Texas instruments TI-89, TI-92 PLUS ME Pro

Page 1
®
ME•Pro
Mechanical Engineeri ng
User’s Manual A software Application
for the TI-89 and TI-92 Plus
Version 1.0 by
da V
inci Technologies Group Inc
Page 2
ME••Pro®
A software Application
For TI-89 and TI-92 Plus
User’s Guide
August 2000
© da Vinci Technologies Group, Inc.
Rev. 1.0
da Vinci Technologies Group, Inc.
1600 S.W. Western Blvd
Suite 250
Corvallis, OR 97333
www.dvtg.com
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Notice

This manual and the examples contained herein are provided “as is” as a supplement to ME
Pro
application software available from Texas Instruments for TI-89, and TI-92 Plus platforms. da Vinci
Te chnologies Gro up, Inc. (“da V inci”) makes no warranty of any kind with r egard t o this manual or the accompanying software, including, but not limited to, the implied warranties of merchantability and fitness for a particular purpose. da Vinci shall not be lia ble for any errors or f or
incidenta l or consequ ent ial dama ges in connec tion with the furnishing, per f ormance, or use of this manual , or the examp l es her ei n .
Copyr ight da Vinci Technologies Gr oup, Inc. 2000. All rights reserved. PocketProfessional and ME•Pro are registered trademarks of da Vinci Technologies Group, Inc. TI-GRAPH LINK is a trademark of Texas Instruments Incorporated, and Acrobat is a registered trademark of Adobe Systems Incorporated.
We welcome your comments on the software and the manual. Forward your comments, preferably by e-mail, to da Vinci at support@dvtg.com.
Acknowledgements
The ME•Pro software was developed by Chris Bunsen, Dave Conklin, Michael Conway, Curtis Gammel, and Megh a Shyam with the gene r ou s support of TI’ s developmen t te a m . The user ’ s gu ide wa s developed by Mi cha e l Conway, Curtis Gammel, Melinda Shaffer, and Megha Shyam. Many helpful comments from the testers at Texas Instrume nts and other locations during β testing phase is gratefully acknowledged.
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Table of Contents

TABLE OF CONTENTS4
HAPTER
C
NTRODUCTION TO
1: I
1.1 Key Featur es of MEPro....................................................................................................12
1.2 Purchasing, Downloading and Installing MEPro...............................................................13
1.3 Ordering a Manual.............................................................................................................13
1.4 Memory Requirements......................................................................................................13
1.5 Differences between TI-89 and TI-92 plus.......................................................................... 13
1.6 Starting MEPro................................................................................................................13
1.7 How to use this Manual......................................................................................................14
1.8 Manual Disclaimer............................................................................................................. 14
1.9 Summary.................................................................................................................... ....... 15
PART I: ANALYSIS……………………………………………………………………………………...16
HAPTER
C
NTRODUCTION TO ANALYSIS
2: I
2.1 Introduction.......................................................................................................................17
2.2 Features of Analysis...........................................................................................................18
2.3 Finding Analysis................................................................................................................18
2.4 Solving a Problem in Analysis ...........................................................................................18
2.5 Tips for Analysis ...............................................................................................................20
2.6 Function keys ....................................................................................................................20
2.7 Session Folders, Variable Names.......................................................................................22
2.8 Overwriting of variable values in graphing......................................................................... 22
2.9 Reserved Variables............................................................................................................22
HAPTER
C
TEAM TABLES
3: S
3.1 Saturated Steam Properties................................................................................................. 23
3.2 Superheated Steam Properties............................................................................................23
3.3 Air Properties ....................................................................................................................23
3.4 Using Steam Tables ...........................................................................................................24
3.5 Validity Range for Temperature and Pressure..................................................................... 25
HAPTER
C
HERMOCOUPLES
4: T
4.1 Introduction.......................................................................................................................26
4.2 Using the Thermocouples Function ....................................................................................26
4.3 Basis for Temperature/Voltage Conversions.......................................................................27
HAPTER
C
APITAL BUDGETING
5: C
5.1 Using Capital Budgeting.................................................................................................... 28
HAPTER
C
FOR MECHANICAL ENGINEERS
6: EE
6.1 Impedance Calculations..................................................................................................... 32
6.2 Circuit Performance...........................................................................................................33
6.3 Wye ↔ ∆ Conversion........................................................................................................ 34
HAPTER
C
7: E
FFLUX
..........................................................................................................................36
7.1 Constant Liquid Level........................................................................................................36
7.2 Varying Liquid Level......................................................................................................... 36
7.3 Conical Vessel................................................................................................................... 37
7.4 Horizontal Cylinder ........................................................................................................... 38
7.5 Large Rectangular Orifice.................................................................................................. 38
7.6 ASME Weirs......................................................................................................................... 39
7.6.1 Rectangular Notch .......................................................................................................... 39
7.6.2 Triangular Weir ..............................................................................................................39
7.6.3 Suppressed Weir .............................................................................................................40
7.6.4 Cipolletti Weir................................................................................................................40
HAPTER
C
ECTION PROPERTIES
8: S
8.1 Rectangle........................................................................................................................... 42
••••
RO
P
ME
............................................................................................12
..........................................................................................17
................................................................................................................23
............................................................................................................26
.......................................................................................................28
...................................................................................32
.....................................................................................................42
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8.2 Hollow Rectangle.............................................................................................................. 43
8.3 Circle.................................................................................................................................43
8.4 Circular Ring.....................................................................................................................44
8.5 Hollow Circle....................................................................................................................45
8.6 1 Section - Uneven.............................................................................................................45
8.7 I Section - Even ................................................................................................................. 46
8.8 C Section........................................................................................................................... 47
8.9 T Section ...........................................................................................................................48
8.10 Trapezoid.........................................................................................................................48
8.11 Polygon ........................................................................................................................... 49
8.12 Hollow Polygon............................................................................................................... 50
HAPTER
C
ARDNESS NUMBER
9: H
.......................................................................................................52
9.1 Compute Hardness Number ...............................................................................................52
PART II: EQUATIONS………………………………………………………………………………… ...54
HAPTER
C
NTRODUCTION TO EQUATIONS
10: I
.....................................................................................55
10.1 Solving a Set of Equations ...............................................................................................55
10.2 Viewing an Equation or Result in Pretty Print ..................................................................56
10.3 Viewing a Result in different units ...................................................................................56
10.4 Viewing Multiple Solutions..............................................................................................57
10.5 when (…) - conditional constraints when solving equations .............................................. 58
10.5 Arbitrary Integers for periodic solutions to trigonometric functions................................... 58
10.7 Partial Solutions...............................................................................................................59
10.8 Copy/Paste.......................................................................................................................59
10.9 Graphing a Function.........................................................................................................59
10.10 St oring a nd re c alling variabl e value s in M EPro-creation of session folders.................... 61
10.11 solve, nsolve, and csolve and user-defined functions (UDF)........................................ 61
10.12 Entering a guessed value for the unknown using nsolve ..................................................61
10.13 Why can't I compute a solution?.....................................................................................62
10.14 Care in choosing a consistent set of equations.................................................................62
10.15 Not e s for the adva nced user in troubleshooting calculations ............................................ 62
HAPTER
C
EAMS AND COLUMNS
11: B
..................................................................................................64
11.1 Simple Beams......................................................................................................................64
11.1.1 Uniform Load...............................................................................................................64
11.1.2 Point Load ....................................................................................................................66
11.1.3 Moment Load ............................................................................................................... 68
11.2 Cantilever Beams.................................................................................................................70
11.2.1 Uniform Load...............................................................................................................70
11.2.2 Point Load ....................................................................................................................71
11.2.3 Moment Load ............................................................................................................... 73
11.3. Columns............................................................................................................................. 75
11.3.1 Buckling.......................................................................................................................75
11.3.2 Eccentricity, Axial Load................................................................................................ 76
11.3.3 Secant Formula ............................................................................................................. 77
11.3.4 Imperfections in Columns .............................................................................................79
11.3.5 Inelastic Buckling .........................................................................................................81
HAPTER
C
12: EE
FOR MES
................................................................................................................83
12.1 Basic Electricity...................................................................................................................83
12.1.1 Resistance Formulas......................................................................................................83
12.1.2 Ohm’s Law and Power..................................................................................................84
12.1.3 Temperature Effect ....................................................................................................... 85
12.2 DC Motors...........................................................................................................................86
12.2.1 DC Series Motor ...........................................................................................................86
12.2.2 DC Shunt Motor............................................................................................................88
12.3 DC Generators..................................................................................................................... 90
12.3.1 DC Series Generator ..................................................................................................... 90
12.3.2 DC Shunt Generator......................................................................................................91
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12.4 AC Motors...........................................................................................................................92
12.4.1 Three φ Induction Motor I............................................................................................. 92
12.4.2 Three φ Induction Motor II............................................................................................ 94
12.4.3 1 Induction Motor......................................................................................................... 96
HAPTER
C
13: G
AS LAWS
.....................................................................................................................98
13.1 Ideal Gas Laws....................................................................................................................98
13.1.1 Ideal Gas Law...............................................................................................................98
13.1.2 Constant Pressure.......................................................................................................... 99
13.1.3 Constant Volume........................................................................................................ 101
13.1.4 Constant Temperature.................................................................................................103
13.1.5 Internal Energy/Enthalpy............................................................................................. 104
13.2 Kinetic Gas Theory........................................................................................................ 106
13.3 Real Gas Laws................................................................................................................... 108
13.3.1 van der Waals: Specific Volume..................................................................................108
13.3.2 van der Waals: Molar form.......................................................................................... 109
13.3.3 Redlich-Kwong: Sp.Vol.............................................................................................. 110
13.3.4 Redlich-Kwong: Molar................................................................................................112
13.4 Reverse Adiabatic.......................................................................................................... 113
13.5 Polytropic Process.......................................................................................................... 115
HAPTER
C
14: H
EAT TRANSFER
.........................................................................................................118
14.1 Basic Transfer Mechanisms ............................................................................................... 118
14.1.1 Conduction ................................................................................................................. 118
14.1.2 Convection..................................................................................................................120
14.1.3 Radiation .................................................................................................................... 121
14.2 1 1D Heat Transfer ................................................................................................................ 122
14.2.1 Conduction..................................................................................................................... 122
14.2.1.1 Plane Wall ............................................................................................................... 122
14.2.1.2 Convective Source ................................................................................................... 123
14.2.1.3 Radiative Source......................................................................................................125
14.2.1.4 Plate and Two Fluids................................................................................................127
14.2.2 Electrical Analogy .......................................................................................................... 128
14.2.2.1 Two Conductors in Series......................................................................................... 129
14.2.2.2 Two Conductors in Parallel ...................................................................................... 131
14.2.2.3 Parallel-Series.......................................................................................................... 132
14.2.3 Radial Systems ...............................................................................................................135
14.2.3.1 Hollow Cylinder....................................................................................................... 135
14.2.3.2 Hollow Sphere ......................................................................................................... 136
14.2.3.3 Cylinder with Insulation Wrap.................................................................................. 137
14.2.3.4 Cylinder - Critical radius..........................................................................................139
14.2.3.5 Sphere - Critical radius.............................................................................................141
14.3 Semi-Infinite Solid.............................................................................................................142
14.3.1 Step Change Surface Temperature............................................................................... 142
14.3.2 Constant Surface Heat Flux......................................................................................... 143
14.3.3 Surface Convection..................................................................................................... 145
14.4 Radiation........................................................................................................................... 146
14.4.1 Blackbody Radiation................................................................................................... 146
14.4.2 Non-Blackbody radiation ............................................................................................148
14.4.3 Thermal Radiation Shield............................................................................................ 149
HAPTER
C
HERMODYNAMICS
15: T
.....................................................................................................152
15.1 Fundamentals................................................................................................................. 152
15.2 System Properties .............................................................................................................. 153
15.2.1 Energy Equations........................................................................................................ 153
15.2.2 Maxwell Relations ...................................................................................................... 155
15.3 Vapor and Gas Mixture...................................................................................................... 157
15.3.1 Saturated Liquid/Vapor ............................................................................................... 157
15.3.2 Compressed Liquid-Sub cooled ...................................................................................159
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15.4 Ideal Gas Properties........................................................................................................... 160
15.4.1 Specific Heat...............................................................................................................160
15.4.2 Quasi-Equilibrium Compression..................................................................................162
15.5 First Law...............................................................................................................................163
15.5.1 Total System Energy...................................................................................................163
15.5.2 Closed System: Ideal Gas...............................................................................................167
15.5.2.1 Constant Pressure.....................................................................................................167
15.5.2.2 Binary Mixture......................................................................................................... 169
15.6 Second Law........................................................................................................................... 171
15.6.1 Heat Engine Cycle .......................................................................................................... 171
15.6.1.1 Carnot Engine..........................................................................................................171
15.6.1.2 Diesel Cycle............................................................................................................. 173
15.6.1.3 Dual Cycle............................................................................................................... 177
15.6.1.4 Otto Cycle................................................................................................................180
15.6.1.5 Brayton Cycle .......................................................................................................... 184
15.6.2 Clapeyron Equation.....................................................................................................186
HAPTER
C
ACHINE DESIGN
16: M
.......................................................................................................188
16.1 Stress: Machine Elements .................................................................................................. 188
16.1.1 Cylinders .................................................................................................................... 188
16.1.2 Rotating Rings............................................................................................................ 189
16.1.3 Pressure and Shrink Fits.............................................................................................. 190
16.1.4 Crane Hook................................................................................................................. 192
16.2 Hertzian Stresses................................................................................................................193
16.2.1 Two Spheres...............................................................................................................193
16.2.2 Two Cylinders............................................................................................................ 195
16.3.1 Bearing Life................................................................................................................ 197
16.3.2 Petroff's law................................................................................................................ 198
16.3.3 Pressure Fed Bearings................................................................................................. 199
16.3.4 Lewis Formula............................................................................................................200
16.3.5 AGMA Stresses.......................................................................................................... 201
16.3.6 Shafts..........................................................................................................................203
16.3.7 Clutches and Brakes ........................................................................................................... 204
16.3.7.1 Clutches....................................................................................................................... 204
16.3.7.1.1 Clutches ................................................................................................................ 204
16.3.7.2 Uniform Wear - Cone Brake.....................................................................................206
16.3.7.3 Uniform Pressure - Cone Brake................................................................................ 207
16.4 Spring Design........................................................................................................................208
16.4.1 Bending.......................................................................................................................... 208
16.4.1.1 Rectangular Plate.....................................................................................................208
16.4.1.2 Triangular Plate........................................................................................................ 209
16.4.1.3 Semi-Elliptical......................................................................................................... 210
16.4.2 Coiled Springs................................................................................................................ 212
16.4.2.1 Cylindrical Helical - Circular wire............................................................................ 212
16.4.2.2 Rectangular Spiral.................................................................................................... 213
16.4.3 Torsional Spring............................................................................................................. 215
16.4.3.1 Circular Straight Bar ................................................................................................215
16.4.3.2 Rectangular Straight Bar...........................................................................................216
16.4.4 Axial Loaded.................................................................................................................. 217
16.4.4.1 Conical Circular Section........................................................................................... 217
16.4.4.2 Cylindrical - Helical..................................................................................................... 219
16.4.4.2.1 Rectangular Cross Section .....................................................................................219
16.4.4.2.2 Circular Cross Section ........................................................................................... 220
HAPTER
C
UMPS AND HYDRAULICS
17: P
............................................................................................222
17.1 Basic Definitions ...........................................................................................................222
17.2 Pump Power .................................................................................................................. 223
17.3 Centrifugal Pumps.............................................................................................................225
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17.3.1 Affinity Law-Variable Speed.......................................................................................225
17.3.2 Affinity Law-Constant Speed......................................................................................226
17.3.3 Pump Similarity.......................................................................................................... 227
17.3.4 Centrifugal Compressor............................................................................................... 228
17.3.5 Specific Speed ............................................................................................................229
HAPTER
C
AVES AND OSCILLATION
18: W
...........................................................................................231
18.1 Simple Harmonic Motion...................................................................................................231
18.1.1 Linear Harmonic Oscillation........................................................................................231
18.1.2 Angular Harmonic Oscillation..................................................................................... 232
18.2 Pendulums.........................................................................................................................233
18.2.1 Simple Pendulum........................................................................................................233
18.2.2 Physical Pendulum...................................................................................................... 235
18.2.3 Torsional Pendulum....................................................................................................236
18.3 Natural and Forced Vibrations.............................................................................................. 236
18.3.1 Natural Vibrations...........................................................................................................236
18.3.1.1 Free Vibration..........................................................................................................236
18.3.1.2 Overdamped Case (ξ>1)........................................................................................... 238
18.3.1.3 Critical Damping (ξ=1) ............................................................................................ 239
18.3.1.4 Underdamped Case (ξ<1)......................................................................................... 241
18.3.2 Forced Vibrations ........................................................................................................... 244
18.3.2.1 Undamped Forced Vibration.....................................................................................244
18.3.2.2 Damped Forced Vibration ........................................................................................ 245
18.3.3 Natural Frequencies...........................................................................................................247
18.3.3.1 Stretched String........................................................................................................ 247
18.3.3.2 Vibration Isolation ................................................................................................... 248
18.3.3.3 Uniform Beams............................................................................................................249
18.3.3.3.1 Simply Supported..................................................................................................250
18.3.3.3.2 Both Ends Fixed.................................................................................................... 251
18.3.3.3.3 1 Fixed End / 1 Free End....................................................................................... 252
18.3.3.3.4 Both Ends Free...................................................................................................... 254
18.3.3.4 Flat Plates.................................................................................................................... 255
18.3.3.4.1 Circular Flat Plate.................................................................................................. 255
18.3.3.4.2 Rectangular Flat Plate............................................................................................257
HAPTER
C
EFRIGERATION AND AIR CONDITIONING
19: R
...................................................................259
19.1 Heating Load.................................................................................................................259
19.2 Refrigeration......................................................................................................................261
19.2.1 General Cycle .............................................................................................................261
19.2.2 Reverse Carnot............................................................................................................ 262
19.2.3 Reverse Brayton.......................................................................................................... 263
19.2.4 Compression Cycle.....................................................................................................264
HAPTER
C
TRENGTH MATERIALS
20: S
...............................................................................................267
20.1 Stress and Strain Basics ..................................................................................................... 267
20.1.1 Normal Stress and Strain.............................................................................................267
20.1.2 Volume Dilation.........................................................................................................268
20.1.3 Shear Stress and Modulus............................................................................................ 269
20.2 Load Problems...................................................................................................................270
20.2.1 Axial Load.................................................................................................................. 270
20.2.2 Temperature Effects....................................................................................................271
20.2.3 Dynamic Load ............................................................................................................272
20.3 Stress Analysis ..................................................................................................................274
20.3.1 Stress on an Inclined Section ....................................................................................... 274
20.3.2 Pure Shear...................................................................................................................275
20.3.3 Principal Stresses........................................................................................................276
20.3.4 Maximum Shear Stress................................................................................................277
20.3.5 Plane Stress - Hooke's Law..........................................................................................278
20.4 Mohr’s Circle Stress..........................................................................................................280
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20.4 Mohr’s Circle Stress.......................................................................................................280
20.5 Torsion.............................................................................................................................. 281
20.5.1 Pure Torsion............................................................................................................... 281
20.5.2 Pure Shear...................................................................................................................283
20.5.3 Circular Shafts............................................................................................................ 284
20.5.4 Torsional Member.......................................................................................................285
HAPTER
C
LUID MECHANICS
21: F
.....................................................................................................288
21.1 Fluid Properties .......................................................................................................... ....... 288
21.1.1 Elasticity..................................................................................................................... 288
21.1.2 Capillary Rise............................................................................................................. 289
21.2 Fluid Statics............................................................................................................. ............. 291
21.2 1 Pressure Variation........................................................................................................... 291
21.2.1.1 Uniform Fluid.......................................................................................................... 291
21.2.1.2 Compressible Fluid..................................................................................................292
21.2.1 Pressure Variation........................................................................................................... 292
21.2.1.3 Troposphere.............................................................................................................292
21.2.1.4 Stratosphere............................................................................................................. 293
21.2.2.1 Floating Bodies........................................................................................................294
21.2.2.2 Inclined Plane/Surface.............................................................................................. 296
21.3 Fluid Dynamics.................................................................................................................297
21.3.1 Bernoulli Equation...................................................................................................... 297
21.3.2 Reynolds Number ....................................................................................................... 299
21.3.3 Equivalent Diameter....................................................................................................300
21.3.4 Fluid Mass Acceleration.................................................................................................. 302
21.3.4.1 Linear Acceleration .................................................................................................. 302
21.3.4.2 Rotational Acceleration ............................................................................................303
21.4 Surface Resistance............................................................................................................. 304
21.4.1 Laminar Flow – Flat Plate ........................................................................................... 304
21.4.2 Turbulent Flow – Flat Plate.........................................................................................306
21.4.3 Laminar Flow on an Inclined Plane..............................................................................309
21.5 Flow in Conduits ................................................................................................................... 311
21.5.1 Lamina r F l ow: Smooth Pipe....................................................................................... 311
21.5.2 Turbulent Flow: Smooth Pi pe..................................................................................... 313
21.5.3 Turbulent Flow: Rough Pipe....................................................................................... 316
21.5.4 Flow pipe Inlet............................................................................................................ 319
21.5.5 Series Pipe System...................................................................................................... 321
21.5.6 Parallel Pipe System .................................................................................................... 322
21.5.7 Venturi Meter.................................................................................................................324
21.5.7.1 Incompressible Flow ................................................................................................ 324
21.5.7.2 Compressible Flow...................................................................................................327
21.6 Impulse/Momentum............................................................................................................... 330
21.6.1 Jet Propulsion ............................................................................................................. 330
21.6.2 Open Jet..........................................................................................................................331
21.6.2.1 Vertical Plate ........................................................................................................... 331
21.6.2.2 Horizontal Plate ....................................................................................................... 332
21.6.2.3 Stationary Blade.......................................................................................................333
21.6.2.4 Moving Blade..........................................................................................................335
HAPTER
C
YNAMICS AND STATICS
22: D
.............................................................................................338
22.1 Laws of Motion ............................................................................................................. 338
22.2 Constant A ccelerat ion........................................................................................................ 340
22.2.1 Linear Motion............................................................................................................. 340
22.2.2 Free Fall .....................................................................................................................341
22.2.3 Circular Motion........................................................................................................... 342
22.3 Angular Motion................................................................................................................. 343
22.3.1 Rolling/Rotation.......................................................................................................... 343
22.3.2 Forces in Angular Motion............................................................................................ 345
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22.3.3 Gyroscope Motion.......................................................................................................346
22.4 Projectile Motion ........................................................................................................... 347
22.5 Collisions..............................................................................................................................349
22.5.1 Elastic Collisions ............................................................................................................ 349
22.5.1.1 1D Collision............................................................................................................. 349
22.5.1.2 2D Collisions........................................................................................................... 350
22.5.2 Inelastic Collisions.......................................................................................................... 351
22.5.2.1 1D Collisions........................................................................................................... 351
22.5.2.2 Oblique Collisions.................................................................................................... 352
22.6 Gravitational Effects.......................................................................................................... 354
22.6.1 Law of Gravitation...................................................................................................... 354
22.6.2 Kepler's Laws ............................................................................................................. 356
22.6.3 Satellite Orbit..............................................................................................................358
22.7 Friction..............................................................................................................................360
22.7.1 Frictional Force........................................................................................................... 360
22.7.2 Wedge........................................................................................................................ 362
22.7.3 Rotating Cylinder........................................................................................................363
22.8 Statics................................................................................................................................364
22.8.1 Parabolic cable............................................................................................................ 364
22.8.2 Catenary cable............................................................................................................365
PART III: REFERENCE ……………………………………………………………………….368
HAPTER
C
NTRODUCTION TO REFERENCE
23: I
...................................................................................369
23.1 Introduction................................................................................................................... 369
23.2 Finding Reference.......................................................................................................... 369
23.3 Reference Screens.......................................................................................................... 370
23.4 Using Reference Tables .................................................................................................370
HAPTER
C
NGINEERING CONSTANTS
24: E
..........................................................................................372
24.1 Using Constants............................................................................................................. 372
HAPTER
C
25: T
RANSFORMS
..............................................................................................................374
25.1 Using Transforms ..........................................................................................................374
HAPTER
C
ALVES AND FITTING LOSS
26: V
.........................................................................................376
26.1 Valves and Fitting Loss Screens.....................................................................................376
HAPTER
C
RICTION COEFFICIENTS
27: F
............................................................................................377
27.1 Friction Coefficients Screens..........................................................................................377
HAPTER
C
ELATIVE ROUGHNESS OF PIPES
28: R
................................................................................378
28.1 Relative Roughness Screens ........................................................................................... 378
HAPTER
C
ATER-PHYSICAL PROPERTIES
29: W
...................................................................................379
29.1 Water-Physical Properties Screens..................................................................................379
HAPTER
C
ASES AND VAPORS
30: G
.....................................................................................................380
30.1 Gases and Vapors S creens.............................................................................................. 381
HAPTER
C
HERMAL PROPERTIES
31: T
................................................................................................382
31.1 Thermal Properties Screens............................................................................................ 382
HAPTER
C
UELS AND COMBUSTION
32: F
............................................................................................383
32.1 Fuels and Combustion Screens.......................................................................................383
HAPTER
C
33: R
EFRIGERANTS
...........................................................................................................385
33.1 Refrigerants Screens ...................................................................................................... 386
HAPTER
C
34: SI P
REFIXES
...............................................................................................................387
34.1 Using SI Prefixes........................................................................................................... 387
HAPTER
C
REEK ALPHABET
35: G
......................................................................................................388
PART IV: APPENDIX AND INDEX…………………………………………………………….…….389
PPENDIX
A
REQUENTLY ASKED QUESTIONS
A F
.................................................................................390
A.1 Questions and Answers...................................................................................................390
A.2 General Questions........................................................................................................... 390
A.3 Analysis Questions.......................................................................................................... 392
A.4 Equations Questions........................................................................................................ 392
A.5 Graphing.........................................................................................................................395
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A.6 Reference........................................................................................................................ 396
PPENDIX
A
B W
ARRANTY
ECHNICAL SUPPORT
, T
................................................................................397
B.1 da Vinci License Agreement............................................................................................397
B.2 How to Contact Customer Support................................................................................... 398
PPENDIX
A
C: TI-89 & TI-92 P
LUS
EYSTROKE AND DISPLAY DIFFERENCES
- K
...................................399
C.1 Display Property Differences between the TI-89 and TI-92 Plus......................................399
C.2 Keyboard Differences Between TI-89 and TI-92 Plus ..................................................... 400
PPENDIX
A
RROR MESSAGES
D E
.......................................................................................................404
D.1 General Error Messages..................................................................................................404
D.2 Analysis Error Messages................................................................................................. 405
D.3 Equation Messages.......................................................................................................... 405
D.4 Reference Error Messages...............................................................................................406
PPENDIX E: SYSTEM VARIABLES AND RESERVED NAMES
A
.............................................................407
INDEX ……………………………………………………………………………………………………408
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Chapter 1: Introduction to ME•••Pro
Thank you for purchasing the MEPro, a m ember of the PocketProfessiona l® Pro software series designed by da Vin ci Technologies Group, Inc., t o meet the porta ble computing needs of student s an d p rofessionals in mechanical engineering. The software is organized in a hierarchical manner so t hat the topics easy to
find. We hope that you will find the MEPro to be a valuable companion in your career as a student and a professional of m echanical engi neering.
Topics in this chapter include:
Key Features of MEPro Purchasing, Download and Installing MEPro Ordering a Manual Memory Requirements Differences between the TI-89 and TI-92 Plus Starting the MEPro How to use this Manual Manual Disclaimer Summary
1.1 Key Features of ME•••Pro
The manual is organized into t hree sections representing the main menu headings of MEPro.
Analysis Equations Reference
Steam Tables Beams and Columns Engineering Constants Thermocouples EE For MEs Transforms Capital Budgeting Gas Laws Valves/Fitting Loss EE For MEs Heat Transfer Friction Coefficients Efflux Thermodynamics Roughness of Pipes Section Properties Machine Design Water Physical Properties Hardness Number Pumps and Hydraulic Machines Gases and Vapors Waves and Oscillation Thermal Properties Refrigeration and Air Conditioning Fuels and Combustion Strength Materials Refrigerants Fluid Mechanics SI Prefixes Dynamics and Statics Greek Alphabet
Thes e main topi c headings a re further divi d ed int o sub- topics. A brief descri ption of the main sections of the software is listed below:
Analysis: Chapt er s 2-9
Analysis is organized into 7 topics and 25 sub-topics. The software tools available in this section incorporate a va riety of an alysis methods used by mech anical en g ineers. Examples include Steam Tables, Thermocouple Calculations, EE for MEs; Efflux, Section Properties, Hardness Number Computations and Capital Budgeting. Wh er e appropriate, dat a en t er ed supports commonly used units.
Equations: Chapters 10-22
This section contains over 1000 equations organized under 12 major subjects in over 150 sub -topics. The equations in each sub-topic have been selected to provide maximum coverage of the subject material. In
MEPro for TI-89, TI-92 Plus Chapter 1 - Introduction to ME-Pro
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addition, the math engine is able to compute multiple or partial solutions to the equation sets. The computed values are filtered to identify results that h ave engineering merit. A powerful built-in unit management feature permits inputs in SI or other customary measurement systems. Over 80 diagrams help clar ify the essentia l na ture of th e problems covered by the equations. Topics covered include, Beams
and Columns; EE for MEs; Gas Laws; Heat Transfer; Thermodynamics; Machine Design; Pumps and Hydraulic Machines; Waves and Oscillation; Strength of Materials; Fluid Mechanics; and, Dynamics and Statics.
Reference: Chapters 23-25
The Reference section contains tables of information commonly needed by mechanical engineers. Topics include, values for Constants used by mechanical engineers; Laplace and Fourier Transform tables;
Valves and Fitting Loss; Friction Coefficient; Roughness of Pipes; Water Physical Properties; Gases and Vapors; Thermal Properties; Fuels and Combustion; Refrigerants; SI prefixes; and the Greek Alphabet.
1.2 Purchasing, Downloading and Installing ME•••Pro
The ME•Pro software can on ly be purch ased on- lin e from the Tex a s Instruments I nc. Online Stor e at
http://www.ti.com/calc/docs/store.htm. Th e software can be installed directly from your computer to your
calculator using TI-GRAPH LINK downloading and installing MEPro software are available from TI’s website.
TM
hardware and software (sold separ at ely). Directions for purchasing,
1.3 Ordering a Manual
Chapters and Appe ndices of the Manual for MEPro can be downloa ded through TI’ s Web Store an d viewed using the free Adobe Acrobat Reader Printed manuals can be purchased separately from da Vinci Technologies Group, Inc. by visiting the website http://www.dvtg.com/ticalcs/docs
TM
that can be downloaded from http://www.adobe.com.
or calling (541) 754-2860, Extension 100.
1.4 Memory Requirements
The MEPro program is installed in the system memory portion of the Flash ROM that is separate from the RAM available to the user. MEPro uses RAM to store some of its session information, including values entered and computed by the user. The exact amount of mem ory requi red depe n d s on the number of user­stored variables and the number of session folders designated by the user. To view the available memory in your TI calculator, use the function

. It is recommended that at least 10K of free RAM be
available for installation and use of MEPro.
1.5 Differences between TI-89 and TI-92 plus
MEPro is designed for two models of graphing calculators from Texas Instruments, the TI-92 Plus and the TI-89. For consistency, keystrokes and symbols used in the manual are consistent with th e TI-89. Equivalent key strokes for the TI-92 Plus are listed in Appendix D.
1.6 Starting ME•••Pro
To begin ME•Pro, start by pressing t he move t he cu rsor bar to press the
key to get to the home screen of MEPro. Alternatively, press
ME•Pro and press the
FlashApps...
key to get to the home screen of MEPro.
/
key. This accesses a pull down menu. Use the $ key to
. an d press
. Then move the highlight bar to MEPro and
/
; then, scroll to
MEPro for TI-89, TI-92 Plus Chapter 1 - Introduction to ME-Pro
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Pull down Menu for
/
(FlashApps...option is at the top of the list)
Pull down Menu on for FlashApps... (MEPro will be in the list)
The MEPro home screen is displayed to the right. The tool bar at the top of the screen lists the titles of the main sections of MEPro which can be activated by pressing the function keys.
b
: Tools: Editing features, information about MEPro in A: About.
c
: Analysis: Accesses the Analysis section of the software.
d
: Equations: Accesses the Equations section of the software.
e
: Reference: Accesses the Reference section of the software.
f
: Info: Helpful hints on MEPro.
To select a topic, use the $ ke y to mov e the highl i ght bar to the d e s ired top ic and pr e s s
, or
alternatively type the number next to the item. The Analysis, Equation and Reference menus are organized in a menu tree of topics and sub-topics. The user can r eturn to a previous level of MEPro by pressing . You can exi t MEPro at any time by pressing the
key. When MEPro is restarted ,
the software returns to its previous location in th e program.
1.7 How to use this Manual
The manual section, chapter heading and page number appear at the bottom of each page. The first chapter in ea ch of the Analysis, Equations and Reference sections gives an overview of succeeding chapters and introduces the navigation and computation features common to each of the main sections. For example, Chapter 2 explains the basic layout of the Analysis section menu and the navigation principles, giving examples of fe atu re s common to all topi cs in Analysis. Each topic in Analysis has a chapter dedicated to describing its functionality in detail. The titles of these chapters correspond to the topic headings in the software menus . They contain example problems and scr een di s p lays of the compu ted solut ions. Troubl e­shooting information, commonly as ked questions , and a bibliograp hy use d to develop the sof tware a re provided in appendixes.
1.8 Manual Disclaimer
The calculator screen displays in the manual were obtained during the testing stages of the software. Some screen displays may appear slightly different due to final chan ges made in the software while the Manual was being comple ted.
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1.9 Summary
The designers of MEPro invite your comments by logging on to our website at http://www.dvtg.com or by e-mail to improvements@dvtg.com easy with the software by providing the following features:
Easy-to-use, menu-based interface .
Computational efficiency for speed and performance.
Helpful-hints and context-sensitive information provided in the status line.
Advanced ME analysis routines, equations, and reference tables.
Comprehensive manual documentation for examples and quick reference.
. We hope that you agree we have made complex computations
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Part I: Analysis
MEPro for TI-89, TI-92 Plus Analysis -
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Chapter 2: Introduction to Analysis

2.1 Introduction
The analysis section contains subroutines and tools designed to perform specific calculations. Computations include estimating thermodynamic properties of water at different temperature and pressure, in Steam
Tables, computing fluid flow rates through different shaped orifices in Efflux, performing Wye to conversions of AC circuits in EE for MEs, and evalueating cash flow for different projects in Capital
Budgeting. The computations are strictly top-down (i.e. the inputs and outputs are generally the same) and the interface for each section guides the user through the solving process. A brief description of some of the diff erent sections in Analys is appear below:
Steam Tables (3 sections): Saturated Steam, Superheated Steam, Air Properties computes the thermodynamic parameters of steam including saturated pressure, enthalpy, entropy, internal energy, and specific volume of the liquid and vapor forms of water given entri es of temperature and/or pressure. This final topic covered computes the thermodynamic properties of dry air at different temperature s.
Thermocouples: This tool converts a specified tempe rature to an emf output in millivolts (mV) or from emf output millivolts (mV) to a specified temperature. The software supports T, E, J, K, S, R and B type thermocouples. These computation algor i t hms result from the IPTS -68 sta ndards adopted in 1968 and modified in 1985.
circuit
Capital Budgeting: This section performs analysis of capital expenditure for a project and compares projects against one another. Four measures of capital budgeting are included in thi s section: Payback period (Payback); Net Present Value (NPV); Internal Rate of Return (IRR); and Profitability Index (PI). This module provides the capability of entering, storing and editing capital expenditures for nine different project s. Project s can be graph ed on NPV vs. k scale.
EE for Mechanical Engineers (3 sections): Performs e valuations on three types of circ u its: Impedance
↔∆
calculations; Circuit Performance; Wye
impedance admittance of a circuit consisting of a resistor, capacitor and inductor connected in Series or Parallel. Performance parameters section computes load voltage and current, complex power delivered, power factor, m axim u m power a va ilabl e to the l oad, a nd the load i mpedance req u ired t o receive the maximum power from a single power source. The final segment of the software converts configurations expressed as a Wye to its ∆∆∆ equivalent. It also performs the reverse computation.
Efflux (6 sections): Constant Liquid level; Varying liquid level; Conical Vessel; Horizontal Cylinder; Large Rectangular Orifice; ASME Weirs (Rectangular notch; Triangular Weir; Suppres s ed Weir; Cipolletti Weir) This section contai ns methods to compute fluid flow via cross sections of different shapes.
Section Properties (1 2 s ect ions): Rectangle; Hollow Rectangle; Circle; Circular Ring (Annulus); Uneven I-section; Even I-section; C section; T section; Trapezoid; Polygon (n-s ided); Hollow P olygon (n-sided, side thickness) Computes are a moment and location of center of mass for diff erent shaped cross sections.
Computed parameters inc lude the cross section area, the pol ar moment o f inertia, the area moment of inertia and radius of gyration on x and y axes.
Hardness Number: A dimensionless number is a measure of the yield of a material from impact. Brinell and Vicker developed two popular methods of measuring the Hardness number. These tests consist of dropping a 10 mm ball of steel with a specified load such as 500 lbf and 3000 lbf. This steel ball results in an indentation in the material. The diameter of indentation indicate s o f the hardness numbe r using e ither the Brinell's or Vicker's formulation.
Circuit conversion
Impedance Calcula tions,
computes the
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2.2 Features of Analysis
Unit Management: Appropriate unit menus for appending units to variable entries or converting computed results are accessible in most sections.
Numeric Computation – Variable entries must consist of real numbers (unless specified). Algebraic expressions must consis t of defined variables so a numeri c val u e can be condensed upon entry.
2.3 Finding Analysis
The following panels illustrate how to start ME•Pro and locate the
Analysis
section.
1. Press the HOME screen to list the applications stored in your calculator.
There are seven sections under desired heading and pressing
/
key in the
Analysis
2. Press 1:FlashApps and press
to display the applications stored in the Flash section of memory.
1
3. HOME screen of ME
Analysis is listed as c on the top function key row.
Pro.
. To sel ect a topic, use t he $ ke y to mov e the highl i ght bar to the
, or alternatively type the number next to the item to select. If a topic contains several sections (Steam Tables, EE for MEs, Efflux, Section properties, an ellipsis (…) will appear next to the title (see below).
From t he home screen of MEPro
Press c
to display
Press for topics in
Tables
Steam
…or, press Z for topics in
Efflux.
the Analysis menu…
2.4 Solving a Problem in Analysis
The following example presents some of the navigational features in Analysis. This example is drawn from Chapter 6: EE for MEs.
Problem - Calculate the performance parameters of a circuit consisting of a current source (10 - 5*i) with a source admittance of .0025 - .0012*I, a load of .0012 + .0034*i. Display the real result of power in kilowatts.
1
Steps 1 and 2 can be combined by pressing and
MEPro for TI -89, TI-92 Plus Chapter 2- Introduction to Analysis
/
.
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1. From the home screen of Pro, Press c to display
ME
the menu of Analysis.
4. While the cursor is
highlighting Load Type, press the right arrow key, or g to display the menu for Load Type.
7. Variable descriptions
beginning with ‘Enter’ require numeric entries.
2. Move the cursor to EE for MEs and press
press Y).
5. In the menu for Load
Type move the cursor to Admittance and press
.
(or
3. Select Circuit Performance from the
submenu in EE for Mes.
6. Admittance is now selected for Load Type and the appropriate variables are displayed.
8. Variable descriptions beginning with the word ‘Resu lt’ are computed fields.
9. When entering a value, press a function key to add the appropriate units (c-h).
10. Following entry of all
input fields, pres s c: Solve to compute the results.
13. To display a result in
different unit s , highlight the variable and press f:Opts move the cursor to 4
:Conv.
MEPro for TI -89, TI-92 Plus Chapter 2- Introduction to Analysis
11. Results: Upper Half 12. Results: Lower Half
14. The unit menu for the variable appears in the top bar. Press t he func tion key
,
15. The computed value for
Real Power, P, is now displaye d in kilowatts (kW).
corresponding to the desired units.
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There are two types of interfaces in
Type 1: Input/Output/Choose Fields (Steam Tables, Thermocouples, EE for MEs, Efflux, Section Properties, and Hardness Number). This input form lists the variables for which a numeric entry is required
and prompts the user to choose a calculation setting if applicable before computing the results. The entries and results are always displayed in the same screen.
Type 2: Multiple Forms/Graphing (Capital Budgeting) This interface includes most of the features of Type 1 with the additional screens used for entering cash flow for individual projects. The graphing
feat ures of t he calculator are enabled i n t his section for v i sualizing the rate of return (Net Present Value vs. discount rate). An example of this interface is described briefly in this chapter, but in mor e detail in Chapter 5: Capital Budgeting.
Capital Budgeting
relevant data such as displays below illustrate the basic user interface.
Input Screen for Capital Budgeting
display Cash Flow for Project
1.
. Press e to
allows the user to compare relative financial performance of several projects with
Interest rate or discount rate (k), IRR, NPV, or Payback period. The screen
Analysis
A separate screen displays the Cash Flow for ‘Project 1’. Press c: Solve to r evert to the previous screen.
:
Press d: Graph. Select ing Multiple Graphs’
overlap of plots for different projects (Projec t 1, Project 2, etc.)
allows the
2.5 Tips for Analysis
The following instructions are useful in the Analysis section:
1. If an ellipsis (…) appears at the end of a menu title, a menu of subtopics exists i n this se ct i on .
2. An arrow ‘→→→’ to the right of a heading, as in Load T ype, indicates an additional menu.
3. Variables ending with an underscore ‘_’, such as Vs_, Zs_, and IL_, allow complex values.
4. Descriptions for variables generally appear in the status line when the variable is highlighted.
5. Variables for which an entr y is required will have a description prefaced by the word ‘Enter’. Computed variables begin with the word, ‘Result’. To convert values from one unit to another, press
6. th e variable at t h e top of the screen. Press the functio n key co rresponding to the appropriate units.
7. To return to the previous level of ME•Pro, press ..
8. To exit ME•Pro, press b: Tools and N: Clear.
9. To return to ME•Pro, p ress
10. To toggle between a graph and ME•Pro in split-screen mode, press
11. To remove the split screen in ME•Pro. 1) Press Full Screen, 5)
: Save.
/
.
f:Opts, and 4:Conv to dis play the unit menu for
/
.
, 2) c: Page 2, 3) ": Split Screen App., 4) :
2.6 Function keys
Analysis
When are s p ecific to the cont ex t of the section. Th ey are listed in Table 2-1:
functions are selected, the function keys in the tool bar access or activate features, which
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Table 2-1 Description of Analysis Function keys
Function Key Description
and
Tools
Solve
Graph
/
b
c
d
e
Labeled " screen level. Th ese fu nctions are: 1: Open – This opens an existing folder to store or recall variables used in an MEPro session.
2: (save as) – Not active in Analysis. 3: New – Creates a new folder for storing variable values used in an MEPro
session. 4: Cut - Removes entered values into the clipboard. Enabled for variables for which t he user ca n enter va l u es.
5: Copy – Copies a highlighted value into the clipboard. 6: Paste – Pastes clipboard contents at cursor location. 7: Delete – Deletes highlighted values. 8: Clear – Return s to the HOME screen of MEPro. 9: (format)-Not active in Analysis. A: About - Displays product name and version number.
Labeled " selected problem and display any resulting output to the user.
Labeled " be represent ed in graphical form. A gr aph can be viewed i n the full s creen or a split screen mode. This can be performed by pressing
Normally labeled as " cursor to be displayed using the entire screen in Pretty Print format. In some cases @ is labeled as "Pict", "Cash". "Pict" is available in the Section Properties or Hardness Number sections and when selected displays a diagram to facilitate better understanding of the problem. "Cash" is used in Capital Budgeting section of the software. Labeled "Opts" - Thi s key displays a po p up me nu l i s ting the options:
1: View - allows the highlighted item to be viewed using Pretty Print. 2: (type) - Not active 3: Units – This activates, or deactivates the unit management feature. 4: Conv – Displays the unit menu for the highlighted variable and allows the
conversion of an entry or result into different units. 5: Icons - Presents a dialog box identifying certain Icons used by the software to display content and context of the information. These icon systems are only used
in equ a t ions. 6: (know)- Not active 7: Want - Not active
Edit” - Brings in a data entry line for the highlighted parameter. “Choose” in Capital Budg et ing enables th e user to select from one of n ine project s. “ Check” requesting the user to press this k ey to select a highlighted parameter for use in an Analysis computation. "Add” Adds a cash flow entr y for a project in Ca pital Budget ing section. (Not active)
" - displays all the functions available on the TI-89 at the Home
" - Pressing this key enables the software to begin solving a
" - This feature is available in input scree ns where the solution can
followed by c. Use
to toggle between the data entry screen and graph window.
View
" - This enables the information highlighted b y the
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2.7 Session Folders, Variable Names
ME•Pro automatically stores its variables in the current folder specified by the user in screen. The cur rent folder name is displayed in th e lower left corner of th e s creen (default is “Main”). To create a new folder to store values for a particular session of ME•Pro, press :/TOOLS, :/NEW and type t he nam e of the new folder (see Gu idebook for the com plete det ail s of crea ting and m anaging fol ders; Chapter 5 for the TI-89 and Chapter 10 for the TI-92 Plus).
There are several ways to display or recall a value:
The conten ts of variables in any folder can be d ispla yed using
variable name and pressing to display the contents of a particular variable.
Variables in a current folder can be recalled in the HOME screen by typing the variable name.
All inputs and calculated results from the Analysis and Equations section are saved as variable
names. Previously cal culat ed , or enter ed values for va ria bl es in a fold er are rep laced wh en equatio ns are solved using new values for inputs.
or the HOME
, moving the cursor to the
2.8 O verwriting of variable values in graphing
When an equation or analysis function is graphed, ME•Pro creates a function for the TI grapher, which expresses the dependent variable in terms of the independent variable. This function is stored under the variable name pro (x). When ME•Pro’s equation grapher is execute d , v alu es are inserted into the independent variable for pro (x) an d va lues for the dependen t value ar e calcul ated. Whatever val u es previously existed in either of the dependent and independent variables in the current folder are cleared. To preserve data under variable names, which may conflict with ME•Pro’s variables, run ME•Pro in a separate folder.
2.9 Reserved Variables
A list of reserved variable names used by the TI operating system, which cannot be used as user variable names or entries are listed in Appendix F.
MEPro
for TI -89, TI-92 Plus
Chapter 2- Introduction to Analysis
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F2: Analysis/1: Steam Tables

Chapter 3: Steam Tables

Steam properties are a complex function of temperature, pressure, volume, critical temperature, critical pressure, and molecular weight of water. Our software takes into account the ranges of temperature and pressure that results in good fits to data gathered over a long periods in many parts of the world.
Steam Tables offer a collectio n of programs organized as a powerful computational engine to calculate thermodynamic properties of steam in a user-friendly environment. Calculations of thermodynamic properties are based on standards and conventions adopted by the International Conventi ons covering properties of saturated and superheated steam.
3.1 Saturated Steam Properties
This section computes the properties of saturated steam at a single temperature or pressure.
Variable Description Units
Ps Saturation pressure MPa Ts Saturation temperature K Vf Specific volume – liquid m3/kg Vg Specific volume – vapor m3/kg Hf Enthalpy liquid kJ/kg Hfg Latent heat of vaporization kJ/kg Hg Enthalpy vapor kJ/kg Sf Entropy liquid Sfg S (g) - S (f) Sg Entropy vapor UF Internal energy liquid UG Internal energy vapor
kJ/(kgK) kJ/(kgK) kJ/(kgK) kJ/(kgK) kJ/(kgK)
3.2 Superheated Steam Properties
The properties of superheated steam require two inputs: temperature and pressure. From the data suppli ed , the program will compute saturated temperature, specific volume, enth alpy and entropy. The data is displayed in a tabular form.
Variable Description Units
Temp Given temperature K Sat Pressure Given pressure MPa Sat Temperature Corresponding temperature K Specific Volume Specific volume m3/kg Enthalpy Enthalpy kJ/kg Entropy Entropy
kJ/(kgK)
3.3 Air Properties
The properties of dry air are computed using the ideal gas law model as the basis. Using temperature as a n input, the softwa re computes a var iety of parameters including specific heats, enthalpy, entrop y, and velocity of sound.
Variable Description Units
Temp Given temperature K CP Specific heat at constant pressure
J/(kgK)
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F2: Analysis/1: Steam Tables
Variable Description Units
CV Specific heat at constant volume H Enthalpy J/kg U Internal energy J/kg E Entropy function IPR Isentropic pressure function unitless IVR Isentropic volume function unitless G Specific heat ratio unitless A Speed of sound m/s
J/(kgK)
J/(kgK)
3.4 Using Steam T ables
Once you have selected STEAM TABLES at the main menu, the first screen displays three subtopics ­Saturated steam, Superheated steam and Air properties. Selecting “saturated steam properties”, allows properti es t o be calcula ted from u ser- entered value of t emperature or pressure. However, p roperties for “superheated steam” req uire v a lues for both tempe ra t ure and pressure. Thermodyna mic properties of air are calculated for dry air conditions only.
Example 3.1:
Calculate the properties of saturated steam at 130 °C. Solution - Select the Saturated S team section. The input screen calls for defining known parameter
(temperature o r pressure) . The default condition is temperature. Move the pointer to the next line and start entering the temperature. The function keys assume unit assignments for the data about to be entered. For our example, enter 130 an d press ? key thereby attaching °C to the value just entered. Press > to solve for the thermodynamic parameters.
Upper Display Lower Display
All the calculated parameters are displayed on the screen w ith SI units attached as shown. If you desire to see the value of saturated pressure parameter Ps, use the ke y to mov e the highl i gh t bar to capture Ps. Press A to display a pull down menu of items to select. Select 4 (Conv). This allows other units for Ps such as Pa, kPa, atm, psi, torr attached to ?, @, A, B, and C respect ively. Pres sing B converts the value of Ps into the new units of psi. The display is refreshed immediately in the units just selected.
Example 3.2:
Calculate the properties of superheated steam at 125 °C and 20 psi. Solution - Select the Supe rhea ted Steam se ction. The input screen calls for e ntering temperat ure and
pressure. Move the pointer to enter the 125 °C and 20 psi for temperature and pressure. Make sure that the appropriate units are attached to the data using the function keys. Press > to solve for the thermodynamic parameters.
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F2: Analysis/1: Steam Tables
Upper Display Lower Display
All the calculated parameters are displayed on the screen with units attached as shown. The displayed results could be converted to other un its as described in the first example described earlier.
Example 3.3:
Calculate the properties of dry air to be -20 °C. Select the Air Properties section. The input screen calls for defining the temperature. Enter a temperature
of – 20 °C. When ent erin g -20 °C, be sure to u se the unary operat or key followed by 20 °C. If you use\ key fo r a negative value this will result in an input error.
Press > to solve for the thermodynamic properties of air.
Input Screen Result screen
All the calculated parameters are displayed on the screen with SI units attached as shown. The parameters computed can be viewed in other units as described in the examples shown here.
3.5 Validity Range for Temperature and Pressure
The computed results are valid only for the following finite ranges of temperature and pressure:
Saturated Superheated
Temperature: 273.16 - 647.3 K Saturated temperature Pressure: 0.006113 - 22.08 MPa Pressure: 0.006113 - 22.08 MPa
References:
1. Lester Haar, John S. Gallagher and George S. Kell, NBS/NRC Steam Tables, Thermodynamic and Transport Properties for Vapor and Liquid States of Water, Hemisphere Publishing Corporation, Washington, DC.
2. Steam Tables, 1967; Thermodynamic properties of Water a nd Steam, The Electric Research Association, Edward Arnold Limited, London England, 1967
3. Th omas F. Levine, Jr., and Peter E. Liley, Steam and Gas Tables with Computer Equations, Academic Press, New York, NY, 1984
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F2: Analysis/ 2: Thermocouples

Chapter 4: Thermocouples

This chapter describes using the software in the Thermocouples menu. Thermocouple parameters are calculated for the class of thermocoupl es in comm on usa g e.
4.1 Introduction
This tool c onverts a specified temperature to an emf output, millivolts (mV), and from an emf output, millivolts (mV), to a specified temperatur e. The software supports Type T, E, J, K , S, R and B thermocouples. The underlying assumption is a reference t emperature of 0 °C. These computation algorithms result from the IPTS-68 standards adopted in 1968 and modified in 1985.
4.2 Using the Thermocouples Function
Select the Thermocouples function from the Analysis menu by highlightin g Thermocouples an d pressing
¸
. This action brings out the main use r inte rf ace scre en. Press B to view the choices availabl e. Seven types of T hermocou ples a re a vail able for com putations. For example, to sele c t a Pt-10%Rh-Pt thermocoup l e referr ed to as S Type ther mocouple, move the hi g h th e highlight bar to Type S and press At this point, the software presents the primary user interface awaiting data entry. At the same time, the mat erial of the ther mocouple a nd the vali d ran g e are displa yed for reference.
¸
or press z.
Thermocouple computations involve computing emf available from a known temperature or computing a temperature from an emf. The latter case is by its very nature not as accurate as the first type of computation.
Screen 1 (for temperature) Screen 2 (for emf)
Example 4.2:
Find the emf for an S type thermocouple at 400 °F. From the value of emf computed, compute t he temperature.
Solution 4.2:
Select Type S thermocouple for t his problem. For temperature, enter 400, t hen press . The computed emf is 1.4777mV. Now return t o Kn own line and select emf for input. Ent er 1.4777 mV for emf to get 403.989 °F for temperature.
Notes: Th e ther mocouple em f ca lcula tion can be ex p anded t o cover the em f produced by th e thermocouple if the reference temp erat ure was di fferent from 0 °C. For example, if t he reference t emperature was 30 °C instead of 0 °C, you compute the resulting emf in two steps; first find the emf (emf 1) for the temperature desired, say 300 °C, and the emf (emf 0) for the reference temperature. The resulting emf for the new refer ence temperature of 30 °C is the difference between the two emf’ s, i.e., "em f 1 - emf 0".
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F2: Analysis/ 2: Thermocouples
4.3 Basis for Temperature/Voltage Conversions
The temperature-to-voltage conversion is based on either a polynomial approximation or a combination of a polynomial coupled with a special sequence. This ensures precise calculations within some prescribed error range. These rages a re displayed on each inpu t screen.
References:
1. Robert L. Powell, William J. Hall, Clyde H. Hyink, Larry L. Sparks, George W. Burns, Margaret Scoger and Harmon H. Plumb, Thermocouple Reference Tables based on IPTS-68, NBS Monograph 125, Omega Press, 1975
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F2: Analysis/3: Capital Budgeting

Chapter 5: Capital Budgeting

This chapter covers the four basic measures of capital budgeting:
Payback Period ♦ Internal Rate of Return Net Present Value ♦ Profitability Index
5.1 Using Capital Budgeting
This section performs analysis of capital expenditure for a project and compares projects against one another. Four measures of capital budgeting are included in this section: Payback period (Payback); Net Present Value (NPV); Internal Rate of Return (IRR); and Profitability Index (PI). This module provides the capability of entering, storing and editing capital expenditures for nine different projects. The following equations are used in calculations:
n
CF
()1
t
CF
t
=
t
+
k
0
Eq. 1
NPV
=
t
=
1
n
CF
+
n
t
=
t
IRR
()1
1
t
=
PI
()1
1
=
CF
CFt: Cash Flow at time t. Payback: The number of time periods it take s a fir m to re cover its original invest ment. NPV: The present values of all future cash flows, discounted at the selected rate, minus the cost of the
investment.
IRR: The discount rate that equates the present value of expected cash flows to the initial co st of the proj ect. PI: The present va lue of the future cash flows, discounted at the selected rate, over the initial cash outlay.
CF
−=
t
=
0
t
CF
+
t
=
t
t
k
0
0
Eq. 2
Eq. 3
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F2: Analysis/3: Capital Budgeting
Field Descripti ons - Input Screen
Project:
k: Payback: npv: IRR: PI: Multiple
Graphs
Full Screen Graph
NAME: t0: t1: tn:
(Project)
Press
to select one of nine unique projects or edit the current name of the project by pressing e for Cash option.
(Discount Rate per Period in %) Enter a real number. (Payback Period) Returns a real number. (Net Present Value) Returns a real number. (Internal Rate of Return) Returns a real number (%). (Profitability Index) Returns a real number. (Graph multiple projects
simultaneously)
(Graph on full or split screen?) Press
Activation of this feature enables the overlay of each success i ve gr a p h (pr oject s) on t he
Press
same axis.
to activate.
to activate.
Field Descriptions - Project Edi t Screen
(Project Name) Enter the name of the project. (Investment at t=0) Enter a real number. (Cash flow at t=1) Enter a positive or negative real number. (Cash flow at t=n) Enter a positive or negative real number.
Example 5.1:
The following projects have been proposed by ACME Consolidated Inc. What are t he Payback period, Net Present Value, Internal Rate of Return, and Profitability Index of each project? Which is the more viable project?
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Table 5-1 Cash Flow for two projects
Name of Project: Plant 1 Plant 2 Investment Outlay:
$75,000 (at t=0) $75,000 (at t =0)
Cost of Capital: 12% 12%
Year Net Cash Flow ($) Net Cash Flow ($)
0 -75,000 -75,000
1 40,000 10,000
2 30,000 20,000
3 20,000 30,000
4 10,000 40,000
F2: Analysis/3: Capital Budgeting
Cash Flow Input: plant1 Cash Flow Input: plant2
Output Scr een: plant1
Output Scr een: plant2
1. With the highlight bar on the Project field, press
not been used. Note: this example uses projects 1 and 2. Press
to select a project to edit. Select a project that has
to return to the Capital Budgeting
screen.
2. Press e to select Cash option; enter the project edit screen; and, edit the cash flows.
3. Enter “plant1” in the Name field. Note: Cash flo w data for this proj ect will be sto r ed in a variable of thi s
name. Therefore the entered name must begin with a letter; be no more than 8 characters in length; and, contain no embedded spaces.
4. Press h 5 times to add 5 time points and enter the cash flows at each time point from the table on the pr e vious page. W hen finished, your s c reen shoul d look like the projec t edi t screen above. Be s ure to en ter 75,000 as a negative number for t0. Press . to save your changes and return to the Capital Budgeting screen.
5. Enter 12 for k.
6. Press c to calculate Payback, NPV, IRR, and PI.
7. Move the highlight bar to Multiple Graphs and press
to en able overlayin g of succes s ive grap hs of
each p roject.
8. Press d to graph the curvilinear relationshi p between the Net Present Value and the Discount Rate.
9. Press followed by
/
to enable the graph editing toolbar.
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F2: Analysis/3: Capital Budgeting
10. T he cur ve i ndicates where k=0, the Net Present Value is simply cash inflows minus cash outflows. The IR R % is sho wn at the point where NPV=0. Using the built-in graphing capabilities o f the TI 89, you can trace the graph to find the values of these two points. The TI 89 will give you the exact coordinates of any point along the graph. Press followed by
/
to return to the Capital Budgeting screen.
11. Repeating steps 1 through 9 for the second project, under the Project field, “plant2” and input the values in Table 5-1. Activating the Multiple Graph feature enables a simultaneous plot of the two projects. This will overlay a second graph on top of a previously plotted function. First plot plant1. After gr ap hing, plot plant2. The first cu rve to ap p ear, is plant1, the second is plant2. The most viable project in terms of discounted cash flows, in this example, is the one with the highest curve.
12. Pressing
c "
resets the display to full screen.
Plot of Project 1 Overlay of Project 2
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F2: Analysis/4: EE for ME’s

Chapter 6: EE for Mechanical Engineers

This chapter describes the software in the AC Circuits section and is organized under three topics. These topics for m the backbo ne of circuit calcu lations of interes t to mechanical en gine e rs.
Impedance Calculations Circuit Performance Wye ↔ ∆ Conversion
6.1 Impedance Calculations

Impedance Calculations

The resistor, capacitor and inductor connected in Series or Parallel. The impedance and admittance values are displayed to the user in r eal or complex form.
topic computes the impedance and admittance of a circuit consisting of a
Field Descripti ons
CONFIG:
Elements:
fr:
R:
L:
C:
ZZ_:
YY_:
Circuit Configuration
Element Combination
Frequency in Hz
Resista n ce in ohms; only appears if RL, RC or RLC is chosen in Elements field
Inductance in Henry; only appears if L, RL, LC or RLC is chosen in Elements field
Capacitance in Farads; only appears if C, LC, RC or RLC is chosen in Elements field
Impedance in ohms
Admittance in Siemens
(Press
_
_
and select Series or Parallel
__
configuration by usin g
_
to display the input screen updated for the new
configuration..
Press
_
_
to display the following circuit elements:
__
L, C, RL, RC , LC an d RLC T he choice of elements determines which of the input fields are available.
Enter a real positive number.
Enter a real positive number.
Enter a real positive number.
Enter a real positive number.
Returns a real or co mp lex number.
Returns a real or comp lex number.
. After choosing, press
Example 6.1:
Compute the impedance of a series RLC circuit consisting of a 10-ohm resistor, a 1.5 Henry inductor and a
4.7 Farad capacitor at a frequency of 100 Hz.
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Input Screen Output Scr een
1. Choose Series for Config and RLC for Elements using th e proce dur e described a bove .
2. Enter 100 Hz for Freq.
3. Enter 10 Ω for R; 1.5 Henry for L; a nd 4.7 F for C.
4. Press
>>>>
to calculate ZZ_ and YY_
.
5. Th e output screen shows the results of computation.
6.2 Ci rcuit Performance
This section shows how to compute the circuit performance of a simple load connected to a voltage or cur rent source. Perform ance parameters include load voltage and current, complex power delivered, power factor, maximum p ower available to the load, and the load impeda nce required to receive the maximum power .
Field Descriptions - Input Screen
Load Type:
Vs_:
Zs_
ZL_:
Type of Load Press
rms Source Voltage in V A real or complex number.
Source Impedance in
Load Impedance in
_
_
to select load impeda nce (Z) or
__
admittance (Y). This w ill de te rmine whether the remaining fields Vs_, Zs_, and ZL_ or Is_, Ys_, and YL_ are di s p layed, respecti vel y.
A real or complex number.
A real or complex number.
Is_:
ys_: yl_:
Vl_: il_: P: Q:
MEPro for TI-89, TI-92 Plus Chapter 6 - Analysis - EE for MEs
rms Source Current in A
Source Admittance in Siemens Enter a real or complex number.
Load Admittance in Siemens Enter a real or complex number.
Load Voltage in V Returns a real, complex number.
Load Current in A Returns a real, complex number.
Real Power in W Return s a real number.
Reactive Power in W Returns a real number.
A real or complex number.
Field Descriptions - Output Screen
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F2: Analysis/4: EE for ME’s
VI_: θθθθ:
Apparent Power in W Returns a complex number.
Power factor Angle in degrees
Returns a real numbe r.
or radian, determined by the
setting
PF: Pmax:
Zlopt_:
Load Power Factor Returns a real number.
Maximum Power Available in W Returns a real number.
Load Impedance for Maximum Power in
- if Impedance is
Returns a real, complex number.
chosen for Load Type at the input screen
Ylopt_:
Load Admittance for Maximum
Returns a real, complex number.
Power in Siemens - if
Admittance, is chosen for Load Type at the input screen
Example 6.2:
Calculate the performance parameters of a circuit consisting of a current source (10 - 5*i) with a source admittance of .0025 - .0012* I, and a load of .0012 + .0034*i.
Input Screen Output: Upper Half Output: Lower Half
1. Ch oose Admittance for Load Type.
2. En ter th e value 10 - 5*i A for
Is_.
3. Enter t he value .0025 - .0012* i siemens for Ys_, and .0012 + .0034* i siemens for a load of YL_.
4. Press ? to calculate the performance parameters.
5. The input and result s of computa tion are shown above.
6.3 Wye ↔↔ ∆∆∆∆ Conversion
The Wye ↔ ∆ Conversion converts three impedances connected in Wye or form to its corresponding or Wye form, i.e., Wye ↔ ∆ or ∆ ↔ Wye
Input Fi elds -
Input Type: Selecti on ch oi ces a re ∆→Wye or Wye→∆. This
determines whether the next 3 fields (input fields) accept or Wye Impedances.
ZZA_: ZZB_: ZZC_:
Impedance Real or complex number.
Impedance Real or complex number.
Impedance Real or complex number.
. Fig. 6.1 Wye Network
ZZ1_: Y Impedance Real or complex number. ZZ2_: Y Impedance Real or complex number.
Z3
Z1
Z2
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ZZ3_: Y Impedance Real or complex number.
Result Fields
ZZA_: ( ZZB_: ( ZZC_: (
Impedance) Real or complex number.
Impedance) Real or complex number.
Impedance) Real or complex number.
ZZ1_: (Y Impedance) Real or complex number. ZZ2_: (Y Impedance) Real or complex number. ZZ3_: (Y Impedance) Real or complex number.
Fig. 6-2 ∆∆∆ Network
ZB
ZC
Example 6.3:
ZA
Compute th e Wye im p edance eq uivalent of a network with impedances 75+12*i, 75-12*i, and 125 ohms.
1. Select ∆ →Y for Input Type. .
2. Enter the values 75+12*i
3. Press
>>>>
to calculate ZZ1_, ZZ2_ and ZZ3_.
, 75 -12*i Ω, 125 Ω for ZZA_, ZZB_ and ZZC_ respectively.
Input Parameters Calculated Output
The computed results are: ZZ1_: 34.0909 - 5.45455 i
ZZ2_: 34.0909 + 5.45455 i
ZZ3_: 20.9782
References:
1. Sanford I. Heisler, The Wiley Engineer's Desk Referen ce, A concise gu ide for t he Pr ofessional Engineer, John Wiley and Sons, New York, NY, 1984
2. James W. Nilsson, Electric Circuits, 2nd Edition, Addison -Wesley Publishing Company, Reading MA, 1987 and later editions.
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F2: Analysis/5: Efflux

Chapter 7: Efflux

This section of Analysis contains methods to compute fluid flow via cross sections of different shapes. Many of th e s e formula s have been d e rived empirically over many years of ex perimental observations.

Constant Liquid Level Varying Liquid Level Conical Vessel ♦ Horizontal Cylinder ASME Weirs

7.1 Constant Liquid Level
This portion of the software computes fluid discharge from an opening of cross sectional area So (m2), and a constant head H (m). The discharge coefficient αααα (unitless) is a unique number (a number less than
1) accounting for t he sh ape of the cross section, edge-rounding effects, an d turbu lence effects. Upon receivi ng these inputs, th e s oftwa re computes the volume d ischa rge, Q, (m V (m/s).
Example 7.1:
Find the discharge parameters for a 1.5 sqft opening subject to a head of 4_m. Assume the discharge coefficient to be 0.85.
3
/s) and the discharge velocity
Input Screen Result Screen
2
To solve the pr oblem, en ter va lues for αααα = 0.85, H = 4 m and so = 1.5 ft
using the appropriate unit keys
acce ssible during data entry. After all the input variables have been entered, press ? key to get the following results.
Given: Results
α = 0.85 So = 1.5 ft
H = 4 m
2
V = 7.52877 m/s
Q = 1.04917 m
3
/s
7.2 Varying Liquid Level
This segment of the software considers fluid discharge from a tank of cross sectional area S (m2), an opening of cross sectional area So (m dischar g e coefficient αααα (unitless) is a unique number (a number less than 1) accounting for the shape of the cross sectio n, edge-rounding effects, and t urbul ence effects. Upon receiving t hese i nput s , th e software computes the time t (s) taken t o see the drop in head heig h t .
Example 7.2:
A large reservoir has an area of 10000 m
2
The drainage opening is 3 m
and has a discharge coefficient of 0.95. How long will this process take?
2
) where the head has dropped from H1 (m) to H2 (m). The
2
ha s hea d level of 4_m that need s to brought down by 0.5_m.
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F2: Analysis/5: Efflux
Input Screen Result Screen
2
To solve the pr oblem, en ter va lues for αααα = 0.95, H1 = 4 m, H2 = 3.5 m and So = 3 m
using the appropriate unit keys accessible when data is being entered. After all the input variables have been entered, press ? key to get the results.
Given: Results
α = 0.95 S = 10000 m So = 3 m
2
H1 = 4 m H2 = 3.5 m
2
t = 204.68 s
7.3 Conical Vessel
This segment considers fluid discha rge from a container that conical in shape. The parameters specified for the computations include cross sectional area, So (m fluid head at t. The discharge coefficient αααα (unitless) is a unique number (a number less than 1) accoun ting for the shape of the cross sectio n, edge-rounding effects, turbu lent effect s , an d ββββ (rad) refers to th e cone an gle. Upon receiving these inpu ts, the software computes the time for the head to \drop from H1 to H2.
Example 7.3:
Find the time taken to dis charge water fro m a 25 ° conical vessel with a disch arge coeff icient of 0.95 for a
2
1.2 cm
opening. Consider the head drops from 24 in to 18 in. The base of the cone has a diameter Dl of
12 in.
2
), and a fluid head H1 (m) at t=0 and H2 (m) the
Input Screen Result Screen
2
To solve the pr oblem, en ter va lues for αααα = 0.95, H1 = 24 in, H2 =18 in, and So = 1.2 cm
using the appropriate unit keys accessible when data is being entered. After all the input variables have been entered, press [F2] key to get the results.
Given: Results
α = 0.95 So = 1.2 cm
2
t = 121.456 s
Dl = 12 in H1 = 24 in H2 = 18 in β = 25 deg
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F2: Analysis/5: Efflux
7.4 Horizontal Cylinder
This segment of the software computes fluid discharge from a cylindrical tank of diameter D (m) and length L (m). The parameters specified for the computations include an opening for di scharge with an area So (m coefficient αααα (unitless) is a unique number (a number less than 1) accounting for the shape of the cross section, edge-rounding effects, a nd turbulent effects, along with Ks (unitless), and a space factor coefficient.
Example 7.4:
Find the time to oil from a cylindrical tank 12 m in diameter and 20 m long vessel with a space factor of
0.8 and orifice discharge coefficient of 0.75. How long does it take f or the oil to drop down 10 cm from an initial height of 5 m. Assume the orifice opening to be 1.2 mm
2
), and a fluid head H1 (m) at t=0 and H2 (m) the fluid head at t. In addition, the discharge
2
.
Input Screen Result Screen
To solve the pr oblem, en ter va lues for αααα = 0.55, Ks = 0.8 m, H1 = 5 m, H2 = 3.5 m, L = 20 m, and So =
2
1.2 cm
using the appropriate unit keys accessible when data is being entered. After all the input
vari ables ha ve been entered, pres s ? key to get the results.
Given: Results
α = 0.75
Q = .090698 _m^3/s H1 = 5 m t = 492061 s H2 = 4.9 m L = 20 m D = 12 m So = 1.2 mm
2
Ks = 0.22 m
7.5 Large Rectangular Orifice
This segment of t he software computes fluid discharge using a large rectangular orifice. The wei r has a width of b (m). The system has a di s charge coefficient αf, fluid h e ights H1 (m) and H2 (m) at the beginning and end.
Example 7.5:
Find the time to water to drop head from 3 m to 4 ft, for a weir 20 ft wide. Assume discharge coefficient of 0.8.
MEPro for TI-89, TI-92 Plus Chapter 7 - Analysis - Efflux
Input Screen Result Screen
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F2: Analysis/5: Efflux
To solve the pr oblem, en ter va lues for ααααf = 0.8, H1 = 3 m, H2 = 4 ft, and b = 20 ft. Enter all the inputs. After all the input variables have been entered, press ? key to get the results.
Given: Results
αf = 0.8
Q = 55.4336 m
3
/s H1 = 3 m H2 = 4 ft b = 20 ft
7.6 ASME Weirs
Weirs are useful devices to meas u re flow of liqu ids in op en ch ann els. A large number of empirica l formulas have been developed in the engineering literature each with its own limitations. A few representative samples are included in this software.
7.6.1 Rectangular Notch
A rectangular weir has a width of b (m), initial static head of H (m), and a velocity of Vo (m/s). Assume a dis charge coefficient αααα (unitless).
Example 7.6.1:
A rectangular weir is 20 ft wide, has water flowing over it at a velocity of 10 ft/s, a static head of 1.25 ft., and a discharge coefficient of .8. F ind t he discharge for the system .
Input Screen Result Screen
Given: Results
α = 0.8
Q = 6.68284 m
3
/s Vo = 10 ft/s H = 1.25 ft b = 20 ft
7.6.2 Triangular Weir
A triangular weir has a water height H (m), a discharge coefficient wi th a stat ic hea d of H (m) and a velocity of Vo (m/s). A ss u me a di scharge coefficient αααα (unitless).
Example 7.6.2:
A triangular weir has an angle of 60 deg, a discharge coefficient of 0.8 and a 2 ft water height. Find the discharge for this condition.
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F2: Analysis/5: Efflux
Input Screen Result Screen
Given: Results
α = 0.8 θ = 60 deg
Q = .316531m
3
/s
H = 2 ft
7.6.3 Suppressed Weir
A suppressed weir helps measure flow dependi ng upon the height above t he weir. The system has a coefficient of disch arge ααααf, a height H (m) over the weir, weir of width b (m), with an initial a velo city of Vo (m/s). Assum e a dischar g e coe fficient αααα (unitless).
Example 7.6.3:
A rectangular weir is 20 ft wide, has water flowing over it at a velocity of 10 ft/s, a static head of 1.25 ft., and the discharge coefficient is .8. Find the discharge for the system.
Input Screen Result Screen
Given: Results
αf = 0.8
Q = 13.4099 m
3
/s H = 1.25 ft b = 20 ft Vo = 10 ft/s
7.6.4 Cipolletti Weir
A Cipolletti weir of wid th b (ft) and static head of H (ft) computes discharge in ft3/s. The formulas used here are determined by experimental observation.
Example 7.6.4:
A rectangular weir is 20 ft wide, has water flowing over it with a 3 ft head. Find the discharge for the system.
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F2: Analysis/5: Efflux
Input Screen Result Screen
Given: Results
H = 3 ft
Q = 9.90832 m
3
/s b = 20 ft
References:
1. Eugene A Avallone and Theodore Baumeister, III, General Edito r s, 9th Editio n, Mark' s Sta ndard Handbook for Mechanical Engineers, McGraw-Hill Book Company, New York, NY
2. Ranald Giles, Fluid Mechanics and Hydraulics, 2nd Edition, Schaum's Outline Series, McGraw­HillBook Company, New York, NY, 1962
3. Michael R. Lindeburg, Mechanical Engineerin g Reference Manual, 8th Edition, Professional Publications Inc., Belmont, CA, 1990
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F2: Analysis/6: Section Properties

Chapter 8: Section Properties

This section of the software computes properties commonly associated with sections. These properties include calculating the area of cross-section, location of the center of mass from the vertical and horizontal axes. Computed area moments, I11 (m vertical and horizontal axes. In cases where it is mean ingful, the polar area moment is computed along with radius of gyration. Twelve standard cross-sections are found in this section. A pictorial description is included where possible.
4
) and I12 (m4), reflect their va l ue with reference to

Rectangle Circle Hollow Circle I Section - Even T Section Polygon

Hollow Rectangle Circular Ring I-Section uneven C-Section Trapezoid Hollow P olygon
8.1 Rectangle
The inpu t screen for the rectangle requires th e user to enter val u es of base, b (m), and height, h (m). In an illustrative example, we use a value of 10 inches for the base and 14 inches for the height. The results are displayed in SI units, however, they can be converted to different units by highlighting the values with the cursor, pressing B: Opts, ;: Conv and pressing a function key (> - D) to display th e desired units .
Entered values
Variable Description Value
b Base 10 in h Height 14 in
Upper Display Lower Display
Solution - To enter th e value of 10 inches for the base, at the data input screen, move the highlight bar to b and press
a
. Type in 10 and press A to append the inch units to the value . Enter a value of 14 in
for h in a similar manner. Press ? to compute the results.
Computed Results
Variable Description Value
Area Area yy1 Distance of center of mass from axis 1 . 127 m yy2 Distance of center of mass from axis 2 . 1778 m I11 Area moment inertia axis 11 I22 Area moment inertia axis 12 Ip Polar area moment
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42
.090322 m
.000952 m .000486 m .001437 m
2
4 4 4
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F2: Analysis/6: Section Properties
Variable Description Value
rg1 Radius of gyration 1 .102653 m rg2 Radius of gyration 2 .073323 m
8.2 Hollow Rectangle
The inpu t screen for the hollow rectangle requires the us er to enter ou ter a nd inner values of the bas e b and bi along with outer and inner heights h and hi. In an illustrative example, we use a value of 10 inches for th e base and 14 inches for the height. The wall thickness is 1.5 in.
Entered values
Label Description Value
b Base 10 in h Height 14 in bi In s ide base (<b) 7 in hi Inside Height (<h) 11 in
Upper Display Lower Display
To enter the value of 10 inches for the base, move the scroll bar to b and press
a
. Type in 10 and
press A to append the inch units to the value. In a like manner, enter 7 in for bi, 14 in for h and 11 in for hi. Press ? to begin the calculations. The calculated results are shown below:
Computed Results
Label Description Value
Area Area
0.040645 m yy1 Distance of center of mass from axis 1 0.127 m yy2 Distance of center of mass from axis 2 0.1778 m I11 Area moment inertia axis 11 I22 Area moment inertia axis 12 Ip Polar area moment
0.000629 m
0.000355 m
0.000983 m rg1 Radius of gyration 1 0.124362 m
2
4 4 4
rg2 Radius of gyration 2 0.093422 m
8.3 Circle
The inpu t screen for the ci rcle requires th e u s er to enter a value of diamet er d. In an illustrative example, we use a value of 50 cm for the diameter.
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F2: Analysis/6: Section Properties
Entered values
Label Description Value
d Diameter 50 cm
Entered Values Computed results
To enter the value of 50 cm for th e diameter, match the scroll bar to d and press
a
. Press ? to
begin the calculations. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
0.19635 m yy1 Distance of center of mass from axis 1 0.25 m I11 Area moment inertia axis 11 Ip Polar area moment
0.003068 m
0.006136 m rg1 Radius of gyration 1 0.125 m
2
4 4
8.4 Circular Ring
The i n p u t scr een for a ci r cu l a r r ing r eq ui res the us er t o en t er values of diam et er d and thickness t. In an illustrative example, we use a value of 1.5 m for a wagon wheel with a ring thickness of 1 in.
Entered values
Label Description Value
d Diameter 1.5 m t Thickness 1 in
Entered Values Computed results
To ent er the valu e of 1.5 m for the diamet er d, at the data input screen match the scroll bar to d and press
a
. Type in 1.5 and press ? to append the meter, m, and units to t he value. Enter a value of 1 in for
t in a similar manner. Press ? to begin the calculations. The calculated results are listed below.
Computed Results
Label Description Value
Area Area yy1 Distance of center of mass from axis 1 0.75 m I11 Area moment inertia axis 11
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0.119695 m
0.033664 m
2
4
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F2: Analysis/6: Section Properties
Label Description Value
Ip Polar area moment
0.067328 m
rg1 Radius of gyration 1 0.53033 m
4
8.5 Hollow Circle
The input screen for the hollow circle requires the user to enter values of outer and inner diameters do and di. As an illustrative example, we use a value of 36 inches for the outer diameter and an i n n er dia meter of 0.8 m.
Entered values
Label Description Value
do Diameter 36 in di Thickness 0.8 m
Entered Values Computed results
To enter the value of 36 inches for the outer diameter, at the data input screen matc h the scroll bar to do and press
a
. Type in 36 and press A to append the inch units to the value. Enter a value of .8 m
for di in a similar mann er. Press ? to begin the calculations. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
1.26449 m yy1 Distance of center of mass from axis 1 0.75 m I11 Area moment inertia axis 11 Ip Polar area moment
0.228399 m
0.456797 m
2
4 4
rg1 Radius of gyration 1 0.425 m
8.6 1 Section - Uneven
The input screen for an uneven I-Section consists of 6 variables needing user input. Pressing A allows a pictorial representation of T- and I-Beams. As shown in the pictu re sh own here b and b1 represent the widths of top and bottom flanges respectively while t and t1 reflect the thickness of top and bottom flanges. The height and thickness of the fin connecting the top and bottom flanges is represented by d and tw.
As an illustrative example, we use an I-Beam; we use a top flange width of 20 in an d 2 in thickness followed by 36 in width and 2.5 in thickness. The fin is 50 cm in height and 4 c m in thickness.
Entered values
Label Description Value
b Width of top flange 20 in t Thickness of top flange 2 in b1 Width of bottom flange 36 in t1 Thickness of bottom flange 2.5 in d Height of fin 50 cm tw Thickness of the fin 4 cm
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F2: Analysis/6: Section Properties
Entered Values Computed results
To enter the value of 20 inches for the base, at the data input screen ma t ch th e scroll bar to b and press
a
. Type in 20 and press A to append the inch units to the value. Enter the remaining values in a
similar fashion. Press ? to begin the calculating process. T he calculated results are listed below.
Computed Results
Label Description Value
Area Area
.103871 m
2
yy1 Dista nce of center of mass from axis 1 . 224422 m yy2 Dista nce of center of mass from axis 2 . 389878 m I11 Area moment inertia axis 11
.006184 m
rg1 Radius of gyration 1 .244002 m
4
8.7 I Section - Even
The input screen for an even I-Section, (sometimes can be looked upon and a H-Section turned on its side), consist of four inputs the top and bottom flange width b and t, and height d of the fin and its thickness tw. Pressing A reveals a pi ct u r e on a n even I­Section.
As an example, we use an even I-Section with a flange width of 100 cm an d a thickn ess of 8 cm, while the fin has a height of 125 cm and a th ic kness of 10 cm .
Entered values
Label Description Value
b Width of top flange 100 cm t Thickness of top flange 8 cm d Height of fin 125cm tw Thickness of the fin 10 cm
Upper Display Lower Display
To enter the value of 100 cm for the flange width, at the data input screen match the scroll bar to b and press
a
. Type in 100 and press B to append the cm units to the value. Use this procedure to enter
8 cm for t, 125 cm for d and 10 cm for tw. Press ? to begin the calculations. The calculated results are listed below.
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F2: Analysis/6: Section Properties
Computed Results
Label Description Value
Area Area
0.285 m yy1 Distance of center of mass from axis 1 0.705 m yy2 Distance of center of mass from axis 2 0.5 m I11 Area moment inertia axis 11
I22 Area moment inertia axis 12
0.087117 m
0.013438 m rg1 Radius of gyration 1 0.552879 m rg2 Radius of gyration 2 0.217138 m
2
4 4
8.8 C Section
The inpu t screen for the C Section, width of flange b and thickness t, the fins have a height d and thickness tw. A pict u re of C Section can be viewed by pressing A. As an example, we will compute the properties of a C Section with a 15 in flange with a thickness of 1 in, and the fin has a height of 6 in and a thickness of 1.5 in.
Entered values
Label Description Value
b Width of top flange 15 in t Thickness of top flange 1 in d Height of fin 6 in tw Thickness of the fin 1.5 in
Upper Display Lower Display
To enter the value of 15 in fo r the f lange width at the data input screen match the scroll bar to b and press
a
. Type in 15 and press A to append the inch units to the value . Enter the values f or the r est of the parameters in a similar manner. Press ? to begin the calculations. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
0.02129 m yy1 Distance of center of mass from axis 1 0.061191 m yy2 Distance of center of mass from axis 2 0.1905 m I11 Area moment inertia axis 11
I22 Area moment iner t i a ax is 12
0.000065 m
0.00046 m rg1 Radius of gyration 1 0.055133 m rg2 Radius of gyration 2 0.146963 m
2
4
4
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F2: Analysis/6: Section Properties
8.9 T Section
The input screen for a T Section requires the user to enter values of top flange width b and thickness t, along height of the fin and its thickness d and tw.
As an example, we compute the properties of a T Section 9 in wide flange with a thickness of 2 cm, and a fin of height 8 in and thickness 1.5 cm.
Entered values
Label Description Value
b Width of top flange 9 in t Thickness of top flange 2 cm d Height of fin 8 in tw T hickness of the fin 1.5 cm
Upper Display Lower Display
To enter the value of 9 in for the base, at the data input screen ma t ch the scroll bar to b and press
a
. Type in 9 and press A to append the inch units to the value. Enter the other values in a similar fashion. Press ? to begin the calculations. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
0.00762 m yy1 Distan ce of center of mass from axis 1 0.05464 m yy2 Distan ce of center of mass from axis 2 0.1143 m I11 Area moment inertia axis 11 0.000033 m I22 Area moment inertia axis 12
0.00002 m rg1 Radius of gyration 1 0.066223 m rg2 Radius of gyration 2 0.05119 m
2
4
4
8.10 Trapezoid
The inpu t screen for a tr apezoid requi res th e u s er to enter va lue of base b, height h, and offset off, and top width c. In an illustrative example, we use a value of 10 in for the base and 6 in for top, a he igh t of 6 in and an offset of 3.1 in.
Label Description Value
b Base length 10 in h Height 6 in off Offset 3.1 in c Top 6 in
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By pressing Ayou can acces s a schema tic of th e tra p ezoid. To enter the valu e of 10 i nches for the bas e, at the data input screen match the scroll bar to b and press
a
. Type in 10 and press A to append the
inch units to the value. Enter the values for other parameters in a like manner. Press ? to begin the calculations and display the results. The calculated results are listed below.
Upper Display Lower Display
Computed Results
Label Description Value
Area Area
0.030968 m yy1 Distance of center of mass from axis 1 0.08255 m yy2 Distance of center of mass from axis 2 0.114194 m I11 Area moment inertia ax is 11 .000059 m I22 Area moment inertia axis 12 0.000115 m Ip Polar area moment
0.000174 m rg1 Radius of gyration 1 0.043533 m rg2 Radius of gyration 2 0.060989 m
2
4
4
4
8.11 Polygon
A solid n-sided polygon with side a forms the basis of this segment. The input screen requires the user to enter values of side a, and number of sides n. In an illustrative example, we use a valu e of 12 in ches for the side of a 6- si d ed polygon. A cross section of the polygon is shown in the screen display here.
Entered values
Label Description Value
a Side Length 12 in n Number of sides 6
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F2: Analysis/6: Section Properties
Upper Display Lower Display
To enter the value of 12 inches for th e base, at the data input screen ma t ch th e scroll ba r to b and press
a
. Type in 10 and press A to append the inch units to the value. Enter the number of sides in a similar manner. Press ? to begin the calculations and display results. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
α ρ1 ρ2
Angle .523599 rad Radius to point 0.3048 m Radius to line 0.263965 m
.241369m
2
yy1 Distan ce of center of mass from axis 1 0.3048 m yy2 Distan ce of center of mass from axis 2 0.263965 m I11 Area moment inertia axis 11 I22 Area moment inertia axis 12
0.004672 m
0.004672 m rg1 Radius of gyration 1 0.139122 m rg2 Radius of gyration 2 0.139122 m
4 4
8.12 Hollow Polygon
A hollow n-sided polygon with sid e a forms the basis of this segment. Th e input screen requires the user to enter values of side a, wall thickness t and number of sides n. In an illustrative example, we use a value of 19 inches for the side, and a thickness of 2 in of a 9-sided pol yg on. A cross se ction of the polygon is s hown in the s creen display here.
Entered values
Label Description Value
a Side Length 19 in n Number of sides 9 t Thickness 2 in
Upper Display Lower Display
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F2: Analysis/6: Section Properties
To enter the value of 19 inches for th e base, at the data input screen match the scroll bar to b and press
a
. Type in 10 and press A to append the inch units to the value. Enter the number of sides in a similar manner. Press ? to begin the calculations and display results. The calculated results are listed below.
Computed Results
Label Description Value
Area Area
α ρ1 ρ2
Angle .349066 rad Radius to point 0.705514 m Radius to line 0.662966 m
0.212191 m
yy1 Dista nce of center of mass from axis 1 0.694796 m yy2 Dista nce of center of mass from axis 2 0.694796 m I11 Area moment inertia axis 11 I22 Area moment inertia axis 12
0.045103 m
0.045103 m
rg1 Radius of gyration 1 0.461039 m
2
4 4
rg2 Radius of gyration 2 0.461039 m
References:
1. Warren C. Young, Roark's Formulas for Stress and Strain, 6th Edition, McGraw-Hill Book Company, New York, NY, 1989.
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F2: Analysis/7: Hardness Number

Chapter 9: Hardness Number

Brinnell and Vicker developed two popular methods of measuring the Hardness number. Brinell composed his tests by dropping a ball of steel onto slab of finite thickness with standard loads such as 500 lbf and 3000 lbf. This steel ball results in a n indentatio n in the material. Vicker had a similar principle to Brinell; instead of steel ball he used a diamond in the shape of square pyramid. By measuring the diameter o f indentation, di, for Brinells test and the diagonal length o f the impressio n, di, for Vickers test, the hardnes s number is computed . By long standing con vention, the diameter di is measured in millimeters.
9.1 Compute Hardness Number
The tests to measure the hardness number allows a 10 mm diameter steel ball or 10mm diagonal length pyramid to be dr opped onto the material surface with a effective force of 500 lbf or 3000 lbf. The formulas used to compute the hardness number for by Brinell's and Vicker's methods give slightly different result s , but do con verge in sever al areas of th e yiel d cu rves.
Upon select i ng th i s topi c, the in pu t scr een (sh own here) presents four options to the use r . You can choose Brinell's method at 500 lbf or 3000 lbf or Vickers' meth od for the same two load values.
Upon making the selection, enter the measured data for di by aligning the highlight bar to Impression size in mm. This number should be less than 10mm. After entering numbers for di, press the solving process. At the end of the computation, the results are displayed on the screen as shown in the ex ampl e bel ow.
Example 9.1:
We choose an example of 4. 2 mm as t he meas u red value of the indentation di and compute BHN and VHN for both 500-lbf loads and the 3000-lbf loads. Input and computed results are shown below.
Brinell’s output at 500 lbf Brinell’s output at 3000 Ibf
a
to start
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F2: Analysis/7: Hardness Number
Given:
di = 4.2 mm
Results:
Vicker’s output at 500 lbf Vicker’s output at 3000 lbf
Test 500 lbf 3000 lbf Brinnell Test 34.4208 206.525 Vicker Test 48.7528 292.517
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Part II: Equations
MEPro for TI-89, TI-92 Plus
- Equations
54
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F3: Equations

Chapter 10: Introduction to Equations

The Equations section of MEPro contains over 1000 equations organized into 12 topic and 185 sub- topic menus.
The user can select to solve equation sets in a pa rticular sub-topic, display all the variables used in
the set of equations, enter the values for the known variables and solve for the unknowns.
The equations in each sub-topic can be solved individually, collectively or as a sub-set.
A unit management feature allows easy entry an d d isplay of resu l t s.
Each equation can be graphed to examine the relationship between to variables in an equation.
Multiple and partial solutions are possible.
Specific parametric information about a particular variable can be displayed by highlighting the
variable, p ress and ©/
The input form accepts valuables for variables that have physical meaning onl y . For example only
positive real values are accepted for va riabl es suc h as r adius , t h ick ness etc.
10.1 Solving a Set of Equations
Equations are accessed from the main level of the ME•Pro by pressing function key ? labeled
"Equations."
display below.
An arrow to the left of the bottom topic ‘ï‘indicates more items are lis ted. Pressing 2 D jumps to
the bottom of the menu list.
Scroll the highlight bar to an item using the arrow key D and press
number appearin g next to subject heading (Heat Transfer is selected for this example).
A dialog box presents more subjects (sub-topics) under the topic heading. For example selecting the
st
1
item (i.e., Basic Transfer Mechanisms) displays a list of subtopics (Conduction, Convection,
Radiation). Select Conduction to display a set of equations for this section. A complete list of topics in the equation section is listed at this end of the chapter.
Use the a rrow key D to move the highlighter a nd press selects all equations in the set. A selected equation is marked with a check (√).
Press to display all of the variables in the selected equations. A brief description of each variable will appear in the status line at the bottom of the screen.
Enter values for the known par ameters, selecting appropriate units for each value using the toolbar menu.
Press to compute values for th e unknown parameters.
Entered and calculated values are distinguished in the display; ‘é‘ for entered values and ‘
computed results.
A pull-down menu listing all the main categories is displayed as shown in the screen
Type: to show a brief description of a variable and its entry parameters.
¸
, or type the su bject
¸
to select an equation, pr essing
‘ for
1. Pressing displays the
Equations menu. Press y to
select ‘Heat Tr ans fer’.
MEPro for TI-89, TI-92 Plus Chapter 10- Intro to equations
2. Press ¨ to display the menu for ‘Basic Transfer
Mechanisms’.
55
3. Press ¨ to display the
equations for ‘Conduction'.
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F3: Equations
4. Equations found under conduction.
5. Select the equations for a calculation by moving the cursor to each equation and pressing mark ‘
¸
. (Check
’’ appears when
selected).
6. Press to display all the variables in the selected equations. Variable description appears on the status line at the bottom of the screen.
7. Enter values for each known variable. Append the units by pressing the function key
8. Once all known variable values are entered, press ? to solve for the unknowns.
9. Note: Comput ed results
‘ are distinguished from
entered values ‘‘. corresponding to the desired units.
Note:
Only values design at ed as known‘ will be used in a computation. Results displayed from an
earlier calculation will not be automatically used unless designated by the user by selecting the variable and pressing
. Press to compute a new result for any input that is changed.
10.2 Viewing an Equation or Result in Pret ty Print
Some times equations and c alculated results exceed the display room of the calculator. The TI-89 and TI 92 Plus include a built-in equation di spla y feature called Pretty Print which is available in many areas of MEPro and can be activated by highlighting a variable or equation and pressing the right arrow key or pressing the function key when it is designated as View. The object can be scrolled us ing the arrow keys . Pressing or reverts to th e p revious scre en.
1. To view an equation in, 2. Scroll features, using the Pretty Print, highlight arrow keys , enable a
and press or . complete view of a large object.
10.3 Viewing a Result in different units
To view a calcula ted resu lt in units which are different from wha t is displa yed, highlight the variable, pres s
/Options and /Conv to dis play the unit tool bar at the top of the screen. Press the appropriate
function key to convert the result to the desired units.
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F3: Equations
1. Highlight the result to be converted q. Press to display
2. The unit menu tool bar is now displayed.
3. Press to convert the result
of q from W to kW. the Options menu, press : conv.
10.4 Viewing Multiple Solutions
When multiple solutions exist, the user is prompted to select the number of a series of computed answers to be displayed. To view additional solutions, press to repeat the calculation and enter another solution number. The user will need to determine which result is most useful to the application. The following example is taken from Ohm’s Law and Power in the EE for MEs section. E quat i ons/E E f or MEs/Basic Electricity/Ohm’s Law and Power.
1. Return to the HOME screen of ME•Pro (: Tools, : Clear
or press repeatedly) and access the equations section by pressing .
Press for ‘Basic Electricity’. Press for ‘Ohm’s Law and
Power.’
1. Select an equation using highlighting the cursor bar and to display variables.
2. Enter know n values f or each variable using the tool bar to designate units. Press to compute the results.
Solution 1: To view another solution, press to
repeat the calculation and enter the number of
another solution to be viewed.
MEPro for TI-89, TI-92 Plus Chapter 10- Intro to equations
57
3. If multiple solutions exis t, a dialogue box will appear requesting the user to enter the number of a solution to view.
Solution 2: Enter a new number for each ‘solve’ to
display a series of multiple solutions.
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F3: Equations
10.5 When (…) - conditional constrain ts when solving equations
In several sections of MEPro, equations are limited to ce rtain variable ranges. An example can be found in Beams and Columns/Simple Beams/Uniform Load (Chapter 11.1.1): A beam, experiencing a uniform load along a distance, a (m ) , from one end, ha s a deflection, v (m), and slope of deflection, v1, located at a distance, x (m), from the end of the beam. Two sets of unique equations compute v and v1 depending on whether th e condition is x≤≤≤a or x≥≥≥≥a. The conditions for th e equations appear in th e ‘when’ clauses preced e the equ ations. ME Pro allow’s the selection of equations under a single when’ clause since the conditions are generally exclusive to each other
Note: The ‘when’ clauses do not serve any mathematical function in the solving process or
for screening variable entry; they are only a guide for selecting equations for a specific circumstance. Additional information for a ‘when’ clause appears in th e status line while it is
highlighted.
2
.
All equat i ons f ollowing a
highlighted ‘when heading
are selected when
is
Only equations under a single
when’ heading can be selected
at a tim e.
Description of constraint
appears in the status line at
the bottom of the screen.
pressed.
Some equation sets do not form a consistent set, which can be solved together. An example occurs in Equations/Fluid Mechanics/Fluid Dynamics/Equivalent Diameter (see Chapter 21.3.3), where each equation represent s flu id flow through a di fferent-shaped cr oss -section. In such a cas e, th e s p ecific conditions for each equation appear on the status line.
10.5 Arbitrary Integers for periodic solutions to trigonometric functions
When a n an g l e val ue i s bein g compu t ed in a tr i g on om et ric fun ct i on su ch as tangen t , cosine an d si n e, MEPro may prompt an entr y for an arbitrary integer (-2, -1, 0, 1, 2…) before displaying a solution. Solutions for angles inside of trigonometric functions are generally periodic, however the solution, which is most often sought, is the principal solution. The principal solution, P, in a periodic trigonometric function, trig (…), is P=trig (θθθθ + a⋅⋅⋅⋅n⋅⋅⋅⋅ππππ) where n is the arbitrary inte g er and a=1 for tan (…), a=2 for sin (…) and cos (…). Sele c ti ng the arbitrary intege r to be 0 gives the principal solution.
2
In at least one known case (Beams and Columns/Simple Beams/Point load), conditions in more than one
when (…) statement can occur simultaneously. A work around is to solve the equation set in two steps, using equations under a single when (…) heading at a time and designating the results fro m o ne calculat i on as the input into the second.
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F3: Equations
Dialog box for entering an arbitrary integer for a periodic solution.
If ‘0’ is entered, the principal solution (P) for an angle is displayed.
If ‘1’ is entered, a non-principal
solution (P+180
o
) is displayed.
10.7 Partial Solutions
"One partial useable solution found" or "Multiple partial solutions found." will be displayed in the status line if values for one or more variables in the selected equation set cannot be computed. This situation can occur if there are more unknowns than equations in th e selected set, the entered values do not form consistent relationships with th e selected equations, or if the selected equations do not establish a closed form relationship between all of the entered values and the unknowns. In such a case, only the calculated variables will be displayed. This section appears in Equations/EE for MEs/Basic Electricity/Resistance Formulas.
Press to select all of the
equations in Resistance
Formulas.
If there are more unknowns Than selected equations or
Relationships between
Variables are not established
From the selected equati o n s...
...a partial solution will be
displayed if one or more of
the unknown variables can
be computed from the
entered inputs.
10.8 Copy/Paste
A computed result and it’s expressed units can be copied and pasted to an appropriate part of the TI operating system using : Tools-5: Copy key sequence to copy a value and : Tools-6: Paste. In a few cases, The TI-89 and TI-92 operating systems, and MEPro use slightly different conventions for displaying units. The unit system in MEPro is designed to conform to t he convention established by SI, however, in order to CUT and PASTE a value and units from ME Pro to another area of the TI operatin g system, MEPro must insert extra characters in the units to match TI’s syntax. The COPY/PASTE function can only work INSIDE of ME•••Pro if the unit feature has been deactivated (press :Opts,
: Units to toggle the unit feature on or off).
10.9 Graphing a Function
The relationship between two variables in an equation can be graphed on a real number scale if the other variables in the equation are defined.
After solving an equation, or entering values for the non-x, y
press /Graph to display the graph settings.
Highlight Eq: and press
to select the equation from the list to graph.
Use the s ame steps as above to select t he i n dependent and dependent variables (Indep: and Depnd:)
from the equation.
Note: all pre-existing values stored in the variables used for Indep: and Depnd : will be cleared
when the graphing function is executed.
for TI-89, TI-92 Plus
MEPro Chapter 10- Intro to equations
59
variables in the equation to be plotted,
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F3: Equations
The graphing unit scal e for each varia bl e reflects the settings in the Equations section of ME•Pro.
Scrolling down the list, specify the graphing rang e s for the x and y variables, whether to graph in f ull
or split screen modes, automatically scale the graph to fit the viewing area, and label the graph.
Press to graph the function.
Once t he gr aph comm and has been executed, MEPro will open a second window to display the plot. All of the TI graphing features are available and are displayed in the toolbar, including Zoom , Trace , Math , etc. All the tools from TI graphics engine are now available to t he user. If the split-screen graphing mode is activated, the user can toggle between the MEPro gr a ph dialogue display and the TI graph by pressing 2 MEPro and the graph by pressing
O
. If the full-screen graphing mode is activated, the user can switch between
O
4:Graph or A: MEPro. To remove the split screen after
graphing, you will need to change the display settings in th e MODE screen of the calculator. To do this:
1) Press to display a pop-up menu. 5) Select FULL. 6) Press
2) Press : Page 2; 3) move the cursor to Split Screen. 4) Press the right arrow key
twice.
1*. Graph an equation by
pressing . Press
choose an equation.
to
2. Select variables f or
Independent (x) and Dependent
(y) variables.
Variable units reflect sett i ngs
Pro.
in ME
4. Select graphing options by pressing
.
5. Split Screen Mode: Toggle Between graph and settings by pressing and
.
6: Full Screen Mode: Pr es s
and
to return to
Pro.
ME
*Before graphing an equation, be sure to specify v alues for var iables in an equation, which are not going to be used as x, and y variables.
Note:
If an error is generated when attempti ng to grap h, be sure that all of the variables in the graphed
equation, which are not specified as the independent, and dependent variables have known values. In the MEPro window, press N to view the equations in the sub-topic, select the equation to be graphed and
press to display the list of variables in the equation and enter values. Only the dependent (y) and independent (x) variables do not have to contain specified values. Press to display the graph dialogue and repea t the above step s to graph the fu nction.
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F3: Equations
10.10 Storing and recalling variable values in ME•••Pro-creation of session
folders
ME•Pro automatically stores its variables in the current folder specified by the user in screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values for a particular session of ME•Pro, press :/TOOLS, :/NEW and type the name of the new folder. There are several ways to display or recall a value:
any
The conten ts of variables in
folder can be displayed using the
cursor to the variable name and pressing to display the contents of a particular variable.
Variables in a current fo ld e r c an b e recalled in the HO M E screen by typing the variable name .
Finally, values and units can be copied and recalled using th e /Tools : COPY and
: PASTE feature. All inputs and calculated results in Analysis and Equations are saved as variable names. Previously calculated, or entered values for variables in a folder are replaced when equations are solved using new values for inputs. To preserve data under variable names, which may conflict with ME•Pro variabl e s , run ME•Pro in a separate folder.
or the HOME
, m oving th e
10.11 solve, nsolve, and csolve and user-defined functions (UDF)
When a set of equations is solved in MEPro, three different functions in the TI operating system (solve, numeric solve, and complex solve) are used to find the most appropriate solution. In a majority of cases, the entered value s are adequate to find numeric solutions using eithe r solve, or csolve functions. However, there are a few instances when UDFs external to equations are incorporated into the solving process. User
defined functions which appear MEPro are th e error functions erfc (x) and erf (x). When all the inputs to a UDF are known, solve or csolve passes a computed result to the equation: however, if the unknown variable is an input to the UDF, solve or csolve are unable to isolate the variable in an explicit fo rm, and the ope rating sy s tem resorts to using nsolve which initiates a trial and error iteration until the s ol ution converges. It s hould be noted that the solution gene rated by nsolve i s not guaranteed to be unique (i.e. this solving process cannot determine if multiple solutions exist.).
Because ns olve is us ed, an equation containing a user-defined f unction ( UDF) cannot be graphed when the dependent variable is contained in the UDF.
Table 15-1 User Defined Functions
2
User-defined Function Topic Sub-topic
erf(x, αd, time) erfc (x, αd, time) erfc (x, αd, h, k, time)
Heat Transfer Step Cha nge in Surface Tem perature Heat Transfer Constant Surface Heat Flux Heat Transfer Surface Convection
10.12 Entering a guessed value for the unknown using nsolve
To accelerat e the con verging process a n d, if multiple solutions exist, enhance the possibility that nsolve res olves the correc t solution, the user can enter a guesse d value for the unk nown which nsolv e w ill use as an initial value in its solving process.
Enter guessed a value for the variable in the input dialogue.
Press /Opts, m/Want.
Press / to compute a solution for the
variable.
erfc(h) is a user-defined function (UDF) that
appears in the ‘Surface Convection’ topic of ‘Heat
for TI-89, TI-92 Plus
MEPro Chapter 10- Intro to equations
Only one variable in a user-defined
function can be specified as an unknown.
61
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F3: Equations
Transfer’. An initial value for the unknown can be
specified for the n-solve process :Opts,
: Want.
Pro posts a notice if the nsolve routine is
ME
used.
The user can enter a value for the
unknown and designate it as a guessed
value to accelerate the nsolve
convergence process.
10.13 Why can't I compute a solution?
If a solution is unable to be computed, you might check the following:
1. Are there at least as ma ny equa tions selected, as there are unkno w n pa ramet ers ?
2. Are the entered values or units for the known parameters reasonable for a specific case?
3. Are the s elected equations consistent in desc ribi ng a pa rticular case? (For example, certain equations
used in t he calcul a t i on of equival ent cir cu l a r cross-secti on d i a m et ers in Fluid Mechanics/Fluid Dynamics/Equivalent Diamete r, ar e only valid for certain sh aped cr oss se ctions) . Check the headings displayed at the bottom of the screen while the equation is highlighted to determine if special restrictions for a particular equation (set) apply.
10.14 Care in choosing a consistent set of equations
The success in obtainin g a useful solution, or a solution at all is strongly dependent on an insight into the problem and care in choosing equations, which describe consistent relationships between the parameters.
The following steps are recommended:
Read the description of each set of equations in a topic to determine which subsets of equations in a
seri es are compa tible a nd consisten t in describing a particular case. Some restrictions or special conditio ns for an equation o r set of equations are listed in the status line while the equation or ‘when’ conditions are highlighted.
Select the equations from a subse t, which describe the relat ionships between al l of the known and
unknown parameters.
As a rule of thumb, sel ect as ma ny equations from the subset as there are unknowns to avoid
redu ndancy or over-specification. Th e eq uati ons have been researched from a variety of sources and use slightly different approximation techniques. Over-specification (selecting too many equations) may lead to an inability of the equation solver to resolve slight numerical differences in different empirical methods o f calculating valu e s for the same variable.
10.15 Notes for the advanced user in troubleshooting calculations
When th ere are no solu tions possible, ME•Pro provides imp ortant clue s via the variables, meinput, meprob, means, and meanstyp. These va riables are defined dur ing the equatio n setup process by the
built-in multiple equation solver. MEPro saves a copy of the problem, it s input s, i ts outputs, and a characterization of the type of solution in the user variables mepr ob, me input, me ans, and meanstyp. For the developer who is curious to know exactly how the problem was entered into the multiple equation solver, or about what the multiple equation solver returne d , a n d to ex a mine relev a n t s t r i ng s . The c ontents of these variables may be viewed and examined by using scroll to the variable name in the current folder and press to view the conten ts of the variable. The
for TI-89, TI-92 Plus
MEPro Chapter 10- Intro to equations
62
. Press
(
followed by |),
Page 63
F3: Equations
string may be recalled to the status line of the home screen, modified and re-executed, if desired. If no
solutions are possible when one should be displayed, try clearing the variables in the current folder, or opening a new folder.
for TI-89, TI-92 Plus
MEPro Chapter 10- Intro to equations
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F3: Equations/1: Beams & Columns

Chapter 11: Beams and Columns

This chapter cov ers the details found in the Beams and Columns section. Three broad areas of common interest are cover ed . Th ey are:

Simple Beams Columns

Cantilever Beams
11.1 Simple Beams
Structural members designed to resist forces acting perpendicular to its longitud inal axis are de fined as Beams. The simplest types of beams can be thought of "planar structures" wherei n all the defl ections occur in the same plan e. T he essential features of a Simple Beam, or Simply Supported Beam, a re a pin support is placed at one end and a roller suppo rt at the oth er. The equ ati on s us ed here co ver Simp l e Beam s where the rollers support are always at the right end of the beam.
Three types of loads a re considered here. They ar e Uniform load, Point load and Moment load. Each of thes e load type s has unique characteristics and is illu strated in the software.
11.1.1 Uniform Load
The equation set here covers problems associated with a un iformly distributed load, p1 (N/m), covering a distance, a (m), from the left end. Equations 1 and 2 yield the slopes at the left and right ends of the beam, θθθθa (rad), a nd, θθθθb (rad), respectively. To com p u te th e d efl ection, v (m), and slope, v1, at any location, x (m) , from the left, use equation s 3 and 4 when x≤≤≤a. When, x, lies between, a. and the beam length, L (m), use equations 5 and 6 to compute, v, and v1. E (Pa), the modulus of elas ticity and I (m
4
), the area moment of inertia, repres ent the material prop erties o f the be am.
2
pa
tanθa
16 1 6
tanθb
16
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
24
=
px
⋅⋅⋅
v
v
24
When a ≤≤≤ x ≤≤≤≤ L, the fol lowin g two equ at ions are applicable
for TI-89, TI-92 Plus
MEPro Chapter 11 - Equations - Beams and Columns
1
=
=
24
24
1
LEI
⋅⋅⋅
pa
1
LEI
⋅⋅⋅
LEI
p
1
LEI
⋅⋅⋅
⋅⋅−
2
2
⋅⋅−
2
27
4 3 22 22 2 3
a a L a L a x aLx Lx=
⋅ −⋅ ⋅+⋅ ⋅ +⋅ ⋅ −⋅⋅⋅ +⋅
4424
27
43 2222 2 3
aaLaLax aLxLx1
⋅ −⋅ ⋅+⋅ ⋅ +⋅ ⋅ − ⋅⋅⋅ +⋅⋅
44 6 12 4
27
2
La
22
La
64
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Page 65
F3: Equations/1: Beams & Columns
2
pa
24
=
24
1
LEI
⋅⋅⋅
2
pa
1
LEI
⋅⋅⋅
v
v
222 23
aL Lxax Lx x=
⋅− ⋅ + ⋅ ⋅ + ⋅ − ⋅ ⋅ + ⋅
27
4126
⋅⋅ + −⋅⋅+⋅
27
462
22 2
La Lx x1
Eq. 5
Eq. 6
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
θa θb
a Load location from left (fixed) end m E Young's modulus Pa I Area moment of inertia
L Length m p1 Load/unit length N/m v Beam deflection m v1 Slope of deflection un itless x Dist. from left end m
Caution: Because the equa tions repres ent a s et wher e several subtopics are covered, the user h a s to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
Example 11.1.1:
A simple beam, 10 meters long, is subject to a uniform load of 1.5 kips/ft, spanning 18 feet from the left end. Find the slopes at left and right ends and deflection at mid-point of the beam. Assume that the Young's modulus of the beam material is 190 GPa and that the area moment i s 170 in4.
Angle at left (fixed) end rad Angle at right (roller supported) end rad
4
m
Upper Display Lower Display
Solution – Select the first three equations to solve th is problem. Select these by highli ghting the equations and pressing parameters and press to solve for the unknown variables. Select 0 to compute the principal solution). The entries and results are shown in the s c reen di splays above.
Given Solution
a = 18 ft E = 190 GPa
I = 170 in L = 10 m
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
. Press to display the variables. Enter the values for the known
4
- Beams and Columns
θa = 2.46317 ° θb = 1.98714 °
v = .12243 m
65
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F3: Equations/1: Beams & Columns
Given Solution
p1 = 1.5 kip/ft x = 5 m
11.1.2 Point Load
This equatio n set c overs problems associated w ith a point load, P (N), located a distance, a (m), from the left (fixed) end of a simple beam. Equations 1 and 2 yield the slope angles, θθθθa (rad), θθθθb (rad) at the left and right ends of the bea m, r es p ectively. Equation 3 computes the length of the beam, L (m), in terms of the load distance from the left, a (m), and ri ght, b (m) ends. T o com p ute the deflection, v (m), and slope v1 at any location, x (m), from the left side, use eq uations 4 and 5 when 0≤≤≤x≤≤≤≤a. Equation 6 yields δδδδc (m), the deflection at the center of the beam when a≥≥≥b. Equatio n 7 computes x1 (m), the location of maximum deflection, δδδδmax (m), calculated by equation 8. The material properties of the beam are represented by, E (Pa) the modulus of elasticity, I (m
4
) the area moment of inertia, and L (m), the length of the beam.
Pab L b
⋅⋅⋅ +
tanθa
=
16
Pab L a
tanθb
Lab=+
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
v
v
When a ≥≥≥ b, the fol lowin g two equ at ions are applicable
δ
=
16
Pbx
⋅⋅
LEI
⋅⋅⋅
6
Pb
=
c
LEI
6
⋅⋅⋅
Pb L b
⋅⋅ ⋅ −⋅
27
=
48
22
Lb
x
1
=
16
LEI
⋅⋅⋅6
⋅⋅⋅ +
16
LEI
⋅⋅⋅6
222
Lb x=
⋅−−
27
22 2
Lb x1
⋅−−
27
22
34
EI
⋅⋅
3
3
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
Eq. 7
15
22
⋅⋅
Pb L b
δ
max =
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
The variable names, description an d applicable default units used in these equations are listed below.
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
27
5
.
⋅⋅⋅
243
LEI
- Beams and Columns
.
66
Eq. 8
Page 67
F3: Equations/1: Beams & Columns
Variable Description Units
δc δmax θa θb
Deflection at mid point m Maximum deflection at x1 m Angle at left (fixed) end r a d
Angle at right (roller supported) end rad a Load location from left (fixed) end m b Dist. from r i ght (roller supported) end m E Young's modulus Pa I Area moment
4
m L Length m P Point load N v Beam deflection m v1 Slope of deflection unitless x Distance from left end m xl Maximum deflection location m
Caution: Because the equa tions repres ent a s et wher e several s ubtopics are covered, the use r has to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
Example 11.1.2:
A si mple beam, 50 ft long, is subject to a point load of 4000 lbf located 10 feet from the left end and a second load of 10 kN located 10 feet from the right end. Find the deflection at mid-beam and the slope at both ends. Assume that the Young's modulus of the beam material is 120 GPa, and the area moment is 1650 in
4
.
Solution - The problem is solved in two stages. First, compute the slopes at both ends, and the deflection at the center for the load of 10 kN. Repeat the calculations using the second load and add the two computed values using the superposition to calculate the final result. S elect an
arbitrary integer of 0 to compute the principal solution (the principal solution, P, in a periodic trigonometric function, trig (…), is P=trig (θ + n⋅π) and n is the arbitrary integer).
First load: Upper Dis play First load: Lower Display
Use Equations 1, 2 and 6 to calculate the results from the first load.
Given Solution
a = 40 ft b = 10 ft E = 120 GPa
I = 1650 in L = 50 ft P = 10 kN
= .005082 m
δc θa = .05167 deg θb = 0.077506 deg
4
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
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F3: Equations/1: Beams & Columns
Second load: Upper Display Second load: Lower Display
Use Equations 1, 2 and 4 to calculate the results from the second load.
Given Solution
a = 10ft b = 40 ft E = 120 GPa
I = 1650 in L = 50 ft P = 4000 lbf x = 25 ft
The final results of the two loads are obtained by invoking the super position principle; thus add the results of the two sets of calculat ions.
Solution
First load .05167 deg .077506 deg .005082 m Second load .091937 deg .13905 deg .005604 m
Total .143607 deg .216556 deg .010686 m
4
θθθθa θθθθb δδδδc/v
θa = .137905 deg θb = .091937 deg
v =.005604 cm
11.1.3 Moment Load
This equatio n set covers pro blems associated w ith a moment load,
MOM (Nm), applied at a distance, a (m), from the left end. Equations 1 and 2 calculate t he slope angles, θθθθa (rad), and, θθθθb
(rad), at the left and right ends of the beam, respectively. The deflecti on d i st a nce, v (m), and slope, v1, at distance, x (m), from the left si d e of t he beam are comp u ted by Equation s 3 and 4. The material properties of the beam are represented by, E (Pa) the modulus of elasticity, I (m of the beam.
4
) the area moment of inertia, and, L (m) the length
tanθa
16
tanθb
16
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
MOM x
v
⋅⋅⋅
6
v
=
6
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
MOM
=
LEI
⋅⋅⋅
6
MOM
=
LEI
⋅⋅⋅
6
LEI MOM
LEI
⋅⋅⋅
- Beams and Columns
aL a L
⋅⋅⋅−⋅ −⋅
632
27
22
aL
⋅⋅ −
3
27
aL a L x=
⋅⋅⋅−⋅−⋅−
632
27
aL a L x1
6323
⋅⋅⋅−⋅ −⋅ −⋅
27
22
222
222
68
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Page 69
F3: Equations/1: Beams & Columns
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Units
θa θb
Angle at left (fixed) end r a d
Angle at right (roller supported) end rad a Distance of load from left (fixed) end m E Young's modulus Pa I Area moment of inertia
4
m L Length m MOM Moment applied to beam
Nm v Beam deflection m v1 Slope of deflection unitless x Distance from left (fixed) end m
Caution: Because the equa tion s repres ent a set wher e s everal subtopics are covered, the use r has to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
Example 11.1.3:
A si mple beam, 10 meters long, i s subject to a moment load of 10kN⋅m at the middle point of the beam. Find the slopes at left and right ends of the beam, and deflection at mid point of the beam. Assume that the Young's modulus of the beam material is 100 GPa, and the area moment to be 125 in
4
.
Upper Display Lower Display
Solution – Select the first, second, and fourth equations to solve this problem. Select these by highlighting the equations and pressing
. Press to display the variables. Enter the values for the known parameters and press to so lve fo r the unkno wn variables. Select an arbitrary integer of 0 to compute the principal solu tion (the principal solution, P, in a periodic trigono me tric function, trig (…), is P=trig (θ + n⋅π) and n is the arbitrary integer) . The entries and re sults are show n in the screen displays above.
Given Solution
a = 5 m E = 100 GPa
I = 125 in L = 10 m
MOM = 10 kN⋅m x = 5 m
θa = .045885 deg θb = -.046885 deg
4
v1 = -.001602
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
69
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F3: Equations/1: Beams & Columns
11.2 Cantilever Beams
Structural members designed to resist forces acting perpendicular to its longitudinal axis are defined as Beams. The essential features of a Cantilever Beam are that it has a fixed support at one end and a free at the other end with no support. The fixed end of the beam is inflexible and does not incur any bending, thus ensuring the deflection and slope to be zero at this end.
Three types of loads are considered here. They are Uniform load, Point load, and Moment load. Each of these load types has unique characteristics and is illustrated in the software.
11.2.1 Uniform Load
This equatio n set covers problems associated with a uniformly distributed load, p1 (N/m), over a distance, a (m), from the left (fixed) end. Equations 1and 2 yield the deflection, δδδδb (m), and slope angle, θθθθb (rad), at the right (free) end of the beam. Deflection v (m), and slope v1, at any location x (m), away from the left can be computed by using equatio ns 3 and 4, when 0≤≤≤x≤≤≤≤a. When x lies between a, and beam length L, use equ a t ions 5 and 6 for the same properties. The material properties of the beam are represented by, the modulus of elasticity E (Pa), and, the area moment of inertia I (m
4
).
3
pa
24
16
px
1
⋅⋅
24
px
=
6
⋅⋅
pa
1
24
pa
1
=
6
⋅⋅
1
EI
⋅⋅
pa
=
⋅⋅
6
2
1
La=
⋅⋅−
4
16
3
EI
22
aaxx=
⋅⋅ −⋅⋅+
64
27
EI
1
22
aaxx1
33
⋅⋅ −⋅⋅+
27
EI
3
EI
⋅⋅
3
xa=
⋅⋅−
4
16
EI
δ
b
tanθb
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
v
v
When a ≤≤≤ x ≤≤≤≤ L, the fol lowin g two equ at ions are applicable
v
v
1
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
δb
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
Deflection at right (free) end m
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F3: Equations/1: Beams & Columns
Variable Description Units
θb
Slope at right (free) end unitless a Distance of load from left (fixed) end m E Young's modulus Pa I Area moment of inertia
4
m L Length m p1 Load/unit length N/m v Beam deflection m v1 Slope of deflection unitless x Distance from left (fixed) end m
Caution: Because the equa tions represent a set wh ere several subtop ics are covered , the user ha s to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
Example 11.2.1:
A cantil ever beam, 10 meters long, is subject to a uniform load of 2.5 k N/m, 18 ft from the fixed end. Find the slope and deflection at the free end of the beam, as well as at the mid-point of the beam. Assume that the Young's modulus of the beam material is 190 GPa and that the area moment is 170 in
4
.
Upper Display Lower Display
Solution – Select equations 1, 2, 5 and 6 to solve this probl em. Select these by highlighting the equations and pressing
. Press to display the variables. Enter the values for the known parameters and press to solve for the unknown variables. Select the integer of 0 to comput e t he pr inc ipal soluti on. The e ntrie s a nd results a re shown in the s c reen displ ays above.
Given Solution
a = 18 ft E = 190 GPa
I = 170 in
δb = .044162 m θb = .293246 deg
4
v = .018571 m
L = 10 m v1 = .005118 p1 = 2.5 kN/m x = 5 m
11.2.2 Point Load
This equatio n set c overs problems associated w ith a point load, P (N), located a distance, a (m), from the le ft (fixed) end. Equations 1 and 2 yield the deflection, δδδδb (m), and slope angle, θθθθb (rad), at the right (free) end of the bea m. To compute the deflect ion, v (m), and slope, v1, at a location x (m) from the left end, use equations 3 and 4, when
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
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x occurs before the location of the load, a (m) (0≤≤≤x≤≤≤≤a), and equations 5 and 6 when x occurs after the load (a≤≤≤x≤≤≤≤L). The ma teri al propert ie of the beam i s represen te d by E (Pa), the modulus of elasticity by E (Pa), the area moment of inertiaby, I (m
4
), and beam length, L (m).
2
Pa L a
⋅⋅⋅−
3
δ
b
=
tanθb
16
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
Px
v
6
v
2
When a ≤≤≤ x ≤≤≤≤ L, the fol lowin g two equ at ions are applicable
Pa x a
v
=
v
1
=
2
16
EI
⋅⋅
6
2
Pa
=
EI
⋅⋅
2
2
EI
⋅⋅
Px
EI
⋅⋅
⋅⋅⋅−
Pa
EI
⋅⋅
ax=
⋅⋅−
3
16
2=
⋅⋅−
16
2
3
16
EI
⋅⋅
6
2
ax1
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
δb θb
a Distance of load from left (fixed) end m E Young's modulus Pa I Area moment
L Length m P Load N v Beam deflection m v1 Slope of deflection unitless x Distance from left (fixed) end m
Deflection at left (fixed) end m Angle at right (free) end of beam rad
4
m
Caution: Because the equa tions repres ent a set where several subtop ics are covered , the user ha s to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
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F3: Equations/1: Beams & Columns
Example 11.2.2:
A cantilever beam, 20 m long, is subject to a 15 kN point load located 18 feet from the fixed end. Find the slope and deflection at the free end of the beam. Assume that the Young's modulus of the beam material is 175 GPa, and that the area moment is 650 in
Solution – Select the first two equations to solve th is problem. Select these by highli ghting the equations and pressing parameters and press to solve for the unknown variables. The entries and results are shown in the scr een di s p l ays above.
Given Solution
a= 18 ft E = 175 GPa I = 650 in L = 20 m P = 15 kN
4
Entered Values Computed results
. Press to display the variables. Enter the values for the known
4
δb = .086643 m θb = .273193 deg
.
11.2.3 Moment Load
This equation set covers problems associated w ith a moment loa d,
MOM (Nm), applied at a distance a (m), from the left (fixed) end. Equations 1 and 2 calculate t he deflect ion, δδδδb (m), and slope, θθθθb
(rad), at the right (free) en d. T he deflection and slope, v (m) and v1, at a distance, x (m), from the left (fixed) side of the beam are defined by equations 3 and 4 for the ca se when 0≤≤≤x≤≤≤≤a, and equat ions 5 and 6 when a<x<L. The materials properti es of the beam a re represented by E (Pa), the modulus of elasticity; I (m inertia; and L (m), the length of the beam.
MOM a
δ
b
MOM a
θ
b
=
When 0 ≤≤≤ x ≤≤≤≤ a, the foll owing t wo equati ons are applicable
MOM x
v
=
MOM x
v
1=
EI
⋅⋅
2
EI
EI
⋅⋅2
EI
⋅⋅−
2
16
2
La=
4
), the area moment of
Eq. 1
Eq. 2
Eq. 3
Eq. 4
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
- Beams and Columns
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F3: Equations/1: Beams & Columns
When a ≤≤≤ x ≤≤≤≤ L, the fol lowin g two equ at ions are applicable
MOM a
v
v
1=
When 0 ≤≤≤ x ≤≤≤≤ L, applies to all equations
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
δb θb
a Dista nce of load from left (fixed) end m E Young's modulus Pa I Area moment
L Length m MOM Applied moment v Beam deflection m v1 Slope of deflection unitless x Distance from left (fixed) end m
EI
⋅⋅
2
MOM a
EI
xa=
⋅⋅−
2
16
Deflection at right (free) end m Angle at right (free) end rad
Eq. 5
Eq. 6
4
m Nm
Caution: Because the equa tions repres ent a s et wher e several s ubtopics are covered, the use r has to select each equation to be included in the multiple equation solver. Pressing will not select all the equations and start the solver.
Example 11.2.3:
A si mple beam, 10 meters long, is subject to a moment load of 1.5 ftkip, 18 feet from the fixed end. Find the slope and deflection at the right end of the beam, and the deflection at mid point of the beam. Assume that the Young's modulus of the beam material is 190 GPa, and that the area moment is 170 in
Solution – Since the load occurs to the right of the midpoint of the beam (x=L/2<a<L, since x=5 m =
16.4 ft <18 ft=a) use equations 1, 2, and 3, to solve this problem. S elect these equations and press to display th e variables. Enter t he values for the known parameters and press to solve for the unknown variables. Select an arbitrary integer of 0 to comp ute the princi p a l solu t i on. The entries and result s a re shown in the screen d i s p l ays above.
4
.
Upper Display Lower Display
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- Equations
- Beams and Columns
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Given Solution
a = 18 ft E = 190 GPa I = 170 in L = 10 m MOM = 1.5 ft⋅kip x =5 m
4
v = .001891 m
δb = .006023 m θb = .047552 deg
11.3. Columns
Structures fail a variety of ways depending upon material properties, loads and conditions of support. In this section, a variety of failures of columns will be considered. For our purpos e, we de fine a column to be long slender structural members loaded axially in compression.
Five type of pr oblems are considered - Buckling, Eccentric Axial load, Secant formula, Column imperfections, and Inelastic buckling.
11.3.1 Buckling
These four equations give an insight into the critical parameters for designing columns. Eq uations 1 and 2 compute the critical load, Pcr (N), in terms of the cross-sectional ar ea of the column, Area (m column length, L (m), ra di us of gyration, r (m), and the area moment of inertia, I
4
). Th e compressive stress, σσσσcr (Pa), is calculated from equatio n 3 and the
(m radius of gyration r is computed in equation 4.
2
), the modulus of elasticity, E (Pa), the
Eq. 1
Eq. 2
Eq. 3
Pcr
Pcr
σ
cr
2
EArea
⋅⋅
π
=
Ke L
r
2
EI
⋅⋅
π
=
Ke L
16
Pcr
=
2
2
Area
Eq. 4
r
I
=
Area
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
σcr Area Area
E Young's modulus Pa I Area moment Ke Effective length factor unitless
Critical stress Pa
m
m
2
4
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- Beams and Columns
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Variable Description Units
L Length m Pcr Critical load N r Radius of gyration m
Example 11.3.1:
A steel column, with an area of 50 in The effective length factor for this column is 1.75, and the area moment is 600 in critical load, the radius of gyration and critical stress.
Solution – Select all of the equations to solve this problem. Press to display the variables. Enter the values for the known parameters and press to solve for the unknown variables. The entries and res u l t s are sh own in the scr e en di s p l a ys above .
Given Solution
Area = 50 in E = 190 GPa Pcr = 925.096 kip I = 600 in Ke = 1.75 L = 20 ft
2
and 20 ft long, has a modulus of elasticity of 190 GPa.
Entered Values Computed results
2
4
σcr = 18.5019 ksi r = 3.4641 in
4
. Find the
11.3.2 Eccentricity, Axial Load
An eccentric load refers to t he sit uati on where th e p oi nt load, P (N), is not loc ated a t th e center of mass of the area, but is offset by a dis tance, xe (m). Equation 1 computes the deflection of the column at the mid­point, δδδδc (m). Equation 2 calculates the maximum bending moment, Mmax (Nm ), for the sp ecified eccentric load. The l oca tion of Mmax is the same as the location of maximum deflection (typically at half the height of the column, L/2).
Eq. 1
Eq. 2
⋅⋅
EI
⋅⋅
P
cos
2
kL
2
Deflection at mid-point m
PxeL
δ
c
=
8
Mmaxxe=
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
δc
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- Equations
- Beams and Columns
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Variable Description Units
E Young's modulus Pa I Area moment of inertia
4
m k Stiffness 1/m L Length m Mmax Maximum bending moment
Nm P Point load N xe Eccentricity offset m
Example 11.3.2:
A 20 kip point load on the column described in Example 11.3.1 is offset 2 inches from the column’s central axis. Find the buckling at the center of the column and the maximum moment if the stif fness is 1 x 10
-5
1/m.
Entered Values Computed results
Solution – Select both equations to solve this problem. Press to display the variables. Enter the values for the known parameters and press to solve for the unknown variables. The entries and res u l t s are sh own in the scr e en di s p l a ys above .
Given Solution
E = 190 GPa I = 600 in
4
δc = .017418 in Mmax = 3.33333 ftkip
k = .00001 1/m L = 20 ft P = 20 kip xe = 2 in
11.3.3 Secant Formula
The e quations in this sect ion focus on further analysi s of eccentric loads. Equation 1 computes the radius of gyration, r (m), from the area moment I (m Equations 2 and 3 show two ways to calculate maximum stress, σmax (Pa), from known pr operties of the column in cluding length, L (m); load, P (N); Area; and the modulus of elasticity, E (Pa). Two key rat ios, the eccent rici ty rati o, ecr and the slenderness ratio, sr, are introduced in equat ions 4 and 5. T he eccentric offset , xe (m), is distance of the load from the central longitudinal axis of the column and c (m) is the distance from the axis to the concave side of the column at the location of m aximum deflection.
r
=
Area
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
4
), and the cross sectional area, Area (m2).
I
- Beams and Columns
Eq. 1
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σ
max
σ
max
ecr
=
sr
=
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
σmax Area Area c Distance of cen troid t o colu mn ed ge m E Young's modulus Pa ecr Eccentricity ratio unitless I Area moment of inertia
L Length m Mmax Maximum bending moment P Point load N r Radius of gyration m S Section modulus
sr Slenderness ratio unitless xe Eccentric offset m
P
= +
Area
P
= ⋅+⋅⋅
Area
xe c
2
r
L r
Mmax
S
 
xe c
1
 
Maximum stress Pa
2
r
cos
 
1
L
rPEArea
2
 
 
Eq. 2
Eq. 3
Eq. 4
Eq. 5
2
m
4
m Nm
3
m
Example 11.3.3:
The column described in example 11.3.1 and 11.3.2 has a 20 kip axial load located at distance of 4 inches from the concave extremum. Find the maximum st r ess, radius of gyration and the slenderness ratio.
MEPro for TI-89, TI-92 Plus Chapter 11
- Equations
Upper Display Lower Display
78
- Beams and Columns
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Solution – Select equations 1, 3, 4, and 5 to solve th is problem . Select these by highlighting the equations and pressing parameters and press to solve for the unknown variables. The entries and results are shown in th e scr een di s p l ays above.
. Press to display the variables. Enter the values for the known
Given Solution
Area = 50 in
c = 4 in E = 190 GPa
I = 600 in
L = 20 ft
P = 20 kip
xe = 2 in
2
4
σmax = 4.61263 MPa
ecr = .666667
r = 3.4641 in
sr = 69.282
11.3.4 Imperfections in Columns
The approaches used in analyzing the impact of imperfections in columns are defined by the seven equations here. A column having a length, L (m), and initial deflection, ao (m), from the centroid (central longitudinal) axis is considered. The ratio αααα (unitless), of the applied load, P (N), to critical load, Pcr (N), for the column is calculated in equation 1. An alternate way to compute αααα is using equation 2, which uses area moment I (m and modulus of elasticity, E (Pa). Equations 3 and 4 compute maximum bending moment Mmax (N⋅m) and the maximum deflection, δδδδmax (Pa), at the mid-point of the column. The three remaining equations solve for the maximum shear stress, σσσσmax (Pa), using alternate methods depending on which variables are known. The variable, r (m), is the ra dius of gyration and, c (m), is the horizontal distance from the centroid axis to the furthest point on the concave side of the column.
P
α
=
4
), column length L (m), applied load P;
Eq. 1
Pcr
2
PL
α
=
M
max =
δ
max =
σ
max
2
π
=+
⋅⋅
EI
Pao
1
α
−ao1
α
Area
σ
max =⋅+
P
Area
MEPro for TI-89, TI-92 Plus
Chapter 11 - Equa tions
- Beams and Columns
max
Mc
P
I
 
ao c
2
⋅−
r11
16
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
 
α
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Eq. 7
 
max
σ
The variable names, d es cription and appli cable defau lt units used i n the eq uations above are listed below.
Variable Description Units
α δmax σmax
ao Deflection w/o load m Area Area
c Centroid offset extremum m E Young's modulus Pa I Area moment of inertia
L Length m Mmax Maximum bending moment P Point load N Pcr Critical load N r Radius of gyration m
P
=⋅+
Area
1
 
1
π
2
ao c
2
r
P
⋅⋅
EAreaLr
Ratio unitless Maximum deflection due to load m Maximum stress Pa
 
2
2
m
4
m Nm
Example 11.3.4:
Find the maximum deflection for the column described in example 11.3.1 and 11.3.2. Assume an initial deflection of 4 inches exists a t the m id- poin t of this column.
Solution – Select equations 2, 3, and 4 t o sol ve t his problem. Select t hese by h i ghli ghting the equations and pressing press to solve for the unknown variables. The entries and results are shown in the screen displays above.
Given Solution
ao = 4 in
E = 190 GPa I = 600 in
L = 20 ft P = 20 kip
Entered Values Computed results
. Press to display the variables. Enter the values for the known parameters and
4
α = .007059 δmax = 4.02844 in Mmax = 6.71406 ftkip
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Chapter 11 - Equa tions
- Beams and Columns
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11.3.5 Inelastic Buckling
Critical load calculations for inelastic buckling are considered in this section. Equation 1 computes the tangenti a l modulus load Pt (N) given the column length L (m), the area moment I (m corresponding critical stress, σσσσt (Pa), is computed in equation 2 in terms of Et, column length, L (m), an d the radius of gyration, r (m). Equations 3 and 4 define a reduced modulus of elasticity, Er, and a reduced stress, σσσσr (Pa), in terms of the known quantities Et, L, r and the Young's modulus of elasticity, E (Pa).
2
Et I
⋅⋅
π
Pt
=
2
π
σ
Er
t
=
L
r
4
=
..
EEt
27
2
L
Et
2
EEt
⋅⋅
55
+
2
4
), and tan gen t modulus Et (Pa). The
Eq. 1
Eq. 2
Eq. 3
2
π
Er
σ
r
The variable names, description an d applicable default units used in these equations are listed below.
Variable Description Units
σr σt
E Young's modulus Pa Er Reduced modulus of elasticity Pa Et Tangent modulus of elasticity Pa I Area moment of inertia
L Length m Pt Tangential load N r Radius of gyration m
Example 11.3.5:
A 25 kip tangential load is applied to the column having the computed properties in example
11.3.1. Find the tangential modulus and stress, in addition to the reduced modulus and stress.
=
2
L
r
Reduced stress Pa Tangential stress Pa
Eq. 4
4
m
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Chapter 11 - Equa tions
- Beams and Columns
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Upper display Lower display
Solution – Select all of the equations to solve this problem. Press to display the variables. Enter the values for the known parameters and press to solve for the unknown variables. The entries and res u l t s are sh own in the scr e en di s p l a ys above .
Given Solution
E = 190 GPa I = 600 in
4
σr = 1.67126 ksi σt = .5 ksi
L = 20 ft Er = 812. 805 ksi Pt = 25 kip Et = 243.171 ksi r = 3.4641 in
References:
1. Mechanics of Materials, 3rd Edition, (1990) James M Gere and Stephen P. Timoshenko, PWS Kent Publishing Company, Boston, MA Specific sections from Chapter 9 and Appendix G.
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Chapter 11 - Equa tions
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ρ

Chapter 12: EE for MEs

This portion of the software deals common electrical engineering problems encountered by mechanical engineers. Four secti ons form the core of the to pic.

Basic Elect rici ty DC Generators

12.1 Basic Electricity
12.1.1 Resistance Formulas
Four equations in this section show the basic relationsh ips between resist ance and cond uctance. The first equation links the resistance, R (), of a bar with a length, len (m), with a uniform cross-sectional area, A (m ρρρρ (Ω⋅m). The second equ a t ion defines the conductance, G (siemen), of the same bar in terms of conductivity, len (m), and A. The 3 recip rocity of condu ctan ce G resistance R as well as the resistivity ρρρρ and conductivity
len
R
=
rd
and 4
th
equations show the
σσσσ
.
A
A
σ
G
=
len
2
), with a resistiv ity,
σσσσ
(Sm/m),
DC Motor AC Motors
Eq. 1
Eq. 2
1
GR=
1
σ
=
ρ
The variable names, d es cription and a pplicable default units used in the equa tions above are listed below.
Variable Description Unit
A Area m2 G Conductance Siemens Len Length m
ρ R Resistance σ
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Equations
Resistivity
Conductivity Siemens/m
83
- EE for ME’s
Eq. 3
Eq. 4
Ω⋅m
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Example 12.1.1:
A copper w ire 1500 m long has a resistivity of 6.5 Ω⋅cm and a cross sectional area of 0.45 cm2. Compute the resistan ce and con d uctance.
Entered Values Computed results
Solution - Exam ining the probl em, two clea r ch oi ces are evident. Either equations 1, 2 and 4 or 1 and 3 can be used to find the solution required . The second choi ce of e quations will be us e d in this example. All the equations can be viewed at the equation screen where in the two equations 1 and 3 c an be selected by using the key to highlight the desired equation and pressing
. On ce bot h equations have been selected , pres s to disp lay all the va riables in the selected eq uation se t. T he sof tware i s now rea dy for receiving the input va ria bl es . Use the ke y to mov e the highl i ght bar to the vari able that needs input. Type the value for the varia ble a nd press
. Repeat to enter all the known variables and press to
solve th e s elected equation set. The comp u ted resu lts are shown in the screen d ispla y sh own here.
Given
A=.45 cm
2
len=1500 m
Solution
G=4.61538E-7 Siemens R=2.16667E6
ρ=6.5 Ω⋅cm
12.1.2 Ohm’s Law and Power
The fu ndamental r elationsh ips between voltage, current and powe r a re presented in t his secti on. Th e first equation is the cla ss ic Ohm 's Law, compu tes the voltage, V (V), in terms of the current, I (A), and the resistan ce, R (). Th e next four equations describe the relationshi p between p ower dissipation, P (W), voltage, V, current, I, resistance, R, and conductance, G (siemens). The final equation represents the reciprocity between resistan ce, R, and conductance, G.
VIR=⋅
PVI=⋅
PIR=⋅
P
=
2
2
V
R
2
PVG=⋅
RG=
MEPro for TI-89, TI-92 Plus Chapter 12 -
1
Equations
- EE for ME’s
Eq. 1
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
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The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
G Conductance Siemens I Current A P Power W R Resistance
V Voltage V
Example 12.1.2:
A 4.7 k load carr i es a current of 275 ma. Calculate th e voltage across the load, power di ssipated an d load conductance.
Entered Values Computed results
Solution - Upon exa min ing t he pr obl em, sever al choices are noted . Ei ther Equations 1, 2 and 6; or 1, 2, 3 and 5; or 2, 3 and 6; or 1, 2 and 5; or, all the equations. C hoose the last option, p ress to open the input screen , en t er al l the known vari ab les a nd press to solve.
Given Solution I = 275 mA G = .000213 siemens
R = 4.7 k
P = 355.438 W V = 1292.5_V
12.1.3 Temperature Effect
This equatio n models the effect o f temp erature on resistance. Electrical r esistance changes from RR1 () to RR2 () when the temperat u re change from T1 (K) to T2 (K) is modified by the temperature
αααα
coefficient of resistan ce
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
(1/K).
()()
α
12112 TTRRRR +=
Eq. 1
α RR1 Resistance, T1 RR2 Resistance, T2
Temperat u re coefficient 1/K
Ω Ω
T1 Temperature 1 K T2 Temperature 2 K
Example 12.1.3:
A 145 resistor at 75 °F reads 152.4 Ω at 125 ºC. F ind t he temp erat u re coefficient of resistance.
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Equations
- EE for ME’s
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φ
φ
φ
Solution
α
= .000505 1/K
-
Solution
unknown variable.
Press to display the i nput screen. En t er th e variable values an d press to solve for the
Given
RR1=145 RR2 = 152.4 T1=75 ºF T2=125_ºC
Entered Values Computed results
12.2 DC Motors
12.2.1 DC Series Motor
These eight equations describe the performance char acteristics of a series DC motor. The first equation links the terminal voltage , Vt (V), to the back emf, Ea (V), defined by the third equation and the IR drop due to armature resistance, Ra (), adjustable resistance, Rd (), and series resistance Rs, (). The
φφφφ
second equation calculates the load torque, TL (Nm), with the machine constant Ke, flux, current, IL (A), and the torque loss, Tloss (Nm). The third equation defines the back emf in the
φφφφ
armature, Ea (V), in terms of Ke, torque generated at the rotor due the magnetic flux, torque generated T as the sum of load torque TL and lost torque Tloss. The last two equations show the
φφφφ
conn ect i on bet ween Ke,
, a field const ant Kef (Wb/A), load current IL, and torque T.
, and mechanical frequency
φφφφ
and current IL. The sixth equation computes the
ωωωω
m (rad/s). The fourth equation shows
(Wb), load
()
ωφ
TlossILKeTL =
ω
= mKeEa
ILKeT =
The fifth equation shows a reciprocal quadratic link between (Nm).
m
Vt
=
ω
Ke
TTlossTL=+
MEPro for TI-89, TI-92 Plus Chapter 12 -
Equations
()
φ
- EE for ME’s
()
Ke
2
φ
ILRdRsRamKeVt +++=
Eq. 1
++
TRdRsRa
86
ωωωω
m, Vt, Ke,
Eq. 2
Eq. 3
Eq. 4
φφφφ
, Ra, Rs, Rd, and torque T
Eq. 5
Eq. 6
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φ
ILKefKe =
Eq. 7
TKfIL=⋅
2
Eq. 8
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
Ea Avera g e emf ind uced in the a rma ture V IL Load current A Ke Machine constant Unitless Kef Field coefficient Wb/A
φ Ra Armature resistance Rd Adjustable resistance Rs Series field resistance T Internal torque TL Load torque Tloss Torque loss
Flux Wb
Ω Ω Ω
Nm Nm Nm
Vt Terminal voltage V ωm
Mechanical radian frequency rad/s
Example 12.2.1:
A series mo to r, wi th a machine constant of 2.4 and rotating at 62 rad/s, is suppl i ed with a terminal voltage of 110 V and produces a torque of 3 Nm. The armature resistance is 10 , the ser ies resistan ce is 5 Ω, and the adjustable resistance is 0.001 . Find the average voltage induc e d in the armature, the flux , and the load current.
Solution 1: Upper Display Solution 2: Upper Display
Solution 1: Lower Display Solution 2: Lower Display
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Equations
- EE for ME’s
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φ
φ
φ
Solution - The first, third and fifth equations are needed to comp ute a solution. Sel ect these by highlighting and pressing press to solve the selected equation set. There are two possible solutions for this example. Type the number of the solution set to be viewed an d press and sel ec t another number. The compute d res ults are s hown in the s c reen displ ays above.
Given Solution Ke = 2.4 Ea = 39.6764 V (70.3236 V)
Ra = 10 Rd = .001 ,
Rs = 5. T = 3. N⋅m Vt = 110. V
ω
m = 62. rad/ s
. Press to display the input scree n, enter all the kno wn variables and
twice. To view another solution set, press to
IL = 4.68793 A (2.64491 A)
φ
= .266642 Wb (.472605 Wb)
12.2 .2 DC Shunt Motor
These seven equations describe the principa l characte ris tics of a DC shun t motor. The first equation expresses the terminal voltage, Vt (V), in terms of the field current, IIf (A), and field resistance, Rf (), along with the external field resistance, Re (). Th e second equation defines the termin al voltage, Vt
φφφφ
(V), in terms of the back emf (expressed in terms of the machine constant, Ke, flux swept,
ωωωω
angular velocity,
m (rad/s), and the IR drop in the armature circuit.
(Wb),
()
The third equation refers to the torque available at the load, TL (Nm), due to the current, Ia (A), in the armature minus the loss of torque, Tloss (Nm), due to friction and other reasons.
The fourth equation gives the definitive relationship between the back emf Ea (V), Ke, (rad/s).
ω
ω
IIfRfReVt +=
Eq. 1
IaRamKeVt +=
TlossIaKeTL =
= mKeEa
Eq. 2
Eq. 3
φφφφ
(Wb), and
Eq. 4
ωωωω
m
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Equations
- EE for ME’s
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φ
ωωωω
The fifth equation displays the reciprocal quadratic relationship between resistance, Ra (), ad just a ble resistance, Rd (), and T (Nm).
m
Vt
=
ω
Ke
The last two equations compute torque T in terms of Tloss, load torque TL, flux
TTlossTL=+
The va ria bl e nam es, descr iption and applicabl e d efa ult units used in the equatio n s above are listed b el ow.
Variable Description Unit
Ea Avera g e emf ind uced in arm ature V Ia Armature current A Iif Field current A Ke Machine constant Unitless
φ Ra Armature resistance Rd Adjustable resistance Re Ext. shunt resistance Rf Field coil resistance T Internal torque TL Load torque Tloss Torque loss Vt Terminal voltage V
ωm
()
φ
()
Ke
IaKeT =
+
TRdRa
2
φ
Flux Wb
Mechanical radian frequency rad/s
m, Vt, K,
φφφφ
, armature
Eq. 5
φφφφ
, Ia (A), and Ke.
Eq. 6
Eq. 7
Ω Ω Ω Ω
Nm Nm Nm
Example 12.2.2:
Find the back emf for a motor with a machine constant of 2.1, rotating at 62 rad/s in a flux of 2.4 Wb.
Entered Values Calculated Results
Solution - Use the fourth equation to solve this problem. Select the equation with the cursor bar and press
. Press
selected equation. The computed result is shown in the screen display above.
Given Solution Ke=2.1 Ea=312.48 V
φ
=2.4 Wb
ω
m=62. rad/ s
MEPro for TI-89, TI-92 Plus Chapter 12 -
Equations
to display the input screen, enter all the known variables and press
- EE for ME’s
89
to solve the
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12.3 DC Generators
12.3.1 DC Series Generator
The two equations in this section describe the properties of a series DC generator. The first equation specifies the field current, IIf (A), and the armature current, Ia (A), to be the same. The seco nd eq u at ion compu tes the terminal voltage, Vt (V), in terms of the induced emf, Ea (V), load current, IL (A), armature resistance, Ra (), and series field windings, Rs ().
Ia IIf=
Vt Ea Ra Rs IL=− +⋅()
Eq. 1
Eq. 2
The varia ble n ames, des c ription a nd a pplicable default units used i n the equa tions above are listed below.
Variable Description Unit
Ea Averag e emf induced in ar matu re V Ia Armature current A IIf Field current A IL Load current A Ra Armature resistance Rs Series field resistance
Ω Ω
Vt Terminal voltage V
Example 12.3.1:
Find the terminal voltage of a series generator with a n arma ture resistance of 0.068 and a series resistance of 0.40 . The generator delivers a 15 A load current from a generated voltage of 17 V.
Entered Values Calculated Results
Solution - Use the second equation to solve this problem. Select this with the highlight bar and press
. Press
to display the input screen, enter all the known variables and press
to solve the
selected equation. The computed result is shown in the screen display above.
Given Solution Ea=17. V Vt = 9.98 V IL=15. A
Ra=.068 Rs=.4_
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φ
12.3.2 DC Shunt Generator
The first equation in this section expresses the induced armature voltage, Ea (V), in terms of the
ωωωω
machin e constant, Ke, th e mechanical angular frequency,
ω
= mKeEa
The second equation defines termi nal voltage, Vt (V), in terms of the field curr ent, IIf (A), external resistance, Re (), and field coil resistance, Rf (). T he third equatio n computes Vt in terms of load current, IL (A), and load resistance, Rl (). The fourth equation expresses Vt as the induced emf, Ea (V), minus armature IR drop, Ra⋅⋅⋅Ia.
m (rad/s), and flux,
φφφφ
(Wb).
Eq. 1
Vt Rf IIf=+⋅()Re Vt IL Rl=⋅
Vt Ea Ra Ia=−⋅
The armature current, Ia (A), is the sum of the load current IL and field current IIf in the fifth equation.
Ia IL IIf=+
The final equation is an alternate form of expression for Ea.
Ea Ra Ia Rf IIf=⋅++ ⋅(Re )
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
Ea Averag e emf induced in ar matu re V Ia Armature current A IIf Field current A IL Load current A Ke Machine constant unitless
φ Ra Armature resistance Re Ext. shunt resistance Rf Field coil resistance Rl Load resistance Vt Terminal voltage V ωm
Flux Wb
Mechanical radian frequency rad/s
Eq. 2
Eq. 3
Eq. 4
Eq. 5
Eq. 6
Ω Ω Ω Ω
Example 12.3.2:
Find the machine constant of a shunt generator running at 31 rad/s and producing 125 V with a 1.8 Wb flux.
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Entered Values Calculated Results
Solution - Use the first equation to solve this problem. Select this by pressing display the input screen, enter all the known variables and press to solve the selected equat i on. T he computed result is shown in the screen display above.
Given Solution Ea= 125. V Ke=2.24014
φ
=1.8 Wb
ω
m=31. rad/ s
. Press to
12.4 AC Motors
12.4.1 Three φφφφ Induction Motor I
These eleven equations define the re lationships amongst key variables used in evaluating the pe rformance of an induction motor. The first equation expresses the relationship between the radian frequency
ωωωω
induced in the rotor, number of poles, p, and the mechanical angular speed,
ωω ω
rspm=−⋅
r (rad/s), the angular speed of the rotating magnetic field, of the stator
2
ωωωω
m (ra d/s).
ωωωω
Eq. 1
s (rad/s),
ωωωω
ωωωω
ωωωω
r,
s,
The second, thir d and fourth equations describe the slip, s, using power per ph ase, Pr (W), and th e power tran sferred to the rotor per phase, Pma (W).
pm
s
=− ⋅1
Pr
ω
s
2
ω
s=
m, p, the induced rotor
Eq. 2
Eq. 3
Pma
rss
ωω
=⋅
Pma is defined in the fifth equation in terms of the rotor current, Ir (A), and the rotor phase voltage, Ema (V).
Pma Ir Ema=⋅ ⋅3
The sixth and seventh equations accoun t for t he mechan ical power, Pme (W),in terms of p, Pma, and torque, T (Nm).
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Eq. 4
92
Eq. 5
ωωωω
m,
ωωωω
s,
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Pme
pm
ω
s
2
ω
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Pma=⋅ ⋅ 3
Eq. 6
Pme T m
=⋅
The eighth eq uation expresse s torque in terms of p, Pma, and
=⋅ ⋅3
pPma
T
2
The last three equations show an equivalent circuit representation of induction motor action and links the power, Pa with rotor resistance, Rr (), rotor current, Ir, slip s, roto r resistance p er phas e, RR1 (),
and the m achi ne const ant , KeM.
Pma Rr Ir
Eq. 7
ω
ωωωω
s.
s
ω
s
1
22
Rr Ir=⋅+−⋅⋅
Eq. 8
Eq. 9
s
Pa
1
⋅⋅
2
Rr Ir=
Eq. 10
s
s
1
Rr =
RR
KeM
2
Eq. 11
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
Ema Phase voltage V Ir Rotor current per phase A KeM Induction motor constant unitless p # poles unitless Pa Mechanical power available W Pma Power in rotor per phase W Pme Mechanical power W Pr Rotor power per phase W RR1 Rotor resistance per phase Rr Equivalent rotor resistance s Slip unitless T Internal torque
ωm ωr ωs
Example 12.4.1:
Find the mechanical power for an induction motor with a slip of 0.95, a rotor current of 75 A, and a resista nce of 1.8 Ω.
Mechanical radian frequency ra d/ s Elect rica l rotor speed rad/s Elect rica l stator sp eed rad/s
Ω Ω
Nm
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Input Values Calculated Results
Solution - Ch oos e equatio n ten to compute the solution. Select by highlighting and pressing Press
to display the input screen, enter all the known variables and press
Given Solution Ir = 75. A Pa = 532.895 W
Rr = 1.8 s = .95
to solve the equation
.
12.4.2 Three φφφφ Induction Motor II
These equations are used to perform equivalent circuit analysis for an induction motor. The first equation shows the power in the rotor per phase, Pma (W), defined in terms of the rotor current, Ir (A), rotor resistance, Rr (), and slip s.
Pma
Rr
Ir=⋅
2
Eq. 1
s
The secon d equation shows the expression for torque, T (N⋅m), in terms of poles p, Pma and radian frequency of the induced voltage in the stator,
representation of torque in terms of the applied voltage, Va (V), stator resistance, Rst (), Rr (), inductive reactance XL (), and
ωωωω
s (rad/s).
ωωωω
s (rad/s). The third equation is an alternate
Pma
3
Tp
=⋅⋅
2
psRr
3
T
=⋅ ⋅ ⋅
2
ω
The fourth equation computes Tmmax (Nm) represents the maximum positive torque available at the rotor, given the parameters of the induction motor stator resistance, Rst, XL, Va, p, and
Tm
max
=⋅ ⋅
The maximum slip, sm, in the fifth equation represents the condition when dT/ds=0.
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s
ω
s
p
3
s
4
ω
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2
Va
2
Rr
Rst
Rst XL Rst
+
+
s
2
Va
22
++
XL
2
94
Eq. 2
Eq. 3
ωωωω
s.
Eq. 4
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sm
=
The sixth equation defines the so-called breakdown torque, Tgmax (Nm), of the motor. The final equation relates, Rr (), with machine constant, KeM, and the rotor resistance per phase, RR1 ().
Tg
max
Rr
=
The variable names, d es cription and a pplicable default units used in the equa tions above are listed below.
Variable Description Unit
Ir Rotor current per phase A KeM Induction motor constant unitless p # poles unitless Pma Power in rotor per phase W RR1 Rotor resistance per phase Rr Equivalent rotor resistance Rs Series field resistance Rst Stator resistance S Slip unitless Sm Maximum slip unitless T Internal torque Tgmax Breakdown torque Tmmax Maximum positive torque Va Applied voltage V ωs XL Inductive reactance
Rr
22
Rs XL
+
3
=− ⋅
4
RR
1
2
KeM
p
s
ω
Rs XL Rst
2
Va
22
+−
Electrical stator speed rad/s
Eq. 5
Eq. 6
Eq. 7
Ω Ω Ω Ω
Nm Nm Nm
Example 12.4.2:
An applied voltage of 125 V is applied to an eight-pole motor rotati ng at 245 rad/s. The stator resistance and reactance is 8 and 12 respectively. Find the maximum torque.
Input Values Calculated Results
Solution - Use the fourth equation to compute the solution. Select by moving the cursor bar, highlighting, and pressing press
to solve the equation.
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Equations
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Given Solution p = 8
Rst = 8. Va = 125. V
ω
s = 245. rad/s
XL = 12._
Tmmax = 17.0658 Nm
12.4.3 1 Induction Motor
These three equations describe the properties of a single-phase induction motor. The first equation
φφφφ
defines the slip for forward flux sf with respect to the forward rotating flux,
ωωωω
frequency of induced is current in the stator, number of poles, p, and the angular mechanical speed of the rotor represent the forward and backward torques, Tf (Nm) and Tb (Nm), for the system with r espect to sf, the number of poles p, t he elect rical stator sp eed, the cu rrents, Isf (A) and Isb (A). The forward torque, Tf, is given by the power dissipated in t he fictitious rotor resistor.
pm
sf
=− ⋅1
ω
s
2
ω
s (rad /s). Other var iables of conseq u ence include the
ωωωω
m (rad/s). The final two equations
ωωωω
s (rad/s), the equivalent rotor resistance, Rr (), and
(Wb). The radian
Eq. 1
p
Tf
1
=⋅ ⋅
ω
Tb
p
2
The variable names, d es cription and a pplicable default units used in the equa tions above are listed b el ow.
Variable Description Unit
Isb Backward stator current A Isf Forward stator current A p # poles unitless Rr Equivalent rotor resistance sf Slip for forward flux unitless Tb Backward torque Tf Forward torque
ωm ωs
Example 12.4.3:
Find the forward slip for an eight-pole induction motor with a stator frequency of 245 rad/s, and a mechanical ra di a n frequency of 62.5 rad/s.
Isf Rr
s
1
=
s
ω
2
Eq. 2
sf
2
2
2
RrIsb
()
22
Eq. 3
sf
Nm Nm
Mechanical radian frequency rad/s Elect rica l stator sp eed rad/s
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Entered Values Calculated Results
Solution - The first equation is needed to compute the solution. Select by highlighting and pressing
. Press to display the input screen, enter all the known variables and press to solve the
equation.
Given Solution p = 8 sf = -.020408
ω
m = 62.5_rad/ s
ω
s = 245._rad/s
References
:
1. Slemon G. R., and Straughen, A., Electric Machines, Addison-Wesley, Reading, MA 1980
2. Stevenson Jr., William D., Elements of Power Systems An a lysis, McGraw-Hill International, New
York, 1982
3. Wildi, Theodore, Ele c t ric al Power Technology, John Wiley and Son, New Jersey, 1981
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Chapter 13: Gas Laws

This section computes pr o perties and state changes for ideal and real gases with an emphasis on ideal gases. Selected topics for real gases include van der Waals, an d Redlich - Kwong models. The id eal gas constant, Rm, 8.31451 J/(mol K) is auto matically inserted into all calculations and doe s not appear in the list of variables in th e calculator.

Ideal Gas Laws Real Gas Laws Polytropic Process

Kinetic Gas Theory Reverse Adiabatic
13.1 Ideal Gas Laws
13.1.1 Ideal Gas Law
The ideal gas law approximate s, to a high degree, the actual properties of a gas at high tempe rature or low pressur e. Equations 1 and 2 define the molar volume, vm (m the gas in terms of the number of moles, N (mol), of gas or total mass, m (kg), occupying a volume, V
3
). Equations 3, 4 and 5 are three alternate forms of the ideal gas relationship between pressure, p
(m (Pa), volume, V, temperature, T (K), molecular weight, MWT (kg/mol), and specifi c vol u me, vs. Th e last equation computes mo lar mass, MWT, in terms of m and N.
V
vm
=
N
V
vs
=
m
pvs
Rm
T⋅=
MWT
3
/mol), and specific volume, vs (m3/kg), of
Eq. 1
Eq. 2
Eq. 3
pV N RmT⋅=⋅ ⋅ pvm RmT⋅= ⋅
MWT
=
m
Eq. 4
Eq. 5
Eq. 6
N
Variable Description Units m MWT N
p Pressure Pa Rm Molar gas constant T Temperature K
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Mass kg Molar mass kg/mol No. moles mol
8.3145 J/(mol⋅K)
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Variable Description Units
V Volume vm Molar volum e vs Specific volume
Example 13.1.1:
A 2-liter container is filled with methane (molecular mass = 16.042 g/mol) to a pressure of 3040 torr at room temperature (25oC). Calculate the number of moles and the total mass of methane.
Entered Values Computed results
Solution - Select the fourth and sixth equations to solve this pr obl em. All the equa tions can be viewed at the equ a t i on s cr een wher ei n th e t wo equat i ons 4 and 6 can be selected by us ing t he key to highlight the desired equation and pressing variables in the selected equation set. Use the ke y to mov e the highl i ght bar to the vari able n e e d in g data. Enter the value for the variable then press have been entered. Press to solve the selected equation set. The computed results are shown in th e screen disp la y shown here.
. Once both equ a tions have been selected, press to display all the
. Rep e at t his until va lues for all known var iable s
m m m
3 3
/mol
3
/mol
Given Solution
MWT = 16.042 g/mol m = 5. 24558 g p=3040 torr N = .32699 mol
o
T=25. V=2. l
C
13.1.2 Constant Pressure
The following e quations describe the changes of the state for a fixed q uantity gas at constant pressure, p. Equation 1 describes Charles’s Law- volume, V (m
the absolute temperature, T (K). Equations 2 and 3 express relationship between W12 (J) in terms of press u re and volume or moles and tem peratures ch ange respect ively. Equations 4 and 5 calculate the change in total entropy, S21 (J/K), and mass-specific entropy (entropy per unit mass), ss21 (J/(kg⋅K)), due to change in temperatur e, T2-T1 (K). Equation 6 calculates the change in entropy per mole of gas, sm21 (J/(mol⋅K)), from ss21, and molecular mass, MWT (kg/mol). Equatio n s 7 and 8 compute the tra nsfer of h ea t to t he system, Q12 (J), due to expansion under constan t pressure. Equ ations 9 and 10 describe the relationships between the specific heat ratio k, to th e s p ecific heats const ant vol u me, cv (J/(kg⋅K)), and constant pressure, cp (J/(kgK)). Equation 11 relates molecular weight, MWT (kg/mol), to the number of moles, N (mol) and total mass of the gas, m (kg).
V
T
2
2
=
V
1
WpVV12 2 1=− ⋅
T
1
16
3
), of a fixed amount of gas is directly proportional to
Eq. 1
Eq. 2
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WNRmTT12 2 1=− ⋅
Smcp
21
=⋅⋅
ss cp
21
=⋅
sm MWT ss21 21=⋅ QmcpTT12 2 1=⋅⋅ −
kW
Q
12
cp cv
k
=
=
k
=+
cp cv
MWT
16
T
2
T
1
T
2
 
T
1
ln
ln
 
16
12 1
Rm
Eq. 3
Eq. 4
Eq. 5
Eq. 6
Eq. 7
Eq. 8
Eq. 9
Eq. 10
MWT
Variable Description Units
cp Specific Heat-constant pressure cv Specific Heat-constant volume k Specific Heat Ratio unitless m Mass kg MWT Molar Mass. kg/mol N No. moles mol p Pressure Pa Q12 Rm Molar Gas constant sm21 ss21 S21 T1 Initial Temperature: 1 K T2 Final Temperature: 2 K V1 Initial Volume V2 Final Volume W12
Example 13.1.2:
Dry air has a molecular mass of 0.0289 kg/mol; see mwa in Reference/Engineering Constants, and a specific heat of 1.0 J/(g¹K) at constant pressure of 1 bar in the temperature range of 200-500K. Air in a 3­m3 cylinder performs work on a frictionless piston, exerting a constant pressure of 1 bar. A heating
m
=
N
J/(kgK) J/(kgK)
Heat Transfer: 1→2 Entropy Ch ange-m ole: 12 J/(mol⋅K)
Entropy Chan ge-mass: 12 J/(kg⋅K) Entropy Chan ge: 1→2
Work Performed: 1→2
J
8.3145 J/(mol⋅K)
J/K
3
m
3
m J
Eq. 11
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