ME451 Kinematics and Dynamics

of Machine Systems

Singular Configurations3.7

October 07, 2013

Radu SerbanUniversity of Wisconsin-Madison

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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

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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

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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:

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Implicit Function Theorem (IFT)

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IFT: Implications for Position Analysis Informally, this is what the Implicit Function Theorem says:

Assume that, at some time tk we just found a solution q(tk) of . If the constraint Jacobian is nonsingular in this configuration, that is

then, 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.

Singular Configurations3.7

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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

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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?

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Physical Singular Configurations[Example 3.7.5]

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Lock-up: PS1[Example 3.7.5]

The mechanism cannot proceed past this configuration “No solution”

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Bifurcation: PS2[Example 3.7.5]

The mechanism cannot uniquely proceed from this configuration “Multiple solution”

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Identifying Singular Configurations

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Example 3.7.5Mechanism Lock-Up (1)

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Investigate rank of augmented Jacobian

Carry out position, velocity, and acceleration analysis ()

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 3.7.5Mechanism Lock-Up (2)

Cannot solve for positions(garbage)

Start seeing convergence

difficulties

Failure to converge

Jacobian ill conditioned

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Example 3.7.5Mechanism Bifurcation (1)

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Use a time step-size Carry out position, velocity, and acceleration analysis ()

Example 3.7.5Mechanism 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

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Use a time step-size Carry out position, velocity, and acceleration analysis ()

Example 3.7.5Mechanism 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

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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.

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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

Dynamics of Planar Systems6

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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

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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

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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)

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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

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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)