engm 541 2011 01 lecture 1 modeling equilibria 2

7/31/2019 ENGM 541 2011 01 Lecture 1 Modeling Equilibria 2

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ENGMENGM 541, ENGM 670541, ENGM 670--X5X5

&& MECE 758MECE 758--X5X5Modeling and Simulation of Engineering SystemsModeling and Simulation of Engineering Systems

WinterWinter 20112011

Lecture 1:Lecture 1:

Introduction; Course Overview;Introduction; Course Overview;

Modeling Physical Systems,Modeling Physical Systems,

LumpedLumped--Parameter Equilibrium SystemsParameter Equilibrium Systems

M.G. LipsettM.G. LipsettDepartment of Mechanical EngineeringDepartment of Mechanical Engineering

University of AlbertaUniversity of Albertahttp://www.ualberta.ca/~mlipsett/ENGM541/ENGM541.htm

MG Lipsett, 2011 2

Department of Mechanical EngineeringEngineering Management Group

ENGM 541, ENGM 670-X5, MECE 758-X5 Modeling and Simulation of Engineering SystemsLecture 1: Course Introduction; Lumped-Parameter Equilibrium Systems

IntroductionIntroduction Engineering systems often comprise complicated assemblies of

components, which can have complex behaviours that are difficult to predict

Internet Sources: www.coolestgadgets.com; www.nasa.gov; www.microway.com.au; www.pbs.org; www.emercedesbenz.com; www.syncrude.com


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MathematicalMathematical AnalysisAnalysis in Engineeringin Engineering

Engineering analysis: formulating governing equations that

describe the behaviour of physical and technologicalsystems, for the purpose of analysis and design

Numerical analysis: solving mathematical equations usingalgorithms

Scientific computing: development of reliable numericalmodels that can be tested in a range of cases (includingknown benchmarks)

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What is Modeling?What is Modeling? A modelis a representation of knowledge

Rules, physical analogs, algebraic equations of physical laws

A systemis a bounded region comprising known elementsthat each interact in understandable ways

Applied numerical modeling has joined empiricalexperimentation and analytical methods for solvingproblems of mathematical physics

The types of systems of interest in this course include:

Models of physical systems Mechanical, electrical, thermal, structural, hydraulic, etc.

Combinations of different physical systems (mixed systems)

Models of material, energy, and information flow forengineering decisions Production systems

Economics

Scheduling

Inventory, and so on, and so on,


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What is Simulation?What is Simulation?

Simulations are solutions of equations that are functions of time

For continuous systems, we develop (and solve) differentialequations

Examples:

Vehicle dynamics

Thermofluid interactions

Industrial processes

Biological processes

Climate change, and so on, and so on,

Often the equations can not be solved in closed-form

Sometimes simulations are based on empirical understanding oftime-varying behaviour that is not expressed as differentialequations (correlations, discrete events, etc.). These are valuablefor systems that are not characterised well by differential equations.

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Scientific Computing at a GlanceScientific Computing at a Glance

(Adapted from A. Quarteroni, Mathematical Models in Science and Engineering,

Notices of the American Mathematical Society Jan 2009)

Interestingproblem

Data fromthe problem

Understandingof the problem

Defining the system

Uncertainty

Sensitivity

Parameter identification

Statistical analysis

Modeling thesystem

Geometry and mesh/network

Governing equations & analysis

Numerical approximation

Algorithms for solving

Computer

simulation &post-processing

Visualisation of results

Validation /Verification Comparison to known results

Benchmark cases

Experiments


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Decision

Engineering Analysis for Design at a GlanceEngineering Analysis for Design at a Glance

Possible

solution

Design

Performance

specifications

Model of

System behaviour

Problem

definition

Assessment of

Performance

of proposed solution

Modeling to predict

how a design will perform

is key to a successfulsolution

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ENGM 541 Course IntroductionENGM 541 Course Introduction Why do engineers need to learn about modeling and

simulation?

Most engineering problems are too complicated or complexto solve analytically

Engineers rely on numerical modeling and simulation toanalyse and design systems that have time-varying aspects

Engineering managers use models of technologies andbusiness processes for decision making

You may want do develop models to solve a technical orbusiness problem, by designing a solution and modelinghow you expect it to perform

You may need to interpret the results of models created byothers


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General Course OutlineGeneral Course Outline

Understanding concepts of formulating mathematical models

based on physics (and other rules of interaction) between theelements of a system

Formulating governing equations and choosing solutionmethods for different types of analyses of physical systems

Understanding advantages and limitations of numericalsolution methods

Understanding simple models for financial decisions andtechnological systems that have event-based dynamics

Using modeling and simulation for design

Presenting and interpreting analysis and simulation results

Analysing engineering systems and processes using generalpurpose programs: MATLABand SIMULINK

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ENGM 541 Course Overview (1)ENGM 541 Course Overview (1) Lecture Room:

Time Slots:

Instructor:

Office:

Office Hours:

TA:

Course Text:

E-Class &Course Web Site:

ETLE 2-001

Lectures: Wednesdays 5:00 pm 8:00 pm

Laboratories: Thursdays 5:00 pm 8:00 pm in ETLE 2-005

(required for ENGM 541 only)

MG Lipsett ([email protected])

Room 5-8J, Mechanical Engineering Building

(5th Floor West)

Wednesdays 1:003:00 pm (other times by appointment)

Masoud Mashkournia

Modeling and Analysis of Dynamic Systems,

by R. Esfandiari & B. Lu (CRC Press)

http://www.ualberta.ca/~mlipsett/ENGM541/ENGM541.htm- Lecture slides

- Assignments

- FAQ and announcements

- Worked examples and sample test questions

CHECK ECLASS & THE WEB SITE OFTEN !!


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ENGM 541 Course Overview (2)ENGM 541 Course Overview (2)

Marks:Marks:

Assignments: 25%Will be due in class and cannot be accepted after solutions are posted

ENGM 541 Labs: 5%

ENGM 541 Project: 15% (ENGM 670 & MECE 758: 20%)Individual, criteria to be announced, due April 6 2011 (before the exam)

Midterm Examination: 20%Wednesday March 2, 2011, 5:00 pm 7:00 pm in ETLE 2-001

Final Examination: 30%Wednesday April 13, 2011, 5:00 pm 7:30 pm in ETLE 2-001

Examinations will be open book & open notes Calculators are allowed but communication features must be

turned off (no computers)

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ENGM 670 & MECE 758 Course OutlineENGM 670 & MECE 758 Course Outline Lectures will be the same for ENGM 541, ENGM 758, and

ENGM 670

But there are additional requirements for grad students:

Supplementary readings

MECE 758: more on physical systems

ENGM 670 more on technological systems

More assignment problems

Additional scope for the individual project Different exam questions


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ENGMENGM 670 & MECE 758670 & MECE 758 Course Overview (2)Course Overview (2)

Marks:Marks:

Assignments: 25%Will be due in class and cannot be accepted after solutions are posted

Lab attendance is not required; but you are responsible forbeing able to do the Matlab coding covered in the labs

Project: 20%Individual, criteria to be announced, due April 6 2011 (before the exam)

Midterm Examination: 25%Wednesday March 2, 2011, 5:00 pm 7:00 pm in ETLE 2-001

Final Examination: 30%Wednesday April 13, 2011, 5:00 pm 7:30 pm in ETLE 2-001

Examinations will be open book & open notes

Calculators are allowed but communication features must beturned off (no computers)

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GeneralGeneral CourseCourse Success FactorsSuccess Factors Keys to success:

Do the homework to master model building

Try the examples in MATLAB

Check E-Class and the web site often

FAQ, worked examples, sample tests

Ask questions! (but think first)

This is a demanding course but you will gain avaluable approach to analysis and design

We have to unlearn some things to do generalsystems analysis correctly

We will also learn by doing

Ill do my best to be interesting


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ENGM 541 Course Overview (4)ENGM 541 Course Overview (4)

University policy: suspected cheating or plagiarism will bereported and investigated

Professional ethics and integrity

Do the right thing.It will gratify some people

and astonish the rest. -Mark Twain

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Your Instructor: MG LipsettYour Instructor: MG Lipsett

Professional Engineer since 1986

Research

Reliability of complex systems (anomalies, machinery diagnostics)

Robotics and automation (excavation, remote embedded sensing)

More sustainable processes for oilsands bitumen production andreclamation

Industrial Experience Atomic Energy of Canada Ltd (R&D in robotic inspection, hazardous

waste site remediation, reliability)

Syncrude Canada Ltd (mining automation & space robotics

teleoperation, extraction process R&D, mine maintenance & reliability)

Seven years in leadership and management roles (Operations, R&D,Projects)


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Engineering AnalysisEngineering Analysis

Types of analysis:

Two means of modeling physical systems:

Once a model has been developed, then numerical

procedures can be used to study system behaviour usingcomputers

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Modeling Physical SystemsModeling Physical Systems Consider a beam:

This is an inherently continuous structure. When weanalyse this beam for deflections, natural frequencies, etc.,we can start from one of two approaches.


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LumpedLumped--Parameter ModelParameter Model

The properties of the continuous system are visualised as

being separate from one another

The beam is modeled as a linkage mechanism

We find a set of algebraic equations from which we candetermine the deflections

The price we pay is one of approximating the physicalsystem at the modeling level.

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Continuous ModelContinuous Model Alternatively, the beam is modeled by deriving differential

equations that represent the continuous system

The solution to the differential equations requires that theybe approximated by algebraic equations (e.g. finitedifference expressions), for almost all non-trivial cases


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Solving Algebraic Equations of the ModelSolving Algebraic Equations of the Model

In either case, we are solving algebraic equations.

After the modeling is complete, we choose the type ofsolution:

We want to have a consistent way to set up problems andto solve them.

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Equilibrium Problems for LumpedEquilibrium Problems for Lumped--Parameter SystemsParameter Systems We are looking for steady-state solutions to problems where

the continuous system has been modeled using lumpedparameters.

We are concerned with systems of interconnectedelementselements. The elements within the problem have propertiesthat we must know before we can proceed.

Elements are connected at nodes.nodes. Here is an example of asystem network:

Loops are paths that start at a particular node, pass througha number of elements, and return to the original node.


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Loop and Node VariablesLoop and Node Variables

A system will have both loop and node variables.

Loop variables describe the path around the loop.Examples:

Node variables describe variables that come together at anode.

Examples:

Loop and node variables:

The loop and node variables are related by the constitutiverelationships of the elements.

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Formulating Constitutive RelationshipsFormulating Constitutive Relationships1. State the variables

2. Describe the element

3. Sketch the constitutive relationship.

4. Use an analytic expression for the relationship


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Constitutive Relationship Example #1Constitutive Relationship Example #1

1. State variables:

2. Describe element:

3. Sketch:

4. Write analytical relationship:

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Constitutive Relationship Example #2Constitutive Relationship Example #21. State variables:

2. Describe element:

3. Sketch:

4. Write analytical relationship:


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Admissibility LawsAdmissibility Laws

The node laws satisfy the admissibility requirement that the

node variable is conserved at a node

The loop laws are similar (but different). Loop variables aregoverned by loop admissibility laws that require the value ofthe loop variable at a node to have only one value

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GeneralisingGeneralising KirchoffsKirchoffs LawLaw We use a general approach for system networks using the

principles of Kirchoffs Laws.

Kirchoffs Laws for electrical circuits use the physical lawsof conservation of charge (node law) and conservation ofenergy added or taken by a potential field (around loops,mesh law), including dissipation. Gain or loss around anentire loop has to be zero (because there is no net changein the location with respect to the field).

For other types of physical systems, we construct ourvariable assignments so that we can exploit similar physicallaws:

Conservation of momentum law (DAlemberts law for forces)

Conservation of mass law for flows, etc., etc.

For non-physical systems, we need similar loop & nodelaws


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Example: LumpedExample: Lumped--Parameter Electrical NetworkParameter Electrical Network

C

L

R1

R2

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Some Equilibrium Element TypesSome Equilibrium Element TypesType Node Variable Loop Variable

Mechanical

Electrical

Fluid Flow

Heat Transfer


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General Procedure for Setting Up A ProblemGeneral Procedure for Setting Up A Problem

1. Choose the variable in which you want your final equations

expressed2. Choose variables so as to satisfy the pertinent admissibility

requirement

3. Choose other variable type & write as many equations asnecessary to check that admissibility is satisfied.

4. Relate the loop and node variables using the constitutiverelationships.

5. Eliminate all but the chosen variables (all of one type) fromthe equations. Substitute in the equations, and group terms.

6. Non-dimensionalise the variables.

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Example: LumpedExample: Lumped--Parameter Mechanical SystemParameter Mechanical System

To model this system, we have two possible approaches:1) Find the forces in the springs (node variables)

2) Find the displacements of the carts (loop variables)

K/6

K/6

K/3 K/2

2P P


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Case 1: Find the Forces in the SpringsCase 1: Find the Forces in the Springs

1) Choose a set of node variables (forces at nodes).

2) Satisfy node admissibility.

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Case 1: Forces in Springs (2)Case 1: Forces in Springs (2)3) Choose loop variables (displacements across elements)

and ensure they satisfy loop admissibility.

K/6

K/6

K/3 K/2

2P P


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Case 1: Forces in Springs (3)Case 1: Forces in Springs (3)

4) Apply constitutive relationships. For linear spring element,

this will be: fi = ki inode variable loop variable

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Case 1: Forces in Springs (4)Case 1: Forces in Springs (4)5) Substitute into the loop equations.


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Case 1: Forces in Springs (5)Case 1: Forces in Springs (5)

6) Try to express in non-dimensional form.

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Case 2: Find the Displacements in the NodesCase 2: Find the Displacements in the Nodes1) Choose a set of loop variables.

2) Satisfy loop admissibility.

K/6

K/6

K/3 K/2

2P P


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Case 2: Displacement of Nodes (2)Case 2: Displacement of Nodes (2)

3) Choose node variables (forces at nodes) and ensure they

satisfy node admissibility.

4) Apply constitutive relationships.

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Case 2: Displacement of Nodes (3)Case 2: Displacement of Nodes (3)5) Substitute into node equations.


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Case 2: Displacement of Nodes (4)Case 2: Displacement of Nodes (4)

Are we done yet? Well, not quite.

From the solution fory1,y2, go back to the definition of the non-dimensional variables to solve for the displacement (theloop variables); then, from their solution, we can find forcesusing the constitutive relationships.

These two methods are called Direct ApproachesDirect Approaches.

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Extremum FunctionsExtremum Functions The other way of formulating the equations governing

systems is to use extremum functions. This includesenergy methods.

We make up a scalar function from the constitutiverelationships of all the elements in the system, and searchfor an extreme value of the function (e.g. minimumpotential energy).

We go back to our original definition of a constitutiverelationship to define two quantities:

1. Content U(energy)

2. Co-Content U* (co-energy)


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EnergyEnergy

Area under the curve is the energy Uin the element:

We writep (which is a node variable) as a function of q(loop variable) and Ubecomes a function of q only.

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CoCo--EnergyEnergy

Similarly to energy, with co-energy U* as a function ofp only

For all sets of state variables satisfying node (loop)admissibility, those also satisfying loop (node) admissibilitywill render the co-energy (energy) an extreme value.


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Break Time: Flexibility of Thinking ProblemsBreak Time: Flexibility of Thinking Problems

8D 24H = 1W

3P = 6

HH & MH @ 12 = N or M

4J+4Q+4K = All the FC

S&M&T&W&T&F&S are D of W

23Y 3Y = 2D

E 8 = Z

Y + 2D = T

C + 6D = NYE

Y S S A = W

NN = GN

N + P + SM = S of C

1 + 6Z = 1M

R = R = R

1B in the H = 2 in the B

Each problem is an equation, which can be solved by substitut ing the appropriate wordsfor the letters. Examples:

3F = 1Y (3 Feet = 1 Yard)4LC = GL (4 Leaf Clover = Good Luck)

Source: A Whack on the Side of the Head, R.von Oech

engm 541 2011 01 lecture 1 modeling equilibria 2

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