mechatronics case study · mechatronics case study k. craig 2 motor rotor inertia with coulomb...
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Mechatronics Case Study K. Craig 1
Mechatronics Case Study
Leonardo da VinciSlider Crank Mechanism
Over 500 Years Ago
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Mechatronics Case Study K. Craig 2
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
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Mechatronics Case Study K. Craig 3
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Mechatronics Case Study K. Craig 4
Design NewsJuly 2011
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Mechatronics Case Study K. Craig 5
Design NewsAugust 2012
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Mechatronics Case Study K. Craig 6
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Mechatronics Case Study K. Craig 7
Design NewsNovember 2010
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Mechatronics Case Study K. Craig 8
Design NewsJune 2011
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Mechatronics Case Study K. Craig 9
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
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Mechatronics Case Study K. Craig 10
Simplifying Assumptions
Diagram of Slider-CrankMechanism
Location ofCenters of Gravity
Constraint Equation
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Mechatronics Case Study K. Craig 11
Position, Velocity, and Acceleration of Sliding Mass
Angular Velocity and Angular Acceleration
ofConnecting Rod
Lagrange EquationFormulation
Generalized Coordinate θ
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Mechatronics Case Study K. Craig 12
T = Kinetic Energyof the Entire System
Crank is in Fixed-Axis Rotation
Connecting Rod is inGeneral Plane Motion
Simplification of T2
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Mechatronics Case Study K. Craig 13
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
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Mechatronics Case Study K. Craig 14
The System has OneDegree of Freedom
Generalized Coordinate is θ
Express T and V in terms of θ
Simplification
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Mechatronics Case Study K. Craig 15
Write T2 in terms of only θ
Write T3 in terms of only θ
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Mechatronics Case Study K. Craig 16
Write T3 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)
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Mechatronics Case Study K. Craig 17
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
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Mechatronics Case Study K. Craig 18
Differential Work Donein terms of
Differential θ
Lots of Differentiationand Algebra!
Summary of Analysis
Equation of Motion
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Mechatronics Case Study K. Craig 19
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));
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Mechatronics Case Study K. Craig 20
Simulink Diagram Based On Equation of Motion
M( ) N( , ) F( )
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Mechatronics Case Study K. Craig 21
SimMechanics Diagram
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Mechatronics Case Study K. Craig 22Simulink Results
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Mechatronics Case Study K. Craig 23SimMechanics Results
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Mechatronics Case Study K. Craig 24
% 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
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Mechatronics Case Study K. Craig 25
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
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Mechatronics Case Study K. Craig 26
Segment 1
Segment 2
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Mechatronics Case Study K. Craig 27
Inverse KineticsSimulink
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Mechatronics Case Study K. Craig 28
Inverse KineticsSimMechanics
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Mechatronics Case Study K. Craig 29
Simulink SimMechanics
Inverse Kinetics: Torque (N-m) Requirement
Time (sec)Time (sec)
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Mechatronics Case Study K. Craig 30
Inverse Kinetics: Speed-Torque Curve
Angular Speed (rad/s)
RequiredMotorTorque(N-m)
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Mechatronics Case Study K. Craig 31
Control Design: PD ControllerSimulink Design Optimization
Why pick PD Control?
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Mechatronics Case Study K. Craig 32
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?
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Mechatronics Case Study K. Craig 33
Desired Step ResponseEnvelope
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Mechatronics Case Study K. Craig 34
Parameters to DefineDesired Step Response
Envelope
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Mechatronics Case Study K. Craig 35
For a PD ControllerOptimize
Kp, Kd, and N
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Mechatronics Case Study K. Craig 36
Step Responsewith Initial Choices
For Kp, Kd, and N
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Mechatronics Case Study K. Craig 37
Optimization Iterations
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Mechatronics Case Study K. Craig 38
Optimized Step ResponseOptimized Control Gains
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Mechatronics Case Study K. Craig 39
Closed-Loop Step Response with Optimized PD Controller
Kp = 1.6656Kd = 0.1319N = 1743.9
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Mechatronics Case Study K. Craig 40
Time (sec)
Theta(rad)
Optimized Unit Step Response
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Mechatronics Case Study K. Craig 41
Optimized Closed-Loop Response to Commanded Motion Profile
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Mechatronics Case Study K. Craig 42
System ResponseTheta (rad)
Commanded MotionTheta (rad)
Time (sec) Time (sec)
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Mechatronics Case Study K. Craig 43
Motion Error = Command - Actual
Time (sec)
Theta(radians)
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Mechatronics Case Study K. Craig 44
Torque Required with PD Control
Time (sec)
Torque(N-m)
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Mechatronics Case Study K. Craig 45
Closed-Loop Torque-Speed Curve
Angular Speed (rad/s)
RequiredMotorTorque(N-m)