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Mechatronics Case Study K. Craig 1
Mechatronics Case Study
Leonardo da VinciSlider Crank Mechanism
Over 500 Years Ago
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
Mechatronics Case Study K. Craig 3
Mechatronics Case Study K. Craig 4
Design NewsJuly 2011
Mechatronics Case Study K. Craig 5
Design NewsAugust 2012
Mechatronics Case Study K. Craig 6
Mechatronics Case Study K. Craig 7
Design NewsNovember 2010
Mechatronics Case Study K. Craig 8
Design NewsJune 2011
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
Mechatronics Case Study K. Craig 10
Simplifying Assumptions
Diagram of Slider-CrankMechanism
Location ofCenters of Gravity
Constraint Equation
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
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
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
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
Mechatronics Case Study K. Craig 15
Write T2 in terms of only θ
Write T3 in terms of only θ
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)
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
Mechatronics Case Study K. Craig 18
Differential Work Donein terms of
Differential θ
Lots of Differentiationand Algebra!
Summary of Analysis
Equation of Motion
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));
Mechatronics Case Study K. Craig 20
Simulink Diagram Based On Equation of Motion
M( ) N( , ) F( )
Mechatronics Case Study K. Craig 21
SimMechanics Diagram
Mechatronics Case Study K. Craig 22Simulink Results
Mechatronics Case Study K. Craig 23SimMechanics Results
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
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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Segment 1
Segment 2
Mechatronics Case Study K. Craig 27
Inverse KineticsSimulink
Mechatronics Case Study K. Craig 28
Inverse KineticsSimMechanics
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Simulink SimMechanics
Inverse Kinetics: Torque (N-m) Requirement
Time (sec)Time (sec)
Mechatronics Case Study K. Craig 30
Inverse Kinetics: Speed-Torque Curve
Angular Speed (rad/s)
RequiredMotorTorque(N-m)
Mechatronics Case Study K. Craig 31
Control Design: PD ControllerSimulink Design Optimization
Why pick PD Control?
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?
Mechatronics Case Study K. Craig 33
Desired Step ResponseEnvelope
Mechatronics Case Study K. Craig 34
Parameters to DefineDesired Step Response
Envelope
Mechatronics Case Study K. Craig 35
For a PD ControllerOptimize
Kp, Kd, and N
Mechatronics Case Study K. Craig 36
Step Responsewith Initial Choices
For Kp, Kd, and N
Mechatronics Case Study K. Craig 37
Optimization Iterations
Mechatronics Case Study K. Craig 38
Optimized Step ResponseOptimized Control Gains
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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Time (sec)
Theta(rad)
Optimized Unit Step Response
Mechatronics Case Study K. Craig 41
Optimized Closed-Loop Response to Commanded Motion Profile
Mechatronics Case Study K. Craig 42
System ResponseTheta (rad)
Commanded MotionTheta (rad)
Time (sec) Time (sec)
Mechatronics Case Study K. Craig 43
Motion Error = Command - Actual
Time (sec)
Theta(radians)
Mechatronics Case Study K. Craig 44
Torque Required with PD Control
Time (sec)
Torque(N-m)
Mechatronics Case Study K. Craig 45
Closed-Loop Torque-Speed Curve
Angular Speed (rad/s)
RequiredMotorTorque(N-m)
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