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ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems November 17, 2011 Finish 6.3 © Dan Negrut, 2011 ME451, UW-Madison “To improve is to change; to be perfect is to change often.” Winston Churchill Before we get started… Last Time Looked at the contribution to the EOM of an external force or torque that is applied at a point P (how to get a generalized force QA) Started discussion about how to get the EOM for a set of bodies (a mechanism) Today: Wrap up EOM discussion after we introduce the Lagrange Multiplier Theorem Example: Pendulum, setting a set of ICs HW due Tuesday, Nov 29: Example: EOM for a slider-crank mechanism Discuss why and how to specify Initial Conditions (ICs) Discussions of TSDA & RSDA 6.1.1 thru 6.1.4, and 6.2.1 MATLAB component will be emailed to you NOTE: Syllabus has been updated, we’ll discuss on Tu 2 Getting Rid of the Internal Forces: Summary Our Goal: Get rid of the constraint forces QC since we don’t know them For this, we had to compromise… We gave up our requirement that holds for any arbitrary virtual displacement Instead, by only selecting virtual displacements ±q that are consistent with the set of constraints present in the system, we saw that we can get rid of QC : …whenever… NOMENCLATURE: Constrained Variational Equations of Motion This is the condition that it takes for a virtual displacement q to be consistent with the set of constraints 3 Short Detour ~ Lagrange Multiplier Theorem ~ Theorem: 4 [Ex. 6.3.3] Example: Lagrange Multipliers First show that any for any x=[x1 x2 x3]T, one has that xTb=0 as soon as Ax=0 Show that there is indeed a vector such that AT + b = 0 5 Going back to EOMs Variational Form of the EOMs: …whenever… NOMENCLATURE: Constrained Variational Equations of Motion This is the conditions that it takes for a virtual displacement q to be consistent with the set of constraints Use this notation and apply the Lagrange Multiplier Theorem introduced two slides ago Lagrange Multiplier Form of the Equations of Motion } 6 Summary of the Lagrange form of the Constrained Equations of Motion Equations of Motion: Position Constraint Equations: The Most Important Slide of ME451 Velocity Constraint Equations: Acceleration Constraint Equations: 7 Some Practical Issues… One relevant question: 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 8 Putting Things in Perspective… Suppose you 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 9 Example 6.3.5: Slider Crank Mechanism Derive the equations of motion (EOM) for the slider crank model in the figure 10 [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? A note on notation t0 or t=0 is the same in ME451 Picture shows the slopes of the unknown function everywhere in plane. You simply have to start somewhere and follow the direction of the slope 11 ODE: Infinite Number of Solutions 12 0.1t ODE Problem: y 0.1y 100e Initial Condition: y0 =[-1000:50:1000] The Need for Initial Conditions (ICs) Suppose that 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: 13 simEngine2D Comment 14 Specifying Initial Conditions (ICs) So you have m equations that must be satisfied by n generalized coordinates. You have m constraints (kinematic and driving, that is) 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… 15 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 IC(q,t0) that will implicitly determine the configuration of the mechanism at time t0 Only at the beginning of the simulation, you specify more IC constraints to determine configuration of mechanism 16 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 17 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) 18 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) 19 ICs: Concluding Remarks This IC issue is actually simple if you remember what we did back when we dealt with the Kinematics problem You only do this at time t=0 (provide a starting configuration for your mechanism). Otherwise, your dynamics problem is underdefined You take care of this IC issue by Specifying in the adm file a good (consistent) initial configuration for your mechanism This is what you should do for simEngine2D; i.e., have consistent guesses for positions and velocities Adding some extra constraints and relying on the Kinematic solver to get the good (consistent) configuration at time t=0 This is what a general purpose solver might do, such as ADAMS 20 Example 6.3.6: Simple Pendulum How do you go about specifying ICs? I’d like the pendulum to start from a vertical configuration (hanging down), and angular velocity to be 2 rad/s. Assume that L=0.2 m Pin joint at point P (between pendulum and ground) 21 End ICs Beginning Constraint Reaction Forces (6.6) 22 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 23 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 24 The Important Observation 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 25 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 6.6.1 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 26 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: 27 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 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 Example: the revolute joint brings along a set of two kinematic constraints and therefore there will be two Lagrange multipliers associated with this joint 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 28 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 29 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 30 End Constraint Reaction Forces 31