1. ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems November 16, 2010 6.2, starting 6.3 © Dan Negrut, 2010 ME451, UW-Madison TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA A AA A A A A AA

2. Before we get started… Last Time Tractor example Looked into inertia properties of 2D geometries Center of mass Parallel axis theorem Mass moment of inertia for composite geometries Started discussion of the concept of generalized force Today Wrap up discussion of the concept of generalized force Look into two types of concentrated forces/torques: TSDA/RSDA Formulating the EOM for an entire mechanism (start discussion) Due on Th: HW: Problem 6.2.1 MATLAB assignment: Kinematics analysis of simple mechanism ADAMS: emailed by the TA 2

3. Generalized Forces: How to Deal with Them The fundamental idea : Whenever some new force shows up, figure out the virtual work that it brings into the picture Then account for this “injection” of additional virtual work in the virtual work balance equation: Caveat: Notice that for rigid bodies, the virtual displacements are  r and . Some massaging of the additional virtual work might be needed to bring it into the standard form, that is 3 (Keep this in mind when you solve Problem 6.2.1)

4. [Review of material from last lecture] Concentrated (Point) Force Setup: At a particular point P, you have a point-force F P acting on the body 4 General Strategy: Step A: write the virtual work produced by this force as a results of a virtual displacement of point P Step B: express this additional virtual work in terms of body virtual displacements End with this: Start with this:

5. Concentrated (Point) Force [Review, cntd.] How is virtual work computed? How is the virtual displacement of point P computed? (we already know this…) 5 The step above: expressing the virtual displacement that the force “goes through” in terms of the body virtual displacements  r and  

6. What’s Left 30,000 Feet Perspective Two important issues remain to be addressed: 1) Elaborate on the nature of the “concentrated forces” that we introduced. A closer look at the “concentrated” forces reveals that they could be Forces coming out of translational spring-damper-actuator elements Forces coming out of rotational spring-damper-actuator elements Reaction forces (due to the presence of a constraint, say between body and ground) 2) We only derived the variational form of the equation of motion for the trivial case of *one* rigid body. How do I derive the variational form of the equations of motion for a mechanism with many components (bodies) connected through joints? Just like before, we’ll rely on the principle of virtual work Where are we going with this? By the end of the week we’ll be able to formulate the equations that govern the time evolution of an arbitrary set of rigid bodies interconnected by an arbitrary set of kinematic constraints. Points 1) and 2) above are important pieces of the puzzle. 6

7. Scenario 1: TSDA (Translational-Spring-Damper-Actuator) – pp. 216 Setup: You have a translational spring-damper-actuator acting between point P i on body i, and P j on body j 7 Translational spring, stiffness k Zero stress length (given): l 0 Translational damper, coefficient c Actuator (hydraulic, electric, etc.) – symbol used “h”

8. Scenario 1: TSDA (Cntd.) General Strategy: Step A: write the virtual work produced by this force as a results of a virtual displacement of point P Step B: express additional virtual work in terms of body virtual displacements 8 Force developed by the TSDA element: Alternatively,

9. Scenario 2: RSDA 9 Rotational spring, stiffness k Rotational damper, coefficient c Actuator (hydraulic, electric, etc.) – symbol used “h” Setup: You have a rotational spring-damper-actuator acting between two lines, each line rigidly attached to one of the bodies (dashed lines in figure)

10. Example 2: RSDA (Cntd.) General Strategy: Step A: write the virtual work produced by this force as a results of a virtual displacement of the body Step B: express this additional virtual work in terms of body virtual displacements 10 Torque developed by the TSDA element: Notation: – torque developed by actuator – relative angle between two bodies

11. End: Discussion regarding Generalized Forces Begin: Variational Equations of Motion for a Planar System of Bodies (6.3.1) 11

12. A Vector-Vector Multiplication Trick… Given two vectors a and b, each made up of nb smaller vectors of dimension 3… …the dot product a times b can be expressed as 12

13. Matrix-Vector Approach to EOMs For body i the generalized coordinates are: Variational form of the Equations of Motion (EOM) for body i (matrix notation): 13 Generalized force, contains all external (aka applied) AND internal (aka reaction) forces… Arbitrary virtual displacement Generalized Mass Matrix:

14. EOMs for the Entire System Assume we have nb bodies, and write for each one the variational form of the EOMs 14 Sum them up to get… Use matrix-vector notation… Notation used:

15. A Word on the Expression of the Forces Total force acting on a body is sum of applied [external] and constraint [internal]: 15 IMPORTANT OBSERVATION: We want to get rid of the constraint forces Q C since we do not know them (at least not for now) For this, we need to compromise…

16. Constraint Forces… Constraint Forces Forces that show up in the constraints present in the system: revolute, translational, distance constraint, etc. They are the forces that ensure the satisfaction of the constraint (they are such that the motion stays compatible with the kinematic constraint) KEY OBSERVATION: The net virtual work produced by the constraint forces present in the system as a result of a set of consistent virtual displacements is zero Note that we have to account for the work of *all* reaction forces present in the system Here is what this buys us: 16 …provided  q is a consistent virtual displacement

17. Consistent Virtual Displacements What does it take for a virtual displacement to be “consistent” [with the set of constraints] at a fixed time t * ? Say you have a consistent configuration q, that is, a configuration that satisfies your set of constraints: Ok, so now you want to get a virtual displacement  q such that the configuration q+  q continues to be consistent: Apply now a Taylor series expansion (assume small variations): 17

18. Getting Rid of the Internal Forces: Summary 18 Our Goal: Get rid of the constraint forces Q C since we don’t know them For this, we had to compromise… We gave up our requirement that holds for any arbitrary virtual displacement, and instead requested that the condition holds for any virtual displacement that is consistent with the set of constraints that we have in the system, in which case we can simply get rid of Q C : provided… This is the condition that it takes for a virtual displacement  q to be consistent with the set of constraints NOMENCLATURE: Constrained Variational Equations of Motion

19. Short Detour ~ Lagrange Multiplier Theorem ~ Theorem: 19

20. [Ex. 6.3.3] Example: Lagrange Multipliers First show that any for any x=[x 1 x 2 x 3 ] T, one has that x T b=0 as soon as Ax=0 Show that there is indeed a vector such that A T + b = 0 20

21. Going back to EOMs Variational Form of the EOMs: 21 …as soon as… This is the conditions that it takes for a virtual displacement  q to be consistent with the set of constraints NOMENCLATURE: Constrained Variational Equations of Motion Use this notation and apply the Lagrange Multiplier Theorem introduced two slides ago Lagrange Multiplier Form of the Equations of Motion

22. Summary of the Lagrange form of the Constrained Equations of Motion Equations of Motion: Position Constraint Equations: Velocity Constraint Equations: Acceleration Constraint Equations: 22 The Most Important Slide of ME451

23. Some practical issues… Before getting overjoyed, how do you know that you can actually compute the acceleration and the Lagrange Multipliers? By coupling the Equations of Motion with the Acceleration Constraint Equations, one ends up with this linear system: Assuming that your Jacobian  q is healthy (that is, has full row rank), it can be proved that because the kinetic energy of a system is always positive the coefficient matrix of the linear system above is nonsingular This means that a solution exists, and not only that, but it is also unique 23

24. Putting Things in Perspective… So you can solve the linear system, and retrieve the accelerations are Lagrange Multipliers What are they good for? Since you have the accelerations, you can integrate them twice and find the velocity and position of each part in your system As for the Lagrange Multipliers, in a next lecture we’ll see that they can be used to compute the reaction force in each joint in the system. These are the “internal” forces that keep the system together They are the forces produced by joints that we just eliminated before 24