ME451 Kinematics and Dynamics of Machine Systems

1. ME451 Kinematics and Dynamics of Machine Systems Singular Configurations 3.7 October 07, 2013 Radu Serban University of Wisconsin-Madison

2. Before we get started… • Last Time: • Numerical solution of systems of nonlinear equations • Newton-Raphsonmethod • Today: • Singular configurations (lock-up and bifurcations) • Assignments: • No book problems until midterm • Matlab 4 and Adams 2 – due October 9, Learn@UW (11:59pm) • Midterm Exam • Friday, October 11 at 12:00pm in ME1143 • Review session: Wednesday, October 9, 6:30pm in ME1152

3. Kinematic Analysis: Stages • Stage 1: Identify all physical joints and drivers present in the system • Stage 2: Identify the corresponding set of constraint equations • Stage 3: Position AnalysisFind the Generalized Coordinates as functions of timeNeeded: and • Stage 4: Velocity AnalysisFind the Generalized Velocities as functions of timeNeeded: and • Stage 5: Acceleration AnalysisFind the Generalized Accelerations as functions of timeNeeded: and

4. Position, Velocity, Acceleration Analysis • The position analysis [Stage 3]: • The most difficult of the three as it requires the solution of a system of nonlinear equations. • Find generalized coordinates by solving the nonlinear equations: • The velocity analysis [Stage 4]: • Simple as it only requires the solution of a linear system of equations. • After completing position analysis, find generalized velocities from: • The acceleration analysis [Stage 5]: • Simple as it only requires the solution of a linear system of equations. • After completing velocity analysis, find generalized accelerations from:

5. Implicit Function Theorem (IFT)

6. IFT: Implications for Position Analysis • Informally, this is what the Implicit Function Theorem says: • Assume that, at some time tk we just found asolution q(tk)of . • If the constraint Jacobian is nonsingularin this configuration, that isthen, we can conclude that the solution is unique, and not only at tk, but in a small interval around time tk. • Additionally, in this small time interval, there is an explicit functional dependency of q on t; that is, there is a function f such that: • Practically, this means that the mechanism is guaranteed to be well behaved in the time interval . That is, the constraint equations are well defined and the mechanism assumes a unique configuration at each time. • Moreover, assuming that is twice differentiable, IFT guarantees that the velocity and acceleration equations hold.

7. 3.7 Singular Configurations

8. Singular Configurations • Abnormal situations that should be avoided since they indicate either a malfunction of the mechanism (poor design), or a bad model associated with an otherwise well designed mechanism • Singular configurations come in two flavors: • Physical Singularities (PS): reflect bad design decisions • Modeling Singularities (MS): reflect bad modeling decisions • Singular configurations do not represent the norm, but we must be aware of their existence • A PS is particularly bad and can lead to dangerous situations

9. Singular Configurations • In a singular configuration, one of three things can happen: • PS1: Mechanism locks-up • PS2: Mechanism hits a bifurcation • MS1: Mechanism has redundant constraints • The important question:How can we characterize a singular configuration in a formal way such that we are able to diagnose it?

10. Physical Singular Configurations[Example 3.7.5]

11. Lock-up: PS1[Example 3.7.5] • The mechanism cannot proceed past this configuration • “No solution”

12. Bifurcation: PS2[Example 3.7.5] • The mechanism cannot uniquely proceed from this configuration • “Multiple solution”

13. Identifying Singular Configurations

14. Example 3.7.5Mechanism Lock-Up (1)

15. Example 3.7.5Mechanism Lock-Up (2) • Investigate rank of augmented Jacobian • Carry out position, velocity, and acceleration analysis () t = 1.90 it = 5 |Jac|= 1.493972e-01 q = [+1.028214e+00 +4.974188e-01 +3.034291e-01] qd = [-8.597511e-01 +2.617994e-01 -1.540014e+00] qdd = [-3.924469e+00 +0.000000e+00 -7.355873e+00] t = 1.95 it = 5 |Jac|= 1.060626e-01 q = [+9.785586e-01 +5.105088e-01 +2.137492e-01] qd = [-1.180227e+00 +2.617994e-01 -2.153623e+00] qdd = [-1.083795e+01 +0.000000e+00 -2.105153e+01] t = 2.00 it = 25 |Jac|= 8.718594e-09 q = [+8.660254e-01 +5.235988e-01 +1.743719e-08] qd = [-1.300238e+07 +2.617994e-01 -2.600476e+07] qdd = [-1.939095e+22 +0.000000e+00 -3.878190e+22] t = 2.05 it = 100 |Jac|= 1.474421e-01 q = [+1.008677e+00 +5.366887e-01 -1.300261e+06] qd = [-8.629133e-01 +2.617994e-01 -1.525969e+00] qdd = [-3.893651e+00 +0.000000e+00 -7.307789e+00] Jacobian ill conditioned Start seeing convergence difficulties Mechanism approaching speed of light Failure to converge Cannot solve for positions (garbage)

16. Example 3.7.5Mechanism Bifurcation (1)

17. Example 3.7.5Mechanism Bifurcation (2) • Use a time step-size • Carry out position, velocity, and acceleration analysis () t = 5.90 it = 3 |Jac| = 2.617695e-02 q = [+5.235390e-02 +1.544616e+00 +2.617994e-02] qd = [-5.234194e-01 +2.617994e-01 -2.617994e-01] qdd = [-3.588280e-03 +0.000000e+00 -1.998677e-13] t = 5.95 it = 3 |Jac| = 1.308960e-02 q = [+2.617919e-02 +1.557706e+00 +1.308997e-02] qd = [-5.235539e-01 +2.617994e-01 -2.617994e-01] qdd = [-1.794293e-03 +0.000000e+00 -1.196983e-12] t = 6.00 it = 3 |Jac| = -2.933417e-15 q = [-2.872185e-15 +1.570796e+00 -2.933417e-15] qd = [-2.563346e-01 +2.617994e-01 +5.464817e-03] qdd = [-2.335469e+13 +0.000000e+00 -2.335469e+13] t = 6.05 it = 12 |Jac| = 1.308960e-02 q = [-9.066996e-16 +1.583886e+00 +1.308997e-02] qd = [+1.815215e-14 +2.617994e-01 +2.617994e-01] qdd = [-7.262474e-13 +0.000000e+00 -7.262474e-13] t = 6.10 it = 2 |Jac| = 2.617695e-02 q = [-2.908636e-18 +1.596976e+00 +2.617994e-02] qd = [-0.000000e+00 +2.617994e-01 +2.617994e-01] qdd = [+2.168404e-19 +0.000000e+00 +0.000000e+00] Jacobian is singular Stepping over singularity and not knowing it We ended up on one of the two possible branches

18. Example 3.7.5Mechanism Bifurcation (3) • Use a time step-size • Carry out position, velocity, and acceleration analysis () t = 5.88 it = 3 |Jac| = 3.141076e-02 q = [+6.282152e-02 +1.539380e+00 +3.141593e-02] qd = [-5.233404e-01 +2.617994e-01 -2.617994e-01] qdd = [-4.305719e-03 +0.000000e+00 +1.678902e-14] t = 5.94 it = 3 |Jac| = 1.570732e-02 q = [+3.141463e-02 +1.555088e+00 +1.570796e-02] qd = [-5.235342e-01 +2.617994e-01 -2.617994e-01] qdd = [-2.153125e-03 +0.000000e+00 +9.807114e-14] t = 6.00 it = 3 |Jac| = 4.163336e-16 q = [+4.775660e-16 +1.570796e+00 +4.163336e-16] qd = [-3.003036e-01 +2.617994e-01 -3.850419e-02] qdd = [+1.610640e+14 +0.000000e+00 +1.610640e+14] t = 6.06 it = 9 |Jac| = -1.570732e-02 q = [-3.141463e-02 +1.586504e+00 -1.570796e-02] qd = [-5.235342e-01 +2.617994e-01 -2.617994e-01] qdd = [+2.153125e-03 +0.000000e+00 +1.112356e-12] t = 6.12 it = 3 |Jac| = -3.141076e-02 q = [-6.282152e-02 +1.602212e+00 -3.141593e-02] qd = [-5.233404e-01 +2.617994e-01 -2.617994e-01] qdd = [+4.305719e-03 +0.000000e+00 -8.836328e-16] Jacobian is singular Stepping over singularity and not knowing it We ended up on the other possible branch

19. Singular Configurations • Remember that you seldom see singularities • Important: The only case when you run into problems is when the constraint Jacobian becomes singular:In this case, one of the following situations can occur: • You can be in a lock-up configuration (you won’t miss this, PS1) • You might face a bifurcation situation (very hard to spot, PS2) • You might have redundant constraints (MS1) • Otherwise, the Implicit Function Theorem (IFT) gives you the answer:If the constraint Jacobian is nonsingular, IFT says that you cannot be in a singular configuration.

20. SUMMARY OF CHAPTER 3 • We looked at the KINEMATICS of a mechanism • That is, we are interested in how this mechanism moves in response to a set of kinematic drivers (motions) applied to it • Kinematic Analysis Steps: • Stage 1: Identify all physical joints and drivers present in the system • Stage 2: Identify the corresponding constraint equations • Stage 3: Position Analysis – Find as functions of time • Stage 4: Velocity Analysis – Find as functions of time • Stage 5: Acceleration Analysis – Find as functions of time

21. 6 Dynamics of Planar Systems

22. Kinematics vs. Dynamics • Kinematics • We include as many actuators as kinematic degrees of freedom – that is, we impose KDOF driver constraints • We end up with NDOF = 0 – that is, we have as many constraints as generalized coordinates • We find the (generalized) positions, velocities, and accelerations by solving algebraic problems (both nonlinear and linear) • We do not care about forces, only that certain motions are imposed on the mechanism. We do not care about body shape nor inertia properties • Dynamics • While we may impose some prescribed motions on the system, we assume that there are extra degrees of freedom – that is, NDOF > 0 • The time evolution of the system is dictated by the applied external forces • The governing equations are differential or differential-algebraic equations • We very much care about applied forces and inertia properties of the bodies in the mechanism

23. Dynamics M&S Dynamics Modeling • Formulate the system of equations that govern the time evolution of a system of interconnected bodies undergoing planar motion under the action of applied (external) forces • These are differential-algebraic equations • Called Equations of Motion (EOM) • Understand how to handle various types of applied forces and properly include them in the EOM • Understand how to compute reaction forces in any joint connecting any two bodies in the mechanism Dynamics Simulation • Understand under what conditions a solution to the EOM exists • Numerically solve the resulting (differential-algebraic) EOM

24. Roadmap to Deriving the EOM • Begin with deriving the variational EOM for a single rigid body • Principle of virtual work and D’Alembert’s principle • Consider the special case of centroidal reference frames • Centroid, polar moment of inertia, (Steiner’s) parallel axis theorem • Write the differential EOM for a single rigid body • Newton-Euler equations • Derive the variational EOM for constrained planar systems • Virtual work and generalized forces • Finally, write the mixed differential-algebraic EOM for constrained systems • Lagrange multiplier theorem (This roadmap will take several lectures, with some side trips)

25. What are EOM? • In classical mechanics, the EOM are equations that relate (generalized) accelerations to (generalized) forces • Why accelerations? • If we know the (generalized) accelerations as functions of time, they can be integrated once to obtain the (generalized) velocities and once more to obtain the (generalized) positions • Using absolute (Cartesian) coordinates, the acceleration of body i is the acceleration of the body’s LRF: • How do we relate accelerations and forces? • Newton’s laws of motion • In particular, Newton’s second law written as

26. Newton’s Laws of Motion • 1st LawEvery body perseveres in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by forces impressed. • 2nd LawA change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. • 3rd LawTo any action there is always an opposite and equal reaction; in other words, the actions of two bodies upon each other are always equal and always opposite in direction. • Newton’s laws • are applied to particles (idealized single point masses) • only hold in inertial frames • are valid only for non-relativistic speeds Isaac Newton (1642 – 1727)