mechatronics case study · mechatronics case study k. craig 2 motor rotor inertia with coulomb...

Mechatronics Case Study K. Craig 1

Mechatronics Case Study

Leonardo da VinciSlider Crank Mechanism

Over 500 Years Ago


Motor Rotor Inertia

with Coulomb &

Viscous Friction

Compliant

Shaft with

Inertia and

Viscous

Damping

Gearbox with

Backlash,

Gear Inertias,

And Friction

Complaint Shafts

and Coupling

with Inertias of

Shafts & Coupling

Pulleys 1 & 2 with

Inertias; Belt Slip

or Backlash

Possible

Belt with Length-

Dependent

Compliance and

Damping; Belt

Inertia may be

included

Rigid Shaft

with Inertia

Rigid Load with

Coulomb & Viscous

Friction acting on it

Rigid Load Moving

with Belt by Coulomb

Friction or Attached

Motion System


Design NewsJuly 2011


Design NewsAugust 2012


Design NewsNovember 2010


Design NewsJune 2011


Initial Physical Model Simplifying Assumptions

• Rigid Homogeneous Bodies – Neglect Compliance• Disk: I = ½ m1r2 Rod: I = (1/12) m2ℓ2

• Friction: Viscous Damping at Crank Pivot• Frictionless Revolute Joints and Prismatic Joint• Neglect External Force Fe


Simplifying Assumptions

Diagram of Slider-CrankMechanism

Location ofCenters of Gravity

Constraint Equation


Position, Velocity, and Acceleration of Sliding Mass

Angular Velocity and Angular Acceleration

ofConnecting Rod

Lagrange EquationFormulation

Generalized Coordinate θ


T = Kinetic Energyof the Entire System

Crank is in Fixed-Axis Rotation

Connecting Rod is inGeneral Plane Motion

Simplification of T2


Sliding Mass is inTranslation

Sum the 3 Kinetic Energies to get T

V = Potential Energy of the System

There is no Elastic Potential Energy,only Gravitational Potential Energy of

the Connecting Rod


The System has OneDegree of Freedom

Generalized Coordinate is θ

Express T and V in terms of θ

Simplification


Write T2 in terms of only θ




Write V in terms of only θ

Final ExpressionsFor System Kinetic Energy

and Potential Energyin terms of θT(θ) and V(θ)


Q is the generalized torqueor force

in the Lagrange formulation

Differential Work done by external forces/torques

Express the Work Done in terms of θ

by writing dx in terms of dθ


Differential Work Donein terms of

Differential θ

Lots of Differentiationand Algebra!

Summary of Analysis

Equation of Motion


I.C. θ = 0ºReleasedfrom Rest

No External ForcesApplied

% System Parameters SI Unitsm1 = 0.232;m2 = 0.332;m3 = 0.600;r = 0.030;l = 0.217;g = 9.81;B = 0.0005; %Crank Viscous Dampingtheta = 0;phi = sin-1((r/l)*sin(θ));


Simulink Diagram Based On Equation of Motion

M( ) N( , ) F( )


SimMechanics Diagram

Mechatronics Case Study K. Craig 22Simulink Results

Mechatronics Case Study K. Craig 23SimMechanics Results


% Reflected Inertia

u = 0:.1:2*pi;

c = (l^2 - r^2*sin(u).^2).^0.5;

I_Reflected = (0.5*m1*r^2) +...

2*m2*(((1/6)*(l^2*r^2*cos(u).^2)./c.^2) + (0.5*r^2*sin(u).^2) + (0.5*r^3*cos(u).*sin(u).^2./c)) +...

2*m3*((0.5*r^2*sin(u).^2) + (r^3*sin(u).^2.*cos(u)./c) + (0.5*r^4*sin(u).^2.*cos(u).^2./c.^2));

Θ (rad)

Ikg-m2

Reflected Inertia to the Crank Axis

Equivalent Kinetic Energies


Motion Profile:Move 0º to 90º in 0.5 sec

Dwell for 0.1 secMove 90º to 0º in 0.2 sec

Dwell for 0.2 sec

5th-Order Polynomial with Zero Velocity & Acceleration at Start & Finish

Time (sec) Time (sec) Time (sec)

radians radians/sec radians/sec2


Segment 1

Segment 2


Inverse KineticsSimulink


Inverse KineticsSimMechanics


Simulink SimMechanics

Inverse Kinetics: Torque (N-m) Requirement

Time (sec)Time (sec)


Inverse Kinetics: Speed-Torque Curve

Angular Speed (rad/s)

RequiredMotorTorque(N-m)


Control Design: PD ControllerSimulink Design Optimization

Why pick PD Control?


M( ) N( , ) F( ) Equation of Motion

M(θ)

θ

slope at operating point dM( )

d

operatingpoint

Linearization

/ 4

operatingpoint

Which oneto use?

Which isthe

worst case?


Desired Step ResponseEnvelope


Parameters to DefineDesired Step Response

Envelope


For a PD ControllerOptimize

Kp, Kd, and N


Step Responsewith Initial Choices

For Kp, Kd, and N


Optimization Iterations


Optimized Step ResponseOptimized Control Gains


Closed-Loop Step Response with Optimized PD Controller

Kp = 1.6656Kd = 0.1319N = 1743.9


Time (sec)

Theta(rad)

Optimized Unit Step Response


Optimized Closed-Loop Response to Commanded Motion Profile


System ResponseTheta (rad)

Commanded MotionTheta (rad)

Time (sec) Time (sec)


Motion Error = Command - Actual

Time (sec)

Theta(radians)


Torque Required with PD Control

Time (sec)

Torque(N-m)


Closed-Loop Torque-Speed Curve

Angular Speed (rad/s)

RequiredMotorTorque(N-m)

mechatronics case study · mechatronics case study k. craig 2 motor rotor inertia with coulomb...

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