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

2. 2 Before we get started… Last Time: Numerical solution of systems of nonlinear equations Newton-Raphson method 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. 3 Kinematic Analysis: Stages

4. 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. 5 Implicit Function Theorem (IFT)

6. 6 IFT: Implications for Position Analysis

7. Singular Configurations 3.7

8. 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. 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. 10 Physical Singular Configurations [Example 3.7.5]

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

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

13. 13 Identifying Singular Configurations

14. 14 Example 3.7.5 Mechanism Lock-Up (1)

15. 15 Mechanism approaching speed of light 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] Example 3.7.5 Mechanism Lock-Up (2) Cannot solve for positions (garbage) Start seeing convergence difficulties Failure to converge Jacobian ill conditioned

16. 16 Example 3.7.5 Mechanism Bifurcation (1)

17. 17 Example 3.7.5 Mechanism Bifurcation (2) 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 We ended up on one of the two possible branches Stepping over singularity and not knowing it

18. 18 Example 3.7.5 Mechanism Bifurcation (3) 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. 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. 20 SUMMARY OF CHAPTER 3

21. Dynamics of Planar Systems 6

22. 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. 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. 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. 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. 26 Newton’s Laws of Motion 1 st Law Every 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. 2 nd Law A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. 3 rd Law To 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)