Dynamics and control of mechanical systems

Transcription

1 Dynamics and control of mechanical systems Date Day 1 (03/05) - 05/05 Day 2 (07/05) Day 3 (09/05) Day 4 (11/05) Day 5 (14/05) Day 6 (16/05) Content Review of the basics of mechanics. Kinematics of rigid bodies - coordinate transformation, angular velocity vector, description of velocity and acceleration in relatively moving frames. Euler angles, Review of methods of momentum and angular momentum of system of particles, inertia tensor of rigid body. Dynamics of rigid bodies - Euler's equation, application to motion of symmetric tops and gyroscopes and problems of system of bodies. Kinetic energy of a rigid body, virtual displacement and classification of constraints. D Alembert s principle. Introduction to generalized coordinates, derivation of Lagrange's equation from D Alembert s principle. Small oscillations, matrix formulation, Eigen value problem and numerical solutions. Modelling mechanical systems, Introduction to MATLAB, computer generation and solution of equations of motion. Introduction to complex analytic functions, Laplace and Fourier transform. PID controllers, Phase lag and Phase lead compensation. Analysis of Control systems in state space, pole placement, computer simulation through MATLAB. 1

2 Contents Focus on Introduction to modeling of mechanical systems Modeling fundamentals of elements of mechanical systems Inertia element (mass) Energy storage elements (spring) Energy dissipation elements (dampers) Examples and exercises 2

3 Introduction A model is an abstraction of the physical world Used for analysis and design, commonly before realization of the physical system Can be obtained from first principles or experimentally The modeling purpose determines level of abstraction Can be complex enough, but no more than the physical object Model derivation from first principles Using physical laws to derive the models Models based on first principles provides better undertanding Such models can also use empirical data to determine parameters, and validate the model Input System Output Give known input (empirical data) and observe output fit model to your data - Applicable for complex systems and blackbox systems 3

4 Introduction Derivation of Differential Equations (Des) is an important part of modeling of mechanical systems. Sources of such DEs Newton s 2 nd Law FF = ma Euler s 2 nd Law MM = Iαα Hooke s Law F = kx Lagrangian Hamiltonian dd dddd LL qq jj - LL qq jj = 0 dddd dddd - HH qq jj = 0 4

5 Introduction Modeling of static and dynamic systems Static systems Relationship is static (not changing with time) Dynamic systems Relationship changes with time, past inputs and initial conditions influence current output ( it is dynamic) Only current input determines the output, which reacts immediately Relationship is represented by algebraic equations Output takes time to respond to inputs Relationship is represented by differential equations Input Output (system response) System Volt, 5v (Ang. Speed = 200 rpm) El-motor Force and motion sources cause elements to respond 5

6 1) Energy storage elements Inertia elements Stiffness (spring) elements 2) Energy dissipation elements Dampers Friction Mechanical systems can be either translational or rotational, for instance, for energy storage elements. Translational system - Inertia element: mass, m - Stiffness element: k For rotational system: - Inertia element: Mass moment of inertia, I - Stiffness element: k t 6 These are passive (non-energy producing) devices

7 1) Inertia elements Inertia elements store energy in the form of KE Commonly components are not added to a mech. system for the purpose of adding inertia Since all materials have mass, however, the mass or inertia element often may represents an undesirable effect in the system. There are some applications in which mass itself serves a useful function, e.g., flywheels. Response of engine with flywheel Flywheels are used as energy-storage devices or as a means of smoothing out speed fluctuations in in mechanical systems such as IC engines or other machines 7 Response of engine without flywheel

8 1) Inertia elements... modelling Inertia elements are modelled as follows Particle (point mass), m Rid body of mass, m Symbol for translational element T, θθ F, x Torque θθ = TT II F, x m Moment of inertia xx = FF mm Mathematical model from Newton s 2 nd Law Note that, for a rigid body in translation, every particle in the body has the same velocity and acceleration, while in flexible bodies the velocities and/or accelerations can vary Symbol for rotational element Mathematical model from Newton s 2 nd Law Note: No real rigid body exists when being accelerated, and the inertia element is a model, not a real/physical object. 8

9 1) Inertia elements... Mass moment of inertia for typical geometries Particle of mass m, rotating at end of a massless rod of length l: I = ml 2 I = mm rr 22 dddd A cylinder (r, m), rotating about center axis: I = 11 mr 2 22 Solid sphere (r, m), rotating about center axis: I = 22 mr 2 33 Uniform rod (l, m), rotating about its center: I = 11 ml Uniform rod (l, m), rotating about its end: I = 11 ml 2 33 Note: moment of inertia of rigid bodies with complex geometry are determined experimentally 9

10 Inertia elements... Stiffness elements (springs) Real springs - Are neither pure nor ideal - Can have inertia and friction Pure spring is a mathematical model, - It is not a real device - It is ideal: massless, no damping and linear For ideal spring element Note: The inverse of stiffness is called compliance (c) / flexibility F k x Force: F = k(x 1 x 2 ) = kx Torque: T = k t (θθ 11 θθ 22 ) = k t θθ Energy: E = kkkk22 10 x c F

11 Inertia elements... Stiffness elements (springs) Real springs are not purely linear - Linearization of a nonlinear spring - Taylor s expansion FF(xx) = FF(xx 0 ) + (xx xx 0 ). dddd dddd 0 + xx xx 0 2! 2 dd 2 FF dddd ! dd 3 FF dddd FF(xx) Ignoring higher order derivatives FF(xx) - FF(xx 0 ) = (xx xx 0 ). F = kx dddd dddd 0 FF(xx 0 ) Nonlinear 11 xx 0

12 Inertia elements... Stiffness elements (springs) Real springs are not purely linear Coil spring Cantilever beam spring Tension rod spring Torsion bar spring clamped-end beam spring ring spring rubber spring (shock mount) hydraulic (oil) spring pneumatic (air) spring kk = GGdd DD 33 kk = LL 33 kk tt = 12 kk = AAAA LL EEEE LL

13 Inertia elements... Stiffness elements (springs) Key steps to modelling of mechanical systems 1) Establish an inertial (fixed) coordinate system 2) Identify discrete mechanical system elements (masses, springs, dampers) and isolate them 3) Determine min. number of variables needed to uniquely define the configuration of system and subtract constraints 4) Draw the free body diagram for each inertia element 5) Write equations that relate each element s loading to deformation 6) Apply Newton s 2nd Law to each element: Translational motion: Newton s 2 nd Law: F = ma Rotational motion: Euler s 2 nd Law: T = Iα 13

14 Inertia elements... Stiffness elements (springs) Example 1: 2 mass and 3 spring system* 14

15 Energy dissipation elements: Damping elements Dampers converts the mechanical energy to thermal energy. Two main forms Friction damping (coulomb dry friction damping) Viscoelastic damping or viscous damping (most common type) F Static friction F Kinetic friction F xx xx xx Linear damper (Assuming small rel. velocity) F = Cxx Coulomb friction F = C Sin( xx) 15 Drag damper F = Cxx 22

16 Energy dissipation elements: Damping elements Characteristics of damping elements Viscoelastic damper The damping force is proportional to the velocity across the damper, The force acts in the direction opposite to the velocity. Friction damping (coulomb damping). Takes place between two surfaces in relative motion Magnitude of damping force is assumed constant not function of the relative velocity at the interface The force is opposite to that of the relative velocity 16

17 Applications: A damper element is used to model a device designed into a system (e.g., automotive shock absorbers) Sprang mass B Unsprang mass 17

18 Energy dissipation elements: Damping elements... Damper force or torque is directly proportional to the relative velocity of its two ends. FF DD = CC dd dddd xx 11 xx 22 = CC ddxx TT DD = CC dd dddd (θθ 1 θθ 2) dddd = CC ddθθ dddd Where C is the damping coefficient = CC xx for translational motion = C θθ - for rotational motion In some textbooks B is used as a damping coeff. C Power dissipated FFxx = CC( xx) 2 Symbol of a damper (translational motion) 18

19 Example 2*: mass spring and damper system C 1 C 2 C 3 19

20 Exercise: Modelling the suspension system of a vehicle k 2 m 2 x 2 m 1 C x 1 m 1 = Unsprung mass m 2 = Spring mass, a quarter of the car s mass C = damping coeff. of shock absorber k 1 = Tire stiffness (damping in tire is neglected) k 2 = Stiffness of suspension coil spring x 0 = motion of the ground stiffness x 1 = coordinate for motion of unsprung mass x 2 = coordinate for motion of sprung mass k 1 x 0 Derive the DEs that describe the motion of the suspension system and formulate it in matrix form 20

21 This lecture has focused on Summary Explanation of the building blocks of modeling mechanical systems: - Inertia elements, - energy storage (spring) elements and - energy dissipation (damper) elements Use of Newton s 2 nd Law and Euler s 2 nd Law to develop mathematical models of the modeling elements Next: Brief Introduction about multibody dynamics. 21