MESB 374 System Modeling and Analysis Translational Mechanical System

Slides:



Advertisements
Similar presentations
Example: Derive EOM for simple 1DOF mechanical system
Advertisements

In the figure, a towing truck (not shown) is about to pull a trailer. The trailer transports a cargo of mass M that is held in place with two elastomeric.
Professor Walter W. Olson Department of Mechanical, Industrial and Manufacturing Engineering University of Toledo Lumped Parameter Systems.
1 Lecture 2 Read textbook CHAPTER 1.4, Apps B&D Today: Derive EOMs & Linearization Fundamental equation of motion for mass-spring- damper system (1DOF).
Lecture 2 Free Vibration of Single Degree of Freedom Systems
Course Outline 1.MATLAB tutorial 2.Motion of systems that can be idealized as particles Description of motion; Newton’s laws; Calculating forces required.
Introduction to Structural Dynamics:
EQUILIBRIUM OF RIGID BODIES. RIGID BODIES Rigid body—Maintains the relative position of any two particles inside it when subjected to external loads.
Scale physical model Mathematical model Numerical model How to Model Physical Systems
© 2011 Autodesk Freely licensed for use by educational institutions. Reuse and changes require a note indicating that content has been modified from the.
ME 440 Intermediate Vibrations Tu, Feb. 17, 2009 Section 2.5 © Dan Negrut, 2009 ME440, UW-Madison.
MESB 374 System Modeling and Analysis Translational Mechanical System
Spring Topic Outline for Physics 1 Spring 2011.
A PPLIED M ECHANICS Lecture 02 Slovak University of Technology Faculty of Material Science and Technology in Trnava.
Feedback Control Systems (FCS) Dr. Imtiaz Hussain URL :
Chapter 7. Free and Forced Response of Single-Degree-of-Freedom Linear Systems 7.1 Introduction Vibration: System oscillates about a certain equilibrium.
When a weight is added to a spring and stretched, the released spring will follow a back and forth motion.
11.3 Principle of Virtual Work for a System of Connected Rigid Bodies Method of virtual work most suited for solving equilibrium problems involving a system.
MOMENT OF INERTIA Today’s Objectives: Students will be able to: 1.Determine the mass moment of inertia of a rigid body or a system of rigid bodies. In-Class.
Lecture 3: Dynamic Models Spring: stores energy Damper: dissipates energy.
MESB 374 System Modeling and Analysis Electrical Systems.
Newton’s 2nd Law: Translational Motion
Today’s Objectives: Students will be able to: a)Apply the three equations of motion for a rigid body in planar motion. b)Analyze problems involving translational.
System Models.
When a weight is added to a spring and stretched, the released spring will follow a back and forth motion.
Lecture 3: Dynamic Models
Dr. Tamer Samy Gaafar Lec. 3 Mathematical Modeling of Dynamic System.
Biomedical Control Systems (BCS) Module Leader: Dr Muhammad Arif muhammadarif Batch: 10 BM Year: 3 rd Term: 2 nd Credit Hours (Theory):
ENERGY UNIT Common Assessment Review. DO NOW Turn in “How Much Energy” lab – remember that I am not taking it late so you need to turn in what you have.
Particle Kinematics Direction of velocity vector is parallel to path Magnitude of velocity vector is distance traveled / time Inertial frame – non accelerating,
Energy Notes Energy is one of the most important concepts in science. An object has energy if it can produce a change in itself or in its surroundings.
Statics. Equilibrium Moves at constant velocity V =C Moves at constant velocity V =C At rest V = 0 At rest V = 0 Constant Velocity.
ME451 Kinematics and Dynamics of Machine Systems Dynamics of Planar Systems November 4, 2010 Chapter 6 © Dan Negrut, 2010 ME451, UW-Madison TexPoint fonts.
CHAPTER 3 MESB 374 System Modeling and Analysis
Course Outline Course Outline Exam topics MATLAB tutorial
Objectives: Write the equation of motion for an accelerating body.
Automatic Control Theory CSE 322
ME375 System Modeling and Analysis
Feedback Control Systems (FCS)
MESB374 System Modeling and Analysis
Introduction to Structural Dynamics
Kinetics of Particles: Newton’s Second Law
Figure 1. Spring characteristics
Mechanics Review – SEMESTER 1
1. Kinetic energy, potential energy, virtual work.
Figure 1. Spring characteristics
Control Engineering ( ) G-14
Physics 111: Mechanics Lecture 5
ME 440 Intermediate Vibrations
Kinematic Analysis (position, velocity and acceleration)
ME375 Handouts - Spring 2002ME375-Fall 2002
Different kinds of energy
Lecture 4: Modeling Mechanical Systems
MAK4041-Mechanical Vibrations
RECTANGULAR COORDINATES
Rolling, Torque, and Angular Momentum
PHYS 211 Exam 1 HKN Review Session
Engineering Mechanics: Statics
POWER AND EFFICIENCY Today’s Objectives: Students will be able to:
Same format as first quiz. Total of 50 points
Engineering Mechanics: Statics
Engineering Mechanics: Statics
POWER AND EFFICIENCY Today’s Objectives: Students will be able to:
Figure 1. Spring characteristics
Control Systems Lecture 5: Mathematical Modeling of Mechanical Systems and State Space Representation Abdul Qadir Ansari, PhD
Control Systems (CS) Lecture-4-5
Chapter 14 : Kinematics Of A Particle – Work and Energy
MULTI DEGREE OF FREEDOM (M-DOF)
DEPARTMENT OF MECHANICAL ENGINEERING
POWER AND EFFICIENCY Today’s Objectives: Students will be able to:
Presentation transcript:

MESB 374 System Modeling and Analysis Translational Mechanical System ME 375 – Spring 2003 MESB 374 System Modeling and Analysis Translational Mechanical System

Translational Mechanical Systems ME 375 – Spring 2003 Translational Mechanical Systems Basic (Idealized) Modeling Elements Interconnection Relationships -Physical Laws Derive Equation of Motion (EOM) - SDOF Energy Transfer Series and Parallel Connections Derive Equation of Motion (EOM) - MDOF In this and next class, we will talk about the modeling and analysis of translational mechanical systems. The basic topics are listed here. By the way, section 3.1 of the textbook gives us a good review of basic concepts of mass, weight, force and systems of units. I wish that you may read it. At first, we would like to introduce some basic, idealized modeling elements. It should be kept in mind that we may use bunches of these elements to model a single object in realty. For example, a series of mass, spring and damper may be used to model a single flexible beam. Then we will discuss the physical laws used to describe the interconnection relationship . By using these physic laws, for example Newton’s second law, we may obtain the equation of motion of system To learn it step by step, we firstly discuss systems with single degree of freedom. System with multiple degree freedom will be discussed later. Also, we will discuss the topic of energy transfer and how to simply to model in case of series or parallel connections.

Key Concepts to Remember ME 375 – Spring 2003 Key Concepts to Remember Three primary elements of interest Mass ( inertia ) M Stiffness ( spring ) K Dissipation ( damper ) B Usually we deal with “equivalent” M, K, B Distributed mass  lumped mass Lumped parameters Mass maintains motion Stiffness restores motion Damping eliminates motion Throughout this course, we will always three basic elements, inertia element, or mass, spring element which is used to model the stiffness of a object, element dissipating energy or damper. Usually we use M to denote mass, K to denote stiffness, B to denote damping. For system with single degree of freedom, M, K, B are scalars. For system with multiple degree of freedom, they are matrices. As we discussed in the first lecture, all the system in realty are distributed system. If we want to model them in lumped models, we need to find the equivalent mass, stiffness and damping. Now let us look at the physical meaning of these three elements. At first, what is mass? How will mass affect the motion? How can we measure the mass? Mass maintains the motion. It means that if there is no external force acting on an object, then it will maintain its current situation. If it moves, then it keep on moving at the same speed. If it original stops, then it will never move until a force acts on it. Stiffness can restore the motion. It is related to potential energy. Damping can eliminate the motion, which means to resist the motion of a object. Does it mean that damping eliminate energy? No. we all know that energy can only be changed from one form to another and never be eliminated. In our case, damping usually change the energy to heat (Kinetic Energy) (Potential Energy) (Eliminate Energy ? ) (Absorb Energy )

Variables x : displacement [m] v : velocity [m/sec] ME 375 – Spring 2003 Variables x : displacement [m] v : velocity [m/sec] a : acceleration [m/sec2] f : force [N] p : power [Nm/sec] w : work ( energy ) [Nm] 1 [Nm] = 1 [J] (Joule) Now let’s look at some basic variables used to describing motion. Displacement is usually defined with respect to a reference frame. Velocity is used to describe how fast the displacement changes with respect to time. Mathematically, it is time derivative of displacement. Acceleration is used to describe how fast velocity changes. Hence, it is Vectors, scalars

Basic (Idealized) Modeling Elements ME 375 – Spring 2003 Basic (Idealized) Modeling Elements Spring Stiffness Element Reality 1/3 of the spring mass may be considered into the lumped model. In large displacement operation springs are nonlinear. K Linear spring  nonlinear spring  broken spring !! Idealization Massless No Damping Linear Stores Energy Hard Spring Potential Energy Soft Spring

Basic (Idealized) Modeling Elements ME 375 – Spring 2003 Basic (Idealized) Modeling Elements Damper Friction Element Mass Inertia Element x2 x1 fD B x f2 M f1 f3 Dissipate Energy fD Stores Kinetic Energy

Interconnection Laws Newton’s Second Law Newton’s Third Law ME 375 – Spring 2003 Interconnection Laws Newton’s Second Law Lumped Model of a Flexible Beam K,M x Newton’s Third Law Action & Reaction Forces x K M K M Massless spring E.O.M.

ME 375 – Spring 2003 Modeling Steps Understand System Function, Define Problem, and Identify Input/Output Variables Draw Simplified Schematics Using Basic Elements Develop Mathematical Model (Diff. Eq.) Identify reference point and positive direction. Draw Free-Body-Diagram (FBD) for each basic element. Write Elemental Equations as well as Interconnecting Equations by applying physical laws. (Check: # eq = # unk) Combine Equations by eliminating intermediate variables. Validate Model by Comparing Simulation Results with Physical Measurements

Vertical Single Degree of Freedom (SDOF) System ME 375 – Spring 2003 Vertical Single Degree of Freedom (SDOF) System B,K,M x xs f Define Problem The motion of the object g Input Output Develop Mathematical Model (Diff. Eq.) Identify reference point and positive direction. M K M B g f x M Draw Free-Body-Diagram (FBD) Write Elemental Equations From the undeformed position From the deformed (static equilibrium) position Validate Model by Comparing Simulation Results with Physical Measurement

Energy dissipated by damper ME 375 – Spring 2003 Energy Distribution EOM of a simple Mass-Spring-Damper System We want to look at the energy distribution of the system. How should we start ? Multiply the above equation by the velocity term v : Ü What have we done ? Integrate the second equation w.r.t. time: Ü What are we doing now ? x K M B f Change of kinetic energy Energy dissipated by damper Change of potential energy

Example -- SDOF Suspension (Example) ME 375 – Spring 2003 Example -- SDOF Suspension (Example) Simplified Schematic (neglecting tire model) Suspension System Minimize the effect of the surface roughness of the road on the drivers comfort. From the “absolute zero” From the path From nominal position g M x K B x x p

Series Connection Û Springs in Series K1 K2 x1 x2 fS KEQ K1 K2 x1 x2 ME 375 – Spring 2003 Series Connection Springs in Series K1 K2 x1 x2 fS KEQ Û K1 K2 x1 x2 fS xj fS

Series Connection Û Dampers in Series x1 x2 x1 fD x2 fD fD fD B1 B2 ME 375 – Spring 2003 Series Connection Dampers in Series x1 x2 x1 fD x2 fD Û fD fD B1 B2 BEQ

Parallel Connection Û Springs in Parallel x1 x2 fS K1 K2 x1 x2 fS fS ME 375 – Spring 2003 Parallel Connection Springs in Parallel x1 x2 fS K1 K2 x1 x2 fS Û fS KEQ

Parallel Connection Û Dampers in Parallel x1 x2 x2 fD x1 BEQ B1 B2 fD ME 375 – Spring 2003 Parallel Connection Dampers in Parallel x1 x2 x2 fD x1 BEQ B1 B2 Û fD fD

Horizontal Two Degree of Freedom (TDOF) System ME 375 – Spring 2003 Horizontal Two Degree of Freedom (TDOF) System DOF = 2 K Absolute coordinates FBD K K Newton’s law

Horizontal Two Degree of Freedom (TDOF) System ME 375 – Spring 2003 Horizontal Two Degree of Freedom (TDOF) System K Static coupling Absolute coordinates Relative coordinates Dynamic coupling

Two DOF System – Matrix Form of EOM ME 375 – Spring 2003 Two DOF System – Matrix Form of EOM K K Input vector Absolute coordinates Output vector Mass matrix Damping matrix Relative coordinates Stiffness matrix SYMMETRIC NON-SYMMETRIC

MDOF Suspension Suspension System ME 375 – Spring 2003 MDOF Suspension Simplified Schematic (with tire model) Suspension System TRY THIS x p

MDOF Suspension Suspension System ME 375 – Spring 2003 MDOF Suspension Simplified Schematic (with tire model) Suspension System Assume ref. is when springs are Deflected by weights Car body Suspension Wheel Tire Road Reference

Example -- MDOF Suspension ME 375 – Spring 2003 Example -- MDOF Suspension Draw FBD Apply Interconnection Laws x2 M2 FS2 FD2 FS2 FD2 FS2 FD2 FS2 FD2 x1 M1 FS1 FD1 FS2 FD2 FS1 FD1

Example -- MDOF Suspension ME 375 – Spring 2003 Example -- MDOF Suspension Matrix Form Mass matrix Damping matrix Stiffness matrix Input Vector