1 / 27

Manufacturing Controls

Manufacturing Controls. FALL 2001 Lecture 3 . Syllabus. DATE TOPIC NOTES 1. Sep. 20 Mechatronics Design Process Ch. 1 2. Sep. 25 System Modeling and Simulation Ch. 2 3. Sep. 27 Laplace Transforms and Transfer Functions Ch. 2 4. Oct. 2 Electrical Examples Ch.2, Notes

casta
Télécharger la présentation

Manufacturing Controls

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Manufacturing Controls FALL 2001 Lecture 3  (C) 2001, Ernest L. Hall, University of Cincinnati

  2. Syllabus • DATE TOPIC NOTES • 1. Sep. 20 Mechatronics Design Process Ch. 1 • 2. Sep. 25 System Modeling and Simulation Ch. 2 • 3. Sep. 27 Laplace Transforms and Transfer Functions Ch. 2 • 4. Oct. 2 Electrical Examples Ch.2, Notes • 5. Oct. 4 Mechanical Examples Ch.2, Notes • 6. Oct. 9 Thermal and Fluid Examples, QUIZ 1 (Take Home) • 7. Oct. 11 Sensors and Transducers Ch. 3 • 8. Oct. 16 Advanced MATLAB • 9. Oct. 18 Analog and Digital Sensing Ch. 3, Notes • 10. Oct. 23 Actuating Devices Ch. 4 • 11. Oct. 25 DC Motor Model Ch. 4, Notes • 12. Oct. 30 Boolean Logic Ch. 5 • 13. Nov. 1 Programmable Logic Controllers Ch. 5, Notes • 14. Nov. 6 Stability and Compensators, P, PI and PD Ch. 6 • 15. Nov. 8 PID Controllers Ch. 7 • 16. Nov. 13 QUIZ 2 (In Class - Open Book) • 17. Nov. 15 Practical and Optimal Compensator Design Ch. 8 • 18. Nov. 20 Frequency Response Methods Ch. 9, Notes • 19. Nov. 22 THANKSGIVING HOLIDAY Ch. 9, Notes • 20. Nov. 27 Optimal Design of a Motion Control System Ch. 9, Notes • 21. Nov. 29 QUIZ 3 (In Class - Closed Book) • 22. Dec. FINAL EXAM (In Class - Closed Book) Comprehensive (C) 2001, Ernest L. Hall, University of Cincinnati

  3. Today’s objective • To continue the introduction to systems theory and the use of Laplace transforms with Matlab by studying a simple example, the rigid link pendulum. • By the end of this class you will be able to describe the linear model of the rigid link pendulum, know the parameters of its model, and simulate the operation of this system. (C) 2001, Ernest L. Hall, University of Cincinnati

  4. The Laplace transform F(s) of f(t) is: (C) 2001, Ernest L. Hall, University of Cincinnati

  5. The Laplace convolution theorem states that (C) 2001, Ernest L. Hall, University of Cincinnati

  6. Convolution reduced to product (C) 2001, Ernest L. Hall, University of Cincinnati

  7. Block diagrams • This transform relationship may be used to develop block diagram representations and algebra for linear systems, which is very useful to simplify the study of complicated systems. (C) 2001, Ernest L. Hall, University of Cincinnati

  8. Basic blocks (C) 2001, Ernest L. Hall, University of Cincinnati

  9. Blocks in cascade (C) 2001, Ernest L. Hall, University of Cincinnati

  10. Blocks in parallel (C) 2001, Ernest L. Hall, University of Cincinnati

  11. Feedback block (C) 2001, Ernest L. Hall, University of Cincinnati

  12. Jd2/dt2 Dd/dt (MgL/2)sin T  (c) Figure 6. Rigid link pendulum structure diagram. (a) Physical diagram; (b) Components of weight vector; (c)Free body torque diagram; L/2 T L/2  Mg Mgcos  (a)  Mg Mgsin  (b) Rigid link pendulum (C) 2001, Ernest L. Hall, University of Cincinnati

  13. Dynamic response for the mechanical system model of the human leg • The model assumes an input torque, T(t), viscous damping, D at the hip joint, and inertia, J, around the hip joint. • Also, a component of the weight of the leg, W = Mg, where M is the mass of the leg and g is the acceleration of gravity, creates a non-linear torque. • Assume that the leg is of uniform density so that the weight can be applied at the centroid at L/2 where L is the length of the leg. (C) 2001, Ernest L. Hall, University of Cincinnati

  14. For definiteness • Let D = 0.01 lb. s • J= 4.27 ft lb. s2 • W = Mg = 40 pounds • L=3 feet • We will use an input torque amplitude of • T(t)= 75 ft lbs. of torque. (C) 2001, Ernest L. Hall, University of Cincinnati

  15. Robot • The pendulum gives us a good model for a robot arm with a single degree of freedom. • With a rigid link, it is natural to drive the rotation by a torque applied to the pinned end and to represent the mass at the center of mass of the link. • Other physical variations lead to different robot designs. For example, if we mount the rigid link horizontally and then articulate it, we reduce the effect of gravity on the motion. (C) 2001, Ernest L. Hall, University of Cincinnati

  16. Apply the rotational form of Newton's second law about the pinned end to balance the torque • The angle of motion is shown with positive direction counter clockwise • The motion is resisted by three torques • the component of weight is (MgL/2)sin q • the damping torque is D(dq/dt) • and the inertial torque is Jd2q/dt2 • For a bob mass at the end of a link, the inertia is J = ML2. However, for a distributed link the inertia is only ML2/12. (C) 2001, Ernest L. Hall, University of Cincinnati

  17. Jd2/dt2 Dd/dt (MgL/2)sin T  (c) Figure 6. Rigid link pendulum structure diagram. (a) Physical diagram; (b) Components of weight vector; (c)Free body torque diagram; L/2 T L/2  Mg Mgcos  (a)  Mg Mgsin  (b) (C) 2001, Ernest L. Hall, University of Cincinnati

  18. Using the small angle approximation, sin q = q gives (C) 2001, Ernest L. Hall, University of Cincinnati

  19. The previous equation is a linear form. Since it is linear, we can take the Laplace transform to obtain the transfer function between the output and input. (C) 2001, Ernest L. Hall, University of Cincinnati

  20. Simplifications • It is also interesting to show how the equations simplify for a pendulum mounted in a horizontal plane rather than a vertical plane. • For a horizontally articulated pendulum or robot, the weight is perpendicular to the motion and does no work so the equation simplifies. (C) 2001, Ernest L. Hall, University of Cincinnati

  21. Horizontal pendulum (C) 2001, Ernest L. Hall, University of Cincinnati

  22. SCARA* Application - assembly & insertion Configuration – RRP Percentage - 15 Advantages - horizontal compliance; high speed; no gravity effect Disadvantage - complex kinematics; variable resolution; limited vertical motion *Selective Compliance Articulated Robot for Assembly (Source for the percent of use: V.D. Hunt, Robotics Sourcebook, New York: Elsevier, 1988.) (C) 2001, Ernest L. Hall, University of Cincinnati

  23. Configurations fit applications • Cartesian • Application – assembly and machine loading • Configuration – PPP • Percentage – 18 • Advantage – equal resolution, simple kinematics • Disadvantage – Poor space utilization (C) 2001, Ernest L. Hall, University of Cincinnati

  24. We can also develop a Matlab m-file solution to this linear differential equation. • J = 4.27; • D = 0.1; • M = 40/32.2; • g = 32.2; • L = 3; • num = [0, 180/3.14159]; % 180/3.14159 is to translate radians into degrees • den = [J, D, M*g*L/2 ]; • t= 0:0.1:10; • impulse(num,den,t); %find impulse response • grid on; • xlabel=(‘Degrees’); • ylabel=(‘Time(seconds)’); • title(‘Unit impulse response of the rigid link pendulum'); (C) 2001, Ernest L. Hall, University of Cincinnati

  25. Impulse response (C) 2001, Ernest L. Hall, University of Cincinnati

  26. Try it! • Copy m-file into Matlab. • Run the m-file • Compare the result • What is the period? • What is the frequency? • Change the time to 100 seconds. • Does the damping effect become more clear? (C) 2001, Ernest L. Hall, University of Cincinnati

  27. Any questions? (C) 2001, Ernest L. Hall, University of Cincinnati

More Related