, November 18, 2010
Initial Conditions
© Dan Negrut, 2010 ME451, UW-Madison
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2. Before we get started… [1]
Last Time
Derived the equations of motion (EOM) for any mechanical system of rigid bodies undergoing 2D motion
The approach is boiler plate: get the generalized mass M, the constraints  present in the model, and the generalized forces QA
Based on these quantities you can write the constrained equations of motion, which constitute a set of differential and algebraic equations
Last lecture contained the most important slide of ME451
Today:
Example: formulating the equations of motion (EOM) for slider-crank
Discuss why and how to specify Initial Conditions (ICs)
HW due Tuesday, Nov 30:
6.3.3, MATLAB component
NOTE: Problem references in its text Prob That’s a typo, it should be (page 116). The dimensions are in meters (in “m”).
MATLAB component will be posted online 2

3. Summary of the Lagrange form of the Constrained Equations of Motion
Position Constraint Equations:
Velocity Constraint Equations:
Acceleration Constraint Equations:
The Most Important Slide of ME451 3

4. Example 6.3.5: Slider Crank Mechanism
Derive the equations of motion (EOM) for the slider crank model in the figure 4

5. [New Topic] The Need for Initial Conditions (ICs)
When dealing with a differential equation, one needs a set of initial conditions to uniquely find the solution of the “problem”
In layman’s words, the “problem” here can be formulated as follows:
I give you at each point in plane the direction (slope) of an unknown function
Can you find me the unknown function if additionally I also give you the value of this function at time t=0?
Picture shows the slopes of the unknown function everywhere in plane. You simply have to start somewhere and follow the direction of the slope 5

6. ODE: Infinite Number of Solutions
ODE Problem:
Initial Condition: y0 =[-1000:50:1000] 6

7. The Need for Initial Conditions (ICs)
You just computed the accelerations, and are ready to apply some numerical formula to integrate the acceleration
What you need though is a set of initial conditions (just like in ME340, if you recall)
How many ICs can/should you specify at time t0?
Recall that you have a set of n generalized coordinates, but they are not totally arbitrary, in that they need to satisfy the set of m constraints present in the system: 7

8. simEngine2D Comment 8

9. Specifying Initial Conditions (ICs)
So you have m equations that must be satisfied by n generalized coordinates.
You need to know the initial configuration of the mechanism (positions and velocities) to be able to start the numerical integration procedure
What does it take to uniquely define the initial configuration of the mechanism?
You can specify an additional set of ndof conditions that your generalized coordinates must satisfy
Recall how we chose the driving constraints…
You have m constraints (kinematic and driving, that is) 9

10. Specifying Position ICs
Important: since you have ndof generalized coordinates that you can choose, it will be up to you to decide the configuration in which the mechanism starts
For a simple pendulum: should it start in a vertical position or in a horizontal position?
You decide based on the problem you solve…
How do you exactly specify the conditions
You do exactly what you did for the kinematic analysis of a mechanism: you had some excess DOFs and prescribed motions (drivers) D(q,t) to “occupy” them
For IC in dynamics you’ll specify ndof initial conditions that will implicitly determine the configuration of the mechanism
Only at the beginning of the simulation, you specify more constraints to determine configuration of mechanism 10

11. Specifying Position ICs
How do you know whether the set of conditions you specified to determine the initial conditions are “healthy”?
Healthy is going to be that IC(q,t0) that leads to a nonsingular constraint Jacobian:
If this is the case, one can apply Newton’s method to solve the system of nonlinear equations below to get the initial configuration of the mechanism q0 11

12. Specifying Velocity ICs
Specifying a set of *position* ICs is not enough
We are dealing with a second order differential problem: We need two sets of ICs
Position ICs (we’ve just done this)
Velocity ICs
Specifying Velocity ICs:
You can play it safe and take the same generalized coordinates that you assigned initial positions and also specify for them some initial speeds (done most often)
You can decide to use a completely new set of generalized coordinates for which you prescribe the value of the velocity at time t0 (strange, but not inconceivable) 12

13. Specifying Velocity ICs
Playing it safe, you find the velocities at time t0 as the solution of the linear system
If you want to be fancy, you replace the matrix qIC with a different matrix D, chosen such that the coefficient matrix continues to be nonsingular (not going to elaborate on this) 13

14. ICs: Concluding Remarks
This IC issue is actually simple if you remember what we did back when we dealt with the Kinematics problem
Note that there is no “dynamics” specifics here (although the need for ICs is rooted in dynamics analysis)
We are in fact using concepts associated with the first part of the course, namely “kinematics” 14

15. Example 6.3.6: Simple Pendulum
How do you go about specifying ICs?
I’d like the pendulum to start from a vertical configuration, and angular velocity to be 2 rad/s.
Pin joint at point P (between pendulum and ground) 15

16. End ICs Beginning Constraint Reaction Forces (6.6)16

17. Reaction Forces: The Framework
Remember that we jumped through some hoops to get rid of the reaction forces that would show up in joints
I’d like to go back and recover them, since they are important
Durability analysis
Stress/Strain analysis
Selecting bearings in a mechanism
Etc.
It turns out that the key ingredient needed to compute the reaction forces in all joints is the set of Lagrange multipliers  17

18. Reaction Forces: The Basic Idea
Recall the partitioning of the total force acting on our mechanical system
Applying a variational approach (principle of virtual work) we ended up with this equation of motion
After jumping through hoops, we ended up with this:
It’s easy to see that 18

19. The Important Observation
Actually, you don’t care for the “generalized” QC flavor of the reaction force, but rather you want the actual force represented in the Cartesian global reference frame
You’d like to have Fx, Fy, and a torque T that is due to the constraint
You report these quantities as they would act at a point P
The strategy:
Look for a force (the classical, non-generalized flavor), that when acting on the body would lead to a generalized force equal to QC 19

20. The Nuts and Bolts
There is a joint acting between Pi and Pj and we are after finding the reaction forces/torques Fi and Ti, as well as Fj and Tj
Figure is similar to Figure out of the textbook
Textbook covers topic well (pp. 234), I’m only modifying one thing:
The book expresses the reaction force/torque Fi and Ti in a body-fixed reference frame attached at point Pi
I didn’t see a good reason to do it that way
Instead, start by deriving in global reference frame OXY and then multiply by AT to take it to the body-fixed reference frame 20

21. The Main Result (Expression of reaction force/torque in a joint)
Suppose that two bodies i and j are connected by a joint, and that the equation that describes that joint, which depends on the position and orientation of the two bodies, is
Suppose that the Lagrange multiplier associated with this joint is
Then, the presence of this joint in the mechanism will lead at point P on body i to the presence of the following reaction force and torque: 21

22. Comments (Expression of reaction force/torque in a joint)
Note that there is a Lagrange multiplier associated with each constraint equation
Number of Lagrange multipliers in mechanism is equal to number of constraints
Each Lagrange multiplier produces (leads to) a reaction force/torque combo
Therefore, to each constraint equation corresponds a reaction force/torque combo that throughout the time evolution of the mechanism “enforces” the satisfaction of the constraint that it is associated with
Example: the revolute joint brings along a set of two kinematic constraints and therefore there will be two Lagrange multipliers associated with this joint
Since each constraint equation acts between two bodies i and j, there will also be a Fj/Tj combo associated with each constraint, acting on body j
According to Newton’s third law, they oppose Fi and Ti, respectively
Note that you apply the same approach when you are dealing with driving constraints (instead of kinematic constraints)
You will get the force and/or torque required to impose that driving constraint 22

23. Reaction Forces ~ Remember This ~
As soon as you have a joint (constraint), you have a Lagrange multiplier 
As soon as you have a Lagrange multiplier you have a reaction force/torque:
The expression of  for all the usual joints is known, so a boiler plate approach renders the value of the reaction force in all these joints
Just in case you want another form for the torque T above, note that 23

24. Example 6.6.1: Reaction force in Revolute Joint of a Simple Pendulum
Pendulum driven by motion:
1) Find the reaction force in the revolute joint that connects pendulum to ground at point O
2) Express the reaction force in the reference frame 24

25. End Constraint Reaction Forces25