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ME451 Kinematics and Dynamics of Machine Systems

ME451 Kinematics and Dynamics of Machine Systems. Dynamics of Planar Systems Tuesday, April 16, 2009 Constraint Reaction Forces – 6.6. Quote of the day: Nobody goes there no more; it’s too crowded. Yogi Berra. © Dan Negrut, 2009 ME451, UW-Madison. Before we get started… . On Tuesday…

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ME451 Kinematics and Dynamics of Machine Systems

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  1. ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems Tuesday, April 16, 2009 Constraint Reaction Forces – 6.6 Quote of the day: Nobody goes there no more; it’s too crowded. Yogi Berra © Dan Negrut, 2009ME451, UW-Madison

  2. Before we get started… • On Tuesday… • No class, I’m out of town • Exam coming up on April 23 • Review session on Wd evening, 7:15 pm • Look at an example • Answer your questions • We’ll meet in 3126ME • I’m open to having the Thursday exam in the evening • Thursday 7:30 – 9:00 PM would be a possibility • If we shift the time of the exam we will use the regular Thursday lecture to cover one more example and review the material 2

  3. 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 3

  4. 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 4

  5. 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 5

  6. 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: 6

  7. 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 7

  8. 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 8

  9. 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 9

  10. End Constraint Reaction Forces 10

  11. ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems April 30, 2009 Elements of the Numerical Solution for Ordinary Differential EquationsChapter 7

  12. 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 12

  13. 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.) 13

  14. 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…) 14

  15. 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 15

  16. 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 16

  17. 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. 17

  18. 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 18

  19. How do you go about solving an Initial Value Problem? A look at: Euler’s Method Predictor-Corrector Method Runge-Kutta Method 19

  20. 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 20

  21. 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 21

  22. 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 22

  23. Example- Integrate in Matlab for 1 second using Euler’s Method- Compare to exact solution • Matlab y0=1; % Euler’s Method dt=.01; t=0:.01:1.; yh=y0; for i=2:length(t) f=-10*yh(i-1); yh(i)=yh(i-1)+f*dt; end % Exact solution y=y0*exp(-10*t); % Plot and compare plot(t,y,'b-',t,yh,'ro'); 23

  24. Euler Method: ~ Effect of Step Size ~ • Solve using step sizes t=0.1, 1 and 5 sec dt=5 sec • Matlab dt=1 sec dt=0.1 sec % Exact solution t=0:.01:50; y=.99*exp(-t/10)+.995*sin(t-1.47); % Numerical solution dt=.1; th=0:dt:50; yh=0; for i=2:length(th) f=-yh(i-1)/10+sin(th(i)); yh(i)=yh(i-1)+f*dt; end plot(t,y,'b-',th,yh,'ro-'); Conclusion: If you use large step-sizes t, the ACCURACY of the solution is very poor (you can’t be too aggressive with size of t) 24

  25. MATLAB Support for solving IVP 25

  26. Ordinary Differential Equations(Initial Value Problem) • An ODE + initial value: • Use ode45 for non-stiff IVPs and ode23t for stiff IVPs (concept of “stiffness” discussed shortly) [t,y] = ode45(odefun,tspan,y0,options) function dydt = odefun(t,y) Initialvlue [initialtime finaltime] • Use odeset to define options parameter 26

  27. IVP Example (MATLAB at work): function dydt = myfunc(t,y) dydt=zeros(2,1); dydt(1)=y(2); dydt(2)=(1-y(1)^2)*y(2)-y(1); • [t,y]=ode45('myfunc',[0 20],[2;0]) Note: Help on odesetto set options for more accuracy and other useful utilities like drawing results during solving. 27

  28. The concept of stiff differential equations, and how to solve the corresponding IVP 28

  29. Example: IVP 0 0.01873075307798 0.03032004603564 0.03681163609403 0.03972896411722 0.04019944117144 - Integrate 5 steps using forward Euler formula: t=0.002, t=0.01, t=0.03 - Compare the errors between numerical and analytical solutions (Algorithm Error) AlgorithmErrorwhen t=0.002: AlgorithmErrorwhen t=0.01: AlgorithmErrorwhen t=0.03: 0 0.36787944117144 0.13533528323661 0.04978706836786 0.01831563888873 0.00673794699909 0 2.04978706836786 -3.99752124782333 8.00012340980409 -15.99999385578765 32.00000030590232 29

  30. Example: Analytical Solution Forward Euler (t=0.03) 30

  31. Concept of stiff IVP’s • IVP’s for which forward Euler doesn’t work well (see example) • In general, the entire class of so called explicit formulas doesn’t work • Forward Euler, Runge-Kutta (RK23, RK45), DOPRI5, Adams-Bashforth, etc. • Stiff IVP’s require a different class of integration formulas • Implicit formulas • Example: backward Euler 31

  32. Forward Euler • Backward Euler Explicit vs. Implicit Formulas(look at Euler family) • Initial Value Problem 32

  33. Example: Exact Solution Backward Euler and Analytical Solution Forward Euler (t=0.03) 33

  34. Other Popular Algorithms for Stiff IVPs • BDF of 1st order: • BDF of 2nd order: • BDF of 3rd order: • BDF of 4th order: • BDF of 5th order: The family of BDF methods (Backward-Difference Formulas): 34

  35. The Two Key Attributes of a Numerical Integrator • Two attributes are relevant when considering a numerical integrator for finding an approximation of the solution of an IVP • The STABILITY of the numerical integrator • The ACCURACY of the numerical integrator 35

  36. Numerical Integration Formula:The STABILITY Attribute • The stability question: • How big can I choose the integration step-size t and be safe? • Tough question answered in a Numerical Analysis class • Different integration formulas, have different stability regions • You’d like to use an integration formula with large stability region: • Example: Backward Euler, BDF methods, Newmark, etc. • Why not always use these methods with large stability region? • There is no free lunch: these methods are implicit methods that require the solution of an algebra problem at each step (we’ll see this on Th) 36

  37. Numerical Integration Formula :The ACCURACY Attribute • The accuracy question: • How accurate is the formula that I’m using? • If I start decreasing t, how will the accuracy of the numerical solution improve? • Tough question answered in a Numerical Analysis class • Examples: • Forward and Backward Euler: accuracy O(t) • RK45: accuracy O(t4) • Why not always use methods with high accuracy order? • There is no free lunch: these methods usually have very small stability regions • Therefore, you are limited to very small values of t 37

  38. ODE solvers in MATLAB 38

  39. End Numerical Methods at LargeBeginning Numerical Methods for ME451 39

  40. ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems November 29, 2007 Elements of the Numerical Solution of the Dynamics ProblemChapter 7

  41. Before we get started… • Exam coming up in one week (Tu, Dec. 4, 9:30 AM) • Exam covers Dynamics (material covered since first exam) • Take home component emailed to you: due at the time of the exam • Preliminary report on Final Project due • Review session at 6:00 PM on Monday, Dec. 3, in this room • Last Time • Discussed how numerical algorithms (methods) play an essential part in finding the time evolution of a mechanism (or any dynamic system for that matter) • Solution of IVP found by either explicit or implicit numerical integration formulas • Key Idea: Differential problem transformed into Algebraic problem through discretization (use of discretization formulas) • Two important attributes associated with a numerical integration algorithm: • Stability • Accuracy 41

  42. 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 42

  43. What’s special about ME451 problems? • Looking at the previous slide there are several things that make the ME451 dynamics problem challenging: • The problem is not in standard form • Furthermore, our problem is not a first order Ordinary Differential Equation (ODE) problem • Rather, it’s a second order ODE problem, due to form of the equations of motion (contain the second time derivative of the positions) • After all, it’s not even an ODE problem • The unknown function q(t); that is, the position of the mechanism, is the solution of a second order ODE problem (first equation previous slide) but it must also satisfy a set of kinematic constraints at position, velocity, and acceleration levels, which are formulated as a bunch of algebraic equations • To conclude, you have to solve a set of differential-algebraic equations (DAEs) • DAEs are much tougher to solve than IVPs • This lecture is about using a numerical method to solve DAEs of multibody dynamics 43

  44. Example: Find the time evolution of the pendulum • Simple Pendulum: • Mass 20 kg • Length L=2 m • Force acting at tip of pendulum • F = 30 sin(2 t) [N] • Although not shown in the picture, assume RSDA element in revolute joint • k = 45 [Nm/rad] & 0=3/2 • c = 10 [N/s] • ICs: hanging down, starting from rest • Stages of the procedure (three): • Stage 1: Derive constrained equations of motion • Stage 2: Indicate initial conditions (ICs) • Stage 3: Apply numerical integration algorithm to discretize DAE problem and turn into algebraic problem 44

  45. Example: Simple Pendulum Stage 1: Deriving EOM (see also posted solution) Stage 2: Indicate Initial Conditions (ICs) 45

  46. Detour: Algorithm for Resolving Dynamics of Mechanisms • If you have the EOM and ICs, how do you go about solving the problem? • This is a research topic in itself • We’ll present almost the simplest algorithm possible • It is based on Newmark’s integration formulas • That is, we are going to use Newmark’s formulas to discretize our differential problem 46

  47. Solution Strategy: Important Slide • This slide explains the strategy through which the numerical solution; i.e., an approximation of the actual solution of the dynamics problem, is produced • Step 1: two integration formulas (Newmark in our case) are used to express the positions and velocities as functions of accelerations • These are also called “discretization formulas” • Step 2: everywhere in the constrained equations of motion, the positions and velocities are replaced using the discretization formulas and expressed in terms of the acceleration • This is the most important step, since through this “discretization” the differential problem is transformed into an algebraic problem • The algebraic problem, which effectively amounts to solving a nonlinear system, is approached using Newton’s method (so again, we need to produce a Jacobian) • Step 3: solve a nonlinear system to find the acceleration and the Lagrange multipliers 47

  48. Overview Newmark Integration Formulas • Newmark method (N.M. Newmark – 1957) • One step method designed to integrate directly second order equations of motion: • The “discretization” formulas that relate position to acceleration and velocity to acceleration are: • The goal is to find the positions, velocities, accelerations and Lagrange multipliers on a grid of time points 48

  49. Newmark (Cntd.) • Newmark Method • 1st Order • Very good stability properties • Choose =0.3025, and =0.6 (these are two parameters that control the behavior of the method) • If we write the equation of motion at each time tn+1 one gets • Now is the time to replace and with the discretization formulas (see previous slide) • You end up with an algebraic problem in which the unknown is exclusively the acceleration 49

  50. The Problem at Hand • Our rigid multibody dynamics problem is slightly more complicated than the Linear Finite Element problem used to introduce Newmark’s discretization formulas • More complicated since we have some constraints that the solution must satisfy • We also have to deal with the Lagrange multipliers that come along with these constraints (from a physical perspective remember that they help enforce satisfaction of the constraints) Linear Finite Element dynamics problem Nonlinear multibody dynamics problem. Newmark algorithm still works as before, problem is slightly messier… 50

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