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# ME451 Kinematics and Dynamics of Machine Systems - PowerPoint PPT Presentation

ME451 Kinematics and Dynamics of Machine Systems. Singular Configurations 3.7 October 07, 2013. Radu Serban University of Wisconsin-Madison. Before we get started…. Last Time: Numerical solution of systems of nonlinear equations Newton- Raphson method Today:

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### ME451 Kinematics and Dynamics of Machine Systems

Singular Configurations

3.7

October 07, 2013

• 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, [email protected] (11:59pm)

• Midterm Exam

• Friday, October 11 at 12:00pm in ME1143

• Review session: Wednesday, October 9, 6:30pm in ME1152

• 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

• 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:

• 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.

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

• 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?

Lock-up: PS1[Example 3.7.5]

• The mechanism cannot proceed past this configuration

• “No solution”

Bifurcation: PS2[Example 3.7.5]

• The mechanism cannot uniquely proceed from this configuration

• “Multiple solution”

Example 3.7.5Mechanism Lock-Up (1)

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)

Example 3.7.5Mechanism Bifurcation (1)

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

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

• 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.

• 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 Systems

• 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

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

• 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)

• 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

• 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)