ME451 Kinematics and Dynamics of Machine Systems

ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems April 7, 2009 6.3 Quote of the day: We learn from history that we learn nothing from history. Bernard Shaw © Dan Negrut, 2009ME451, UW-Madison

Before we get started… • Last Time • Discussed how to determine the generalized force acting on a body • This quantity is key in determining the value of the acceleration: ma=F • Once you have the acceleration, you integrate twice and get the velocity and position of each body • Today • Derive the EOM (Equations of Motion) for a collection of articulated bodies • Keep in mind that up to this point we only looked at the EOM for *one* body • Next Time: report to 1235ME for ADAMS tutorial, time is 11:15 am 2

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 3

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): Arbitrary virtual displacement Generalized force, contains all external (aka applied) AND internal (aka reaction) forces… Generalized Mass Matrix: 4

EOMs for the Entire System • Assume we have nb bodies, and write for each one the variational form of the EOMs Sum them up to get… Use matrix-vector notation… Notation used: 5

A Word on the Expression of the Forces • Total force acting on a body is sum of applied [external] and constraint [internal]: • IMPORTANT OBSERVATION: We want to get rid of the constraint forces QC since we do not know them (at least not for now) • For this, we need to compromise… 6

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: …providedqis a consistent virtual displacement 7

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): 8

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, 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 QC : 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 9

Short Detour ~ Lagrange Multiplier Theorem ~ • Theorem: 10

[AO (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 11

Going back to EOMs • Variational Form of the EOMs: provided… 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 12

Summary of the Lagrange form of the Constrained Equations of Motion • Equations of Motion: • Position Constraint Equations: • Velocity Constraint Equations: • Acceleration Constraint Equations: The Most Important Slide of ME451 13

Some practical issues… • Before getting overexcited, 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 14

What’s the big deal? • 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 15

[AO]Example 6.3.5: Slider Crank Mechanism • Derive the equations of motion (EOM) for the slider crank model in the figure 16

End Deriving EOMBeginning Initial Conditions (ICs) 17

ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems April 14, 2009 Initial Conditions - 6.3.4 Constraint Reaction Forces – 6.6

Before we get started… • No class on Th (Thanksgiving Break) • Next Tu we’ll have an Engineer from Oshkosh Truck making a presentation in second part of lecture • The syllabus was updated at Learn@UW, please take a quick look • 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: the most important slide of ME451 • We started to discuss the need for initial conditions (ICs) 19

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 20

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: 21

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 • Remember the MATLAB assignment with the predator/prey problem, where you needed the initial number of predators and prey to be able to carry out analysis of the dynamics of the system • 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) 22

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… • 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 23

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 24

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) 25

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) 26

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” 27

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. 28

End ICsBeginning Constraint Reaction Forces (6.6) 29

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  30

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 31

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 produced in that joint during the motion of the mechanism • The strategy: • Look for a force (the classical, non-generalized flavor), that when acting at the location of the joint would lead to a generalized force equal to QC 32

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 very well, 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 33

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 on body i to the presence of the following reaction force and torque: 34

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 • Now, to each Lagrange multiplier corresponds 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 constraints • 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 35

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 For this to make sense you have to have in front of your eyes the picture on slide 75 36

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 37

End Constraint Reaction Forces 38

ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems November 27, 2007 Elements of the Numerical Solution for Ordinary Differential EquationsChapter 7

Before we get started… • Exam coming up in one week (Tu, Dec. 4, 9:30 AM) • Exam draws on Dynamics Analysis topic (material covered since first exam) • Take home component emailed to you over the weekend • Rough guidelines provided for what’s expected from you in terms of the preliminary report on the Final Project • Review session at 7:15 PM on Monday, Dec. 3, in this room • Last Time • Discussed about initial conditions (ICs) • Second order differential equations needs ICs for positions and velocities • These ICs should be consistent with the joint constraints associated with your mechanism • Learned how to compute the constraint reaction force that appears in the joints that connect bodies in a mechanism 40

Today’s Lecture… • Looking at algorithms to find an approximation of the solution of the equations that govern the time evolution of a dynamic system • Finding an exact solution within pen/paper framework impossible even for the swinging motion of a pendulum in gravitational field • We need to resort to numerical methods (algorithms) to produce an approximation of the solution • We’ll continue this discussion on Th when we focus on an ME451 specific method • Chris Hubert from Oshkosh Truck will talk about use of kinematics/dynamics analysis in their work • Chris is with Modeling and Simulation group at Oshkosh Truck • Analysis goes beyond kinematics and dynamics (CFD, FEA, etc.) 41

Numerical Method(also called Numerical Algorithm, or simply Algorithm) • Represents a recipe, a succession of steps that one takes to find an approximation the solution of a problem that otherwise does not admit an analytical solution • Analytical solution: sometimes called “closed form” or “exact” solution • The approximate solution obtained with the numerical method is also called “numerical solution” • Examples: • Evaluate the integral • Solve the equation • Solve the differential equation that governs time evolution of simple pendulum • Many, many others (actually very seldom can you find the exact solution of a problem…) 42

Where/How are Numerical Methods Used? • Powerful and inexpensive computers have revolutionized the use of numerical methods and their impact • Simulation of a car crash in minute detail • Formation of galaxies • Folding of a protein • Finding the electron distribution around nuclei in a nanostructure • Numerical methods enable the concept of “simulation-based engineering” • You use computer simulation to understand how your mechanism (mechanical system) behaves, how it can be modified and controlled 43

Numerical Methods in ME451 • In regards to ME451, one would use numerical method to solve the dynamics problem (the resulting set of differential equations that capture Newton’s second law) • The particular class of numerical methods used to solve differential equations is typically called “numerical integrators”, or “integration formulas” • A numerical integrator generates a numerical solution at discrete time points (also called grid points, station points, nodes) • This is in fact just like in Kinematics, where the solution is computed on a time grid • Different numerical integrators generate different solutions, but the solutions are typically very close together, and [hopefully] closed to the actual solution of our problem • Remember: In 99% of the cases, the use of numerical integrators is the only alternative for solving complicated systems described by non-linear differential equations 44

Numerical Integration~Basic Concepts~ • Initial Value Problem: (IVP) • So what’s the problem? • You are looking for a function y(t) that depends on time (changes in time), whose time derivative is equal to a function f(t,y) that is given to you (see IVP above) • In other words, I give you the derivative of a function, can you tell me what the function is? • Remember that both y0 and the function f are given to you. You want to find y(t). • In ME451, the best you can hope for is to find an approximation of the unknown function y(t) at a sequence of discrete points (as many of them as you wish) • The numerical algorithm produces an approximation of the value of the unknown function y(t) at the each grid point. That is, the numerical algorithm produces y(t1), y(t2), y(t3), etc. 45

Relation to ME451 • When carrying out Dynamics Analysis, what you can compute is the acceleration of each part in the model • Acceleration represents the second time derivative of your coordinates • Somewhat oversimplifying the problem to make the point across, in ME451 you get the second time derivate • This represents a second order differential equation since it has two time derivatives taken on the position q 46

How do you go about solving an Initial Value Problem? A look at: Euler’s Method Predictor-Corrector Method Runge-Kutta Method 47

Find solution of this Initial Value Problem: Numerical Integration:Euler’s Method Euler’s Method (t is the step size): • The idea: at each grid point tk, turn the differential problem into an algebraic problem by approximating the value of the time derivative: This step is called “discretization”. It transforms the problem from a continuous ODE problem into a discrete algebraic problem 48

Example:- Integrate 5 steps using Euler’s Method- Compare to exact solution f(t,y) = -10y (note no explicit dependency on time t for f) k=0 y0 = 1.0 k=1 y1 = y0+f(t0,y0)t = 1.0 + (-10*1.0 )*0.01 = 0.9 k=2 y2 = y1+f(t1,y1)t = 0.9 + (-10*.9 )*0.01 = 0.81 k=3 y3 = y2+f(t2,y2)t = 0.81 + (-10*.81 )*0.01 = 0.729 k=4 y4 = y3+f(t3,y3)t = 0.729 + (-10*.729 )*0.01 = 0.6561 k=5 y5 = y4+f(t4,y4)t = 0.6561 + (-10*.6561)*0.01 = 0.5905 Exact solution: y(t) = e-10t Solution: y(0) =1.0000 y(0.01)=0.9048 y(0.02)=0.8187 y(0.03)=0.7408 y(0.04)=0.6703 y(0.05)=0.6065 49

Predictor-Corrector Methods • Instead of computing f(y,t) at one point, you can average over multiple points (two in this case) • To implement, must predict yk+1 using forward Euler Method The predictor The corrector 50

ME451 Kinematics and Dynamics of Machine Systems