Mechanic imp

Copyright © 1993-2001, Hugh Jack

Engineer On a Disk

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This section last updated: April 28, 2002

Copyright © 1993-2002, Hugh Jack

email: [email protected]: (616) 771-6755fax: (616) 336-7215

1. TABLE OF CONTENTSTABLE OF CONTENTS.......................................................................................................... 2MECHANICAL DESIGN ........................................................................................................ 9

BASIC PHILOSOPHY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 9TYPICAL MECHANICAL FUNCTIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - 9REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 10

FORCES ................................................................................................................................. 11SOME BASIC CONCEPTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12VECTOR AND SCALAR FORCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 13

EQUILLIBRIUM.................................................................................................................... 38THE BASIC EQUATIONS OF STATICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - 39FREE BODY DIAGRAMS (FBD) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 44PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 51

MOMENTS ............................................................................................................................ 53CALCULATING SCALAR AND VECTOR MOMENTS - - - - - - - - - - - - - - - - - 56FORCE COUPLES TO MAKE CENTERLESS MOMENTS - - - - - - - - - - - - - - - 64

MECHANISMS...................................................................................................................... 79REACTIONS AND SUPPORTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 79EQUILLIBRIUM OF FORCES AND MOMENTS - - - - - - - - - - - - - - - - - - - - - - 82SPECIAL CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 90STATICALLY INDETERMINATE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 91

TRUSSES AND FRAMES..................................................................................................... 93WHAT ARE TRUSSES? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 93STABILITY OF TRUSSES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 99THE METHOD OF JOINTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 102THE METHOD OF SECTIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 114METHOD OF MEMBERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 122SUMMARY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 132

DRY STATIC FRICTION.................................................................................................... 132THE BASIC PHYSICS OF FRICTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - 132APPLICATIONS OF FRICTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 137

FORCES ............................................................................................................................... 150INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 150SOME BASIC CONCEPTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 151VECTOR VS. SCALAR QUANTITIES - - - - - - - - - - - - - - - - - - - - - - - - - - - - 152MATH REVIEW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 153RECTANGULAR FORM OF VECTORS - - - - - - - - - - - - - - - - - - - - - - - - - - - 158POLAR FORM OF VECTORS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 159

EQUILLIBRIUM.................................................................................................................. 187INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 187THE BASIC EQUATIONS OF STATICS - - - - - - - - - - - - - - - - - - - - - - - - - - - 187FREE BODY DIAGRAMS (FBD) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 196PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 205

STRESS ................................................................................................................................ 208INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 208

TYPES OF STRESS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 210STRESS ANALYSIS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 214STRAIN CAUSED BY AXIAL LOADS - - - - - - - - - - - - - - - - - - - - - - - - - - - 219STRESS STRAIN CURVES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 220ANALYSIS OF MEMBERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 222GENERALIZED STRESS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 225REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 229STRESS ON OBLIQUE PLANES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 229SHEAR STRAIN - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 232POISSON’S RATIO - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 236GENERALIZED HOOKES LAW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 238REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 239

MOMENTS .......................................................................................................................... 240INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 240CALCULATING SCALAR MOMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - 244CALCULATING VECTOR MOMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - 247MOMENTS ABOUT AN AXIS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 251EQUILLIBRIUM OF MOMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 256FORCE COUPLES TO MAKE CENTERLESS MOMENTS - - - - - - - - - - - - - - 260

TORSION ............................................................................................................................. 276INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 276THE RELATIONSHIP BETWEEN STRESS AND STRAIN IN TORSION - - - - 280PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 285REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 286

MASS PROPERTIES ........................................................................................................... 287INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 287CENTRE OF MASS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 290CENTROIDS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 293

FORCES AND MOMENTS ON RIGID BODIES .............................................................. 311INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 311REACTIONS AND SUPPORTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 311EQUILLIBRIUM OF FORCES AND MOMENTS - - - - - - - - - - - - - - - - - - - - - 314SPECIAL CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 323STATICALLY INDETERMINATE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 324

TRUSSES AND FRAMES................................................................................................... 331INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 331WHAT ARE TRUSSES? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 331STABILITY OF TRUSSES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 336THE METHOD OF JOINTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 339THE METHOD OF SECTIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 352ADDITIONAL TOPICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 361

STRESS FAILURE .............................................................................................................. 363FACTOR OF SAFETY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 363REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 367

STRAIN FAILURE .............................................................................................................. 368POISSON’S RATIO - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 369

GENERALIZED HOOKES LAW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 372STRESS CONCENTRATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 373TORSION STRESS CONCENTRATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - 379REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 380

METHOD OF MEMBERS................................................................................................... 381INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 381REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 391SUMMARY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 391REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 391

INTERNAL FORCES IN MEMBERS................................................................................. 392INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 392PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 405REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 409

MOMENTS OF INERTIA ................................................................................................... 410INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 410STRESSES IN BEAMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 410MOMENT CURVATURE IN BEAMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 412EVALUATING THE SECOND MOMENT OF INERTIA - - - - - - - - - - - - - - - 414MOMENTS OF INERTIAS B COMPOSITE AREAS - - - - - - - - - - - - - - - - - - 415POLAR MOMENT OF INERTIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 419REVIEW OF BASIC CALCUATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - 421PRODUCT OF INERTIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 433REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 435

PURE BENDING ................................................................................................................. 436INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 436TRANSVERSE SHEAR - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 440REVIEW OF TRANSVERSE LOADING - - - - - - - - - - - - - - - - - - - - - - - - - - - 445REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 447

DRY STATIC FRICTION.................................................................................................... 448INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 448THE BASIC PHYSICS OF FRICTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - 448APPLICATIONS OF FRICTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 454

MASS PROPERTIES ........................................................................................................... 468CENTRE OF MASS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 471CENTROIDS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 474MOMENTS OF INERTIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 486PRODUCT OF INERTIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 496

INTERNAL FORCES IN MEMBERS................................................................................. 498STRESS ................................................................................................................................ 507

TYPES OF STRESS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 508STRESS ANALYSIS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 512STRESS ON OBLIQUE PLANES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 517GENERALIZED STRESS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 519FACTOR OF SAFETY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 522

STRAIN ................................................................................................................................ 526STRAIN CAUSED BY AXIAL LOADS - - - - - - - - - - - - - - - - - - - - - - - - - - - 526

STRESS STRAIN CURVES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 527ANALYSIS OF MEMBERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 529POISSON’S RATIO - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 533GENERALIZED HOOKES LAW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 535SHEAR STRAIN - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 535STRESS CONCENTRATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 540TORSION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 543TORSION STRESS CONCENTRATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - 546PURE BENDING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 548TRANSVERSE LOADING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 551

MECHANISM DYNAMICS................................................................................................ 555INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 555PLANAR - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 556PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 563REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 564

VIBRATION ........................................................................................................................ 565VIBRATION MODELLING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 565CONTROL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 572VIBRATION CONTROL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 577VIBRATION MEASUREMENT - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 584VIBRATION SIGNALS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 586VIBRATION TRANSDUCERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 587DEALING WITH VIBRATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 591RESOURCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 592PRACTICE QUESTIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 592SOUND/VIBRATIONS TERMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 598REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 599

INTERNAL COMBUSTION ENGINES............................................................................. 600POWER - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 600KINEMATICS AND DYNAMICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 603REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 605

SOUND CONTROL............................................................................................................. 606BASIC PROPERTIES OF SOUND - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 606SOUND MEASUREMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 612THE HUMAN EFFECTS OF SOUND - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 614NOISE CONTROL REGULATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 626SOUND ANALYSIS INSTRUMENTS/TECHNIQUES - - - - - - - - - - - - - - - - - 632EQUIPMENT GENERATED NOISE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 642ROOM ACCOUSTICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 646ENCLOSURES, BARRIERS AND WALLS - - - - - - - - - - - - - - - - - - - - - - - - - 654MATERIALS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 663MUFFLERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 664SOUND AND VIBRATION CONTROL STUDIES - - - - - - - - - - - - - - - - - - - - 671AKNOWLEDGEMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 671PRACTICE QUESTIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 671

INTRODUCTION TO KINEMATICS OF MECHANISMS .............................................. 685

SOME POPULAR MECHANISMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 687SKELETON DIAGRAMS FOR MECHANISMS - - - - - - - - - - - - - - - - - - - - - - 694DOF AND THE KUTZBACH/GRUEBLER CRITERION - - - - - - - - - - - - - - - 695KINEMATIC/GEOMETRIC INVERSION - - - - - - - - - - - - - - - - - - - - - - - - - - 698GRASHOF’S LAW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 700MECHANICAL ADVANTAGE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 702PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 703REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 704

POSITIONS/DISPLACEMENTS OF POINTS AND MECHANISMS.............................. 705MATHEMATICAL TOOLS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 705DEFINING POSITIONS AND DISPLACEMENTS - - - - - - - - - - - - - - - - - - - - 706CLOSED LOOP MECHANISMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 710SOLVING FOR POSITIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 710GRAPHING OF POSITIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 716DISPLACEMENT, TRANSLATION AND ROTATION - - - - - - - - - - - - - - - - 718PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 721REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 721

MECHANISM VELOCITY ................................................................................................. 723THE BASIC RELATIONSHIPS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 723CALCULATION TECHNIQUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 725INSTANTANEOUS CENTERS OF ROTATION - - - - - - - - - - - - - - - - - - - - - 730PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 740REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 742

MECHANISM ACCELERATION ...................................................................................... 743THE BASIC DEFINITION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 743INSTANT CENTERS OF ACCELERATION - - - - - - - - - - - - - - - - - - - - - - - - 751PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 751REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 751

LINKAGE SYNTHESIS ...................................................................................................... 752SELECTION OF MECHANISM TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - 752DESIGN METHODS - SYNTHESIS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 753DESIGN METHODS - DIMENSIONAL - - - - - - - - - - - - - - - - - - - - - - - - - - - 755REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 761

SPATIAL KINEMATICS .................................................................................................... 762BASICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 762HOMOGENEOUS MATRICES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 764SPATIAL DYNAMICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 776DYNAMICS FOR KINEMATICS CHAINS - - - - - - - - - - - - - - - - - - - - - - - - - 781REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 784PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 785

MECHANICAL COMPONENTS........................................................................................ 798CAM DESIGN...................................................................................................................... 799

CAM TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 799CAM MOTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 801USING CAMS AS JOINTS IN MECHANISMS - - - - - - - - - - - - - - - - - - - - - - 812PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 812

REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 814GEARS ................................................................................................................................. 815

SPUR GEARS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 815HELICAL GEARS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 826BEVEL GEARS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 832WORM GEARS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 834REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 838

DESIGN OF MECHANISMS .............................................................................................. 838SIMPLE GEAR TRAINS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 838LINKAGES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 848PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 848REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 848

STATIC ANALYSIS OF GEARS ....................................................................................... 848INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 848ANALYSIS OF GEARS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 849PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 850REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 850

MECHANICAL COOKBOOK ............................................................................................ 851TRANFORMING DEVICES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 851REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 852

A MECHANICAL COOKBOOK ........................................................................................ 853CONNECTORS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 853MOTION/FORCE TRANSMISSIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 866POWER TRANSMISSION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 871

Mechanical Engineering Topics

2. MECHANICAL DESIGN

DI:2.1 BASIC PHILOSOPHY

• Philosophy is an important concept, because design is basically making inexact decisions, in an unpredictable environment.

• Some basic rules to follow are,- use standard materials, processes, components, fasteners etc. where possible- use looser tolerances where possible- minimize parts- only design what the specifications/function requires-

• Typical mechanical design problems can be classified as,- selection design - given certain criteria select a suitable component.- configuration design - how are given components arranged to satisfy given criteria.- parametric design - given a known relationship (equations, graphs, tables, etc.) and con-

strain (limits, objectives, etc) we can calculate an exact dimension (or other param-eter values) for a design.

- original design - new concepts must be generated.- redesign - a design must be changed conceptually to obtain better performance.

2.2 TYPICAL MECHANICAL FUNCTIONS

• This list below can be used to jog thoughts about possible mechanical concepts. [based on Ull-man]

2.3 REFERENCES

Ullman, D.G., The Mechanical Design Process, McGraw-Hill, 1997.

absorbactuateamplifyassembleavoidchangechannelclearcollectconductcontrolconvertcouple

decreasedisassembledirectdissipatedrivefastenguideholdincreaseinterruptjoinliftlimit

locatemoveorientpositionprotectreleaserectifyremoverotatesecureseparateshieldstart

stopsteerstoresupplysupporttransformtranslateverify

3. FORCES

• WHAT? - We look at mechanical structures, and determine the distribution of forces and moments.

• WHY? - a fundamental subject for every form of Mechanical Engineering (and every other branch of engineering that has ever existed)

egr20935.jpg

• The difference between statics and dynamics, in brief,Statics - does nothing, just sits thereDynamics - moving things

EX1.WM

• Consider some of the applications of statics design techniques,egr20921.jpgegr20932.jpgegr20926.jpg

e.g., an electric transmission tower

When the weight of the transmission line (a force) is applied, how much force does each part (beam) of the tower carry. How much support is needed on the ground

If we do the analysis we can then determine how large the beams must be, but this will be taught in Mechanics of Materials.

3.1 SOME BASIC CONCEPTS

• Mass and force - A mass can exert forces through gravity and other effects. Forces can also be exerted by other phenomenon, such as magnetism.

• We can emphasize the relationship between mass as an absolute and gravity as a local. The effects of gravity are dealt with as forces in most statics problems.

• Force has magnitude and direction. Therefore it is well suited to vectors.

• many forces can also operate on the same object, we can replace these with equivalent forces, called resultants.

g 9.81 NKg-------

=

F

MassM = 10kg

(This gravity vector is the averagefor ground level on earth)

F gM=

F 9.81 NKg-------

10 Kg( ) 9.81 10( ) N Kg×Kg

----------------- 98.1N= = =

Consider, the same mass on the moon,

g 1 NKg-------

???=

F 1 NKg-------

10 Kg( ) 1 10( ) N Kg×Kg

----------------- 10N= = =

*Note: the same mass is exerting a different force. Please recall from basic physics that the mass is a grouping of matter. The force is the attraction it feels when it approaches another body. Also beware, when using the Imperial system of units, gravity is often measured in slugs, as a result you must be VERY careful with units.

FGM1M2

r2

-------------------- M1g= = g∴GM2

r2

------------=

g 9.78 NKg-------

= (At the equator)

g 9.83 NKg-------

= (At the north pole)

• We have both action and reaction forces as well. As we apply action forces, there are forces that will resist, these are called reaction forces.

• Some approximations,- we are pretending the forces are applied at points, but in reality a force must be distrib-

uted,

- we generally assume there are no deflections. This is known as the RIGID BODY assumption.

- we often use particle approximations that assume bodies have no size. This simplifies calculations significantly.

- Transmissibility - a force can be moved along a line of action.- Parallelogram law - a method for adding two forces to get a resultant vector.

3.2 VECTOR AND SCALAR FORCES

• definitions,

FP

Fg

each part of the thread on the screw will trans-mit a bit of the force, and each will be a dif-ferent amount (a very hard problem that is dealt with by finite element methods).

F

• Recall that vectors can be added or subtracted using the paralleloram law. This is a variation or the triangle law. In both cases we are putting vectors head to tail. These methods favor drafting solutions to porblems that are not really necessary with calculators, but they are still very use-ful for understanding.

• vectors can be added to get resultant forces in vector (rectangular component) form.

VECTOR - a magnitude and a direction

SCALAR - a simple quantity (no direction)

e.g. gravity is a vector

e.g. mass is 10 kg

9.81 NKg-------

10 kg

A

B

R=A+B A

B

R=A+B

• We can also represent the same forces as scalar magnitudes, and direction,

e.g.

x

y

oil tanker

2 tug boats

F1 = (5000N, 5000N)

FR

F2 = (14000N, 0N)

14000N 5000N

5000NFR

FR F1 F2+ 5000N 5000N,( ) 14000N 0N,( )+ 19000N 5000N,( )= = =

rectangularnotation

• we could have also solved this problem using trigonometry.

e.g.

x

y

oil tanker

FR 19000N( )25000N( )2

+ 386 106N

2× 19646N= = =

θFR

5000N19000N-------------------

atan 14.7°= =

θFR

θF1

F1

F2

FR

F1 5000N( )25000N( )2

+ 50 106N

2× 7071N= = =

θF1

5000N5000N----------------

atan 45°= =

F2 14000N( )20N( )2

+ 14000N= =

θFR

0N14000N-------------------

atan 0°= =

19646N 14.7°∠polar notation

7071N 45°∠

14000N 0°∠

3.2.1 Cartesian Vector Notation

• Remember from before,

FR

F2

F1

180°-45°=135°

F2 14000N=

F1 7071N=

FR2

F12

F22

2 F1 F2 135°cos–+=

FR∴ 70712

140002

2 7071( ) 14000( ) 135°cos–+ 19646N= =

Given,

We can find the magnitude of FR, using the cosine law

θFR

We can also find the angle of FR, using the sine law,

F1

θFRsin----------------

FR

135°sin-------------------=

θFR∴

F1 135°sin

FR-----------------------------

asin 7071 135°sin14000

-------------------------------- asin 14.7°= = =

19646N 14.7°∠

3.2.2 Scalar Notation

• It can also be useful to keep the forces in scalar values, but the direction should still be defined on paper, instead of by convention, as is done with vectors. The most common method is to use x-y-z components, or forces relative to a given direction.

• For example,

x

y

oil tanker

2 tug boats

F1 = (5000N, 5000N)

FR

F2 = (14000N, 0N)

we can also write the vectors in the form,

F1 5000i 5000j+( )N=

F2 14000i( )N=

here the x-component is indicated with ‘i’, and the y-components areindicated with ‘j’. If there was a z-component it would be representedwith a ‘k’. I will underline them to minimize the potential for confusion.

Next, we can add the vectors to find the resultant,

FR F1 F2+ 5000Ni 5000Nj 14000Ni+ += =

Also, notice that I am keeping the units in the calculations, as if theyare variables. I will do this more for illustrative purposes, but this methodwill reduce the chance for unit based calculation mistakes.

FR∴ 19000Ni 5000Nj+ 19000i 5000j+( )N= =cartesian vector notation

• Scalar notation is often made obvious by using ‘x’, and ‘y’, or similar subscripts.

• direction, location, signs, etc. are all defined by convention, and very compact mathematical methods can be used.

• These problems can also be solved using cosine and sine law force additions on force triangles. Considering the last example,

F1

F2

FR

70° 20°

x

y

F1 = 5NF2 = 2N

where,

FRxF1 20°cos F2 180° 70°–( )cos+ 4.0N= =

FRyF1 20°sin F2 180° 70°–( )sin+ 3.6N= =

FRx4.0N=

FRy3.6N=

Scalar notation

• Consider the large pendulum below as an example where a force tringle could be used to find the tensions in the cables.

egr20928.jpgegr20927.jpg

FR

F2=2N

F1=5N

20°

φ

90°

use cosine law (we could also use pythagoras) to find FR,

FR2

F12

F22

2F1F2 90°cos–+=

FR2

52

22

2 5( ) 2( ) 0( )–+=∴ 29=

Next, you can try using the sine law to find the direction angles,

• Consider the example below,

3.2.3 3D Vectors

• We will use right-handed coordinates

α

β

A

B

C

TA

TB

uv

Find an equation that relates all of the tensions and angles.

• Consider the following conditions,

x

y

z

z is upy is to the rightx is towards you

i

j

k

** How to remember: on the right hand the thumb points along the positive z-axis, and when you curl your fingers into a fist it should push the positive x-axis 90° into the positive y-axis. Try this with the coordinate axes above. If you need to push the x-axis 270° you have left handed coordinates.

ASIDE: these axes could also be shown in different orientation, but the right hand rule will still apply.

x

y

z

NOTE: 3D axis are typically drawn in a manner similar to the oblique views used in traditional drafting. The axis drawn horizontal and vertical lie on the plane of the page, while the axis on an angle is

- out of the page if it points downward- into the page if it points upwards

θα

θγ

θβ

x

y

z

F 3i 4j 5k+ +( )N=

F 3N( )24N( )2

5N( )2+ + 7.07N= =

First, find the magnitude of the vector (using the 3D pythagorean theorem),

Next, we can find a unit vector representation,

F 7.07N7.07N-------------- 3Ni 4Nj 5Nk+ +( ) 7.07N 3

7.07----------i 4

7.07----------j 5

7.07----------k+ +

= =

F∴ 7.07N 0.42i 0.57j 0.71k+ +( )=

A unit vector of length 0.422

0.572

0.712

+ + 1.0= =

We can find the angles between the axis, and the force, using direction cosines

θαcosFx

F------ 3

7.07---------- 0.42= = = θα∴ 65.2°=

θβcosFy

F------ 4

7.07---------- 0.57= = = θβ∴ 55.2°=

θγcosFz

F------ 5

7.07---------- 0.71= = = θγ∴ 44.8°=

NOTE: We will sometimes use θx, θy, θz as a replacement for θα, θβ, θγ respectively, these are often referred to as α, β, γ in other places.

θα 65.2°= θβ 55.2°= θγ 44.8°=F 7.07N=

(direction) cosine notation

• To emphasize the main relations,

• An example to illustrate this technique is, ([Hibbeler, 1992], prob. 2-56, pg. 49)

ASIDE:

F F λxi λyj λ zk+ +( )=

1 λx2 λy

2 λz2

+ +=

1 λx2 λy

2 λ z2

+ +=

1Fx

F------

2 Fy

F------

2 Fz

F------

2+ + θαcos( )2 θβcos( )2 θγcos( )2

+ += =

you can use this to checkresults.

1 θαcos( )2 θβcos( )2 θγcos( )2+ +=

F F θαcos i θβcos j θγcos k+ +( )=

F Fx2

Fy2

Fz2

+ +=

θαFx

F------

acos= θβFy

F------

acos= θγFz

F------

acos=

x

y

z

120°

60°

45°

45°

60°

F1 300N=

F2 500N=

Find the resultant force for the two forces shown in vector projection (F1) and cosine notation (F2).

NOTE: this problem has two different, but common ways of representing vec-tors. F2 is represented using direction cosines, but F1 is represented by a technique known as vector projection. The common way to deal with this representation is to find the force component that lies on the projection plane, and then use that for some of the calculations.

NOTE: Forces represented with vector projection are similar to forces mea-sured with azimuth and elevation. Surveyors commonly use these angles to specify direction.

For F2:F2 F2 60°icos 45°jcos 120°kcos+ +( )=

F2∴ 500N 60°icos 45°jcos 120°kcos+ +( )=

F2∴ 250i 354j 250k–+( )N=

For F1:

F1xyF1 60°cos 150N= =

First, find the component of the force projected onto the x-y plane.

Now, find the components of the force in the x-y plane,

F1xF– 1xy

45°sin 106.1– N= =

F1yF1xy

45°cos 106.1N= =

Finally, find the z-component,

F1zF1 90° 60°–( )cos 259.8N= =

And, put the forces together into a vector form,

F1 106i– 106j 260k+ +( )N=

• Consider the case below, where we know positions, and forces, but we want to find the resultant force,

Finally add the two forces,

FR F1 F2+ 250i 354j 250k– 106i– 106j 260k+ + +( )N= =

FR∴ 144i 460j 10k+ +( )N=

FR 1442

4602

102

+ + 482N= =

And find the magnitude, and angles of the resultant,

θαFRx

FR---------

acos 72.6°= =

θβFRy

FR---------

acos 17.5°= =

θγFRz

FR---------

acos 88.8°= =

NOTE: One significant figure has been lost.

Boeing Balloon Co.

xy

z

(-2, 3, 0)

(-15, 8, 0)

(-4, 13, 0)

(-10, 10, 10)

F1

F2

F3

If we have an inflated hot air balloon that has a buoyancy force keeping it aloft, and three people of the ground resisting it’s rise by holding ropes. The position of the ends of the ropes is shown. Find the resultant force on the balloon. What will happen?

FB

F1 200N=

F2 300N=

F3 500N=

FB 1000N=

we know that,Basket

to start, we should recognize that the ropes are actually force vectors, and the direction of the rope is the direction of the force vectors. And, knowing the endpoints of the ropes, allows us to calculate direction vectors, and from this we can find force vec-tors for each rope.

Lets first label the points,

P1 = (-2, 3, 0)udP2 = (-15, 8, 0)udP3 = (-4, 13, 0)udPB = (-10, 10, 10)ud

Next, find the relative displacements between the people, and the bottom of the balloon,

d1 P1 PB– 2– 3 0, ,( )ud 10– 10 10, ,( )ud– 8 7– 10–, ,( )ud= = =

d2 P2 PB– 15– 8 0, ,( )ud 10– 10 10, ,( )ud– 5– 2– 10–, ,( )ud= = =

d3 P3 PB– 4– 13 0, ,( )ud 10– 10 10, ,( )ud– 6 3 10–, ,( )ud= = =

Now, use the vectors to find the forces, by multiplying the force magnitude by

F1 F1 ud1F1

P1 PB–

P1PB------------------ 200N

8i 7j– 10k–( )

82

72

102

+ +------------------------------------

110i 96j– 137k–( )N= = = =

F2 300N5– i 2j– 10k–( )

52

22

102

+ +---------------------------------------

132– i 53j– 264k–( )N= =

F3 500N6i 3j 10k–+( )

62

32

102

+ +------------------------------------

249i 125j 415k–+( )N= =

We can also write out the equation for the buoyancy force,

FB 1000k( )N=

the direction unit vector,

NOTE: the distance units are missing in this prob-lem, so ‘ud’ is used. These cancel out

• As a practical example of where 3D vectors might be required, consider the power line pole. It uses a tension cable anchored in the ground to resist the forces exerted by the power lines.

egr20933.jpg

3.2.4 Dot (Scalar) Product

• We can use a dot product to find the angle between two vectors

• We can use a dot product to project one vector onto another vector.

Finally, we want to sum the forces, and find the magnitude,

FR F1 F2 F3 FB+ + +=

FR∴ i 110 132– 249+( ) j 96– 53– 125+( ) k 137– 264– 415– 1000+( )+ +=

FR∴ 227i 24j– 184k+( )N=

FR 2272

242

1842

+ + 293N= =

It is obvious from the results that the balloon will move up, and out of the page as drawn because the z (or k) component is positive. The other two components suggest the balloon will fly forward and the left.

F1 2i 4j+=

F2 5i 3j+=

x

y

θcosF1 F2•F1 F2------------------=

θ∴ 2( ) 5( ) 4( ) 3( )+

22

42

+ 52

32

+-------------------------------------------

acos=

θ∴ 224.47( ) 6( )

----------------------- acos 32.5°= =

θ

• The use for the dot product will become obvious in later sections.

F1 3i– 4j 5k+ +( )N=

V 1j 1k+=

x

y

z

We want to find the component of force F1 that projects onto the vector V. To do this we first convert V to a unit vector, if we do not, the component we find will be multiplied by the magni-tude of V.

λVVV------

1j 1k+

12

12

+--------------------- 0.707j 0.707k+= = =

F1VλV F1• 0.707j 0.707k+( ) 3i– 4j 5k+ +( )N•= =

F1V∴ 0( ) 3–( ) 0.707( ) 4( ) 0.707( ) 5( )+ + 6N= = V

F1

F1V

ASIDE: UNIT VECTORS AND DOT PRODUCTS

Unit vectors are useful when breaking up vector magnitudes and direction. As an exam-ple consider the vector, and the displaced x-y axes shown below as x’-y’.

x

y

x’y’

45°

60°

F 10N=

We could write out 5 vectors here, relative to the x-y axis,

x axis 2i=

y axis 3j=

x‘ axis 1i 1j+=y‘ axis 1i– 1j+=

F 10N 60°∠ 10 60°cos( )i 10 60°sin( )j+= =

None of these vectors has a magnitude of 1, and hence they are not unit vectors. But, if we find the equivalent vectors with a magnitude of one we can simplify many tasks. In particular if we want to find the x and y components of F relative to the x-y axis we can use the dot product.

λx 1i 0j+= (unit vector for the x-axis)

Fx λx F• 1i 0j+( ) 10 60°cos( )i 10 60°sin( )j+[ ]•= =

∴ 1( ) 10 60°cos( ) 0( ) 10 60°sin( )+ 10N 60°cos= =

This result is obvious, but consider the other obvious case where we want to project a vector onto itself,

λFFF------

10 60°icos 10 60°jsin+

10--------------------------------------------------------- 60°icos 60°jsin+= = =

Incorrect - Not using a unit vector

FF F F•=

10 60°cos( )i 10 60°sin( )j+( ) 10 60°cos( )i 10 60°sin( )j+( )•=10 60°cos( ) 10 60°cos( ) 10 60°sin( ) 10 60°sin( )+=

100 60°cos( )260°sin( )2

+( ) 100= =

Using a unit vector

FF F λF•=

10 60°cos( )i 10 60°sin( )j+( ) 60°cos( )i 60°sin( )j+( )•=

10 60°cos( ) 60°cos( ) 10 60°sin( ) 60°sin( )+=

10 60°cos( )260°sin( )2

+( ) 10= = Correct

Now consider the case where we find the component of F in the x’ direction. Again, this can be done using the dot product to project F onto a unit vector.

ux' 45°icos 45°jsin+=

Fx' F λx'• 10 60°cos( )i 10 60°sin( )j+( ) 45°cos( )i 45°sin( )j+( )•= =

10 60°cos( ) 45°cos( ) 10 60°sin( ) 45°sin( )+=

10 60° 45°coscos 60° 45°sinsin+( ) 10 60° 45°–( )cos( )= =

Here we see a few cases where the dot product has been applied to find the vector pro-jected onto a unit vector. Now finally consider the more general case,

V1

V2

θ1

θ2

x

y

V2V1

V2V1V2 θ2 θ1–( )cos=

Next, we can manipulate this expression into the dot product form,

First, by inspection, we can see that the component of V2 (projected) in the direction of V1 will be,

V2 θ1 θ2coscos θ1 θ2sinsin+( )=

V2 θ1icos θ1jsin+( ) θ2icos θ2jsin+( )•[ ]=

V2V1

V1---------

V2

V2---------• V2

V1 V2•V1 V2------------------

V1 V2•V1

------------------ V2 λV1•= = = =

Or more generally,

V2V1V2 θ2 θ1–( )cos V2

V1 V2•V1 V2------------------= =

V2∴ θ 2 θ1–( )cos V2V1 V2•V1 V2------------------=

θ2 θ1–( )cos∴V1 V2•V1 V2------------------=

*Note that the dot product also works in 3D, and similar proofs are used.

3.2.5 Summary• the basics of statics as a topic were covered• engineering units and calculations• representations covered in this section were,

- scalar values- vector values

rectangularpolarcartesiandirection cosinesvector projectiondirection vectors

• The dot product was shown as a way to project one vector onto another, or final angles between them.

3.2.6 Practice Problems

1. Four forces act on bolt A as shown. Determine the resultant of the forces on the bolt.

2. Find the tension in cable A and B if the tension in cable C is 100N.

F2=80N

F1=150N

F4=100N

F3=110N

30°

15°

20°

A

x

y

ANS. Fx=199N, Fy=14N

3. A disabled automobile is pulled by two ropes as shown below. If the resultant of the two forces must be 300lb, parallel to the forward roll of the car, find (a) the tension in each of the ropes, knowing that α = 30°, (b) the value of α such that the tension in rope 2 is minimum.

4. A force F acts at the end of a pipe. Determine the magnitudes of the components that act per-pendicular to, and along the axis of the end of the pipe. (the pipe lies in the y-z plane)

A

B

C

M=3kg

F=100N

45°

ANS. TA=71N,TB=71N

T1

T2

α

20°A

ANS.a) TA=b) 70°

x

y

z

F 25i 30j– 10k+{ } N=

F F1 F2+=

F1

F2

F

4”

3”

5”

ans. 18N along, 36.1N across

5. Convert between the representations given on the left, and the results requested on the right.

x

y10dN

45°

Polar Rectangular

( x , y ) = ( __________ , __________ )

a)

ANS.(7.1dN,7.1dN)

x

yPolarCartesian

M = { -10i - 20j } lb.

b)

θ =

M =

ANS. 22.3lb, 243°

y

zDirection Cosines Cartesian

F = i(________)

c)

+j(________)

+k(________)

x

θy 70°=

θz 60°=F 5.0m=

ANS. (4.0m,1.7m,2.5m)

6. An F-117A stealth fighter is supposed top be flying N20°E, but a strong wind from West to East is pushing it off course. If the plane is pointed N20°E, but is actually moving N23°E, and its 22,000 lb engine is at full thrust, a) what force is the wind exerting on the plane? b) What is the answer in newtons?

7. Given the system of vectors pictured, a) give the resultant force using cartesian notation b) find the magnitude of the resultant force in metric units. c) Then then using cosine angles, and

y

ProjectedCartesian

L = {20i + 5j + 10k}Pa

d)

x

θxy 76°=

θz _________=L 22.9Pa=

z

θxy _________=

L __________=

θz 64°=

3°

20°

N (North)

E (East)a)

b)

Fw= 1137lb

Fw= 5070N

finally d) projected onto the x-y plane.

3.2.7 References

Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 6th edition, MacMillan Publish-ing Co., New York, USA, 1992.

4. EQUILLIBRIUM

• Put simply equillibrium describes the condition where all forces are balanced (no acceleration). Static equillibrium describes the state where all forces are balanced, and the object is not in motion.

• For an object to be in equillibrium, all the forces and moments must be balanced for,- each particle in a rigid body

50°

50°

115°

30°

20°

x

y

z

(10,10,10)

F1 10N=

F2 20N=

F3 10lb=

a) i(41.5) + j(37.8) + k(25.3) N

b) 61.5N

c) θx = 47.6°, θy = 52.1°, θz = 65.7°

d) Fxy = 56.1N

each rigid body- each object made up of rigid bodies

4.1 THE BASIC EQUATIONS OF STATICS

• There are two basic balances that must exist in any statics problems.

• if the sum of the forces is not zero, then the system will undergo translation, and the problem cannot be solved with statics methods.

• if the sum of moments is not zero, then the system will undergo rotation, and the problem cannot be solved with statics methods.

• At least one of these two equations will appear in every statics problem.

• A simple example,

• Let’s consider a force balance problem, ([Hibbeler, 1992]prob 3-40, pg. )

EQUILLIBRIUM OF FORCES

EQUILLIBRIUM OF MOMENTS (more later)

F∑ 0=

M∑ 0=

Fx∑ F1xF2x

F3x+ + 0= =

Looking at the three ropes pullingthe ring, each rope exerts a forcewith components in the x and ydirections.

F1

Fy∑ F1yF2y

F3y+ + 0= =

F2

F3

REVIEW: what type of force representation is this?

A

B

C

1.5 ft.

1.5 ft.

3 ft.

2.5 ft.4.5 ft.

3 ft.y

z

20 lb.

x

D

a 20 lb. bucket is suspended with three ropes that are joined at a point (D). each rope is connected to an anchored hook. Hook A is on the x-z plane, B is on the x-axis, and C is on the y-z plane. Find the ten-sions in each of the ropes in Newtons.

First, find the position vectors of the points in the problem,

D 1.5 1.5 0, ,( )ft=

A 4.5 0 3, ,( )ft=

B 1.5 0 0, ,( )ft=

C 0 2.5 3, ,( )ft=

Next, find the displacement vectors of the ropes,

DC C D– 0 1.5–( )i 2.5 1.5–( )j 3 0–( )k+ + 1.5i– j 3k+ +( )ft= = =

DC 1.52

12

32

+ + 3.5ft·= =

DB B D– 1.5 1.5–( )i 0 1.5–( )j 0 0–( )k+ + 1.5j–( )ft= = =

DB 1.52

1.5ft·= =

DA A D– 4.5 1.5–( )i 0 1.5–( )j 3 0–( )k+ + 3i 1.5j– 3k+( )ft= = =

DA 32

1.52

32

+ + 4.5ft·= =

DFDB

FDC

FDA

20 lb.

Now, using the direction vectors, let’s find force vectors,

FDC FCDCDC-----------

1.5 FC–

3.5--------------------

iFC

3.5---------

j3 FC

3.5-------------

k+ += =

FDB FBDBDB-----------

0 FB

1.5------------

i1.5– FB

1.5--------------------

j0 FB

1.5------------

k+ += =

FDB∴ FB–( )j=

FDA FADADA-----------

3 FA

4.5------------

i1.5 FA–

4.5--------------------

j3 FA

4.5------------

k+ += =

FDA∴ 0.667 FA( )i 0.333– FA( )j 0.667 FA( )k+ +=

Finally, the mass of the bucket is,

FBUCKET 20lb–( )k 20lb 4.448N1lb

----------------- –

k 88.96N–( )k= = =

Now, having all of the force vectors, we can write out the general equation,(keep in mind because we are using vectors, we don’t need to define thepositive direction),

F∑ 0= Fx∑ Fy∑ Fz∑ 0= = =

Fx∑ 0.428 FC– 0.667 FA+ 0= =

Fy∑ 0.286 FC FB– 0.333– FA+ 0= =

Fz∑ 0.857 FC 0.667 FA 88.96N–+ 0= =

Basically there arethree equations, andthree unknowns, sowe can do a simpleparametric solution.

An algebraic solution of the three equations above leads to the results,

FA 44N=

FB 5N=

FC 69N=

FDC∴ 0.428– FC( )i 0.286 FC( )j 0.857 FC( )k+ +=


1. Two cables (AB and AC) and a force (P) act on the top of a flag pole (AD). Find the magnitude of force P required to keep the flag pole standing? Assume that cable AB is under 5 KN of ten-sion.

PROBLEM SOLVING PHILOSOPHY- NUMBER OF UNKNOWNS AND EQUATIONS:

An important constraint when solving problems is that there must be as many (or more) equations as there are unknowns. If there are more unknowns than equations, the problem is unsolvable at that point, so you should number all equations with the intention of using then later, and having to refer to them. Keep in mind that while it is nice to get numbers right away, in many prob-lems you will have to solve equations in parametric or matrix forms - trying to avoid these problems will only make life complicated.

x

y

z

PA

B

C=(28.5,0,15)D

5

80°

50°O

ANS

Given,

αOA 50°=OAz 5= γOA 80°=

Find the missing direction cosine angle, and magnitudes,

1 αOAcos( )2 βOAcos( )2 γOAcos( )2+ += βOA 41.7°=∴

OA γOAcos 5= OA 28.8=∴

OAx OA αcos 18.5= =

OAy OA βcos 21.5= =

In summary, the points are,

A 18.5 21.5 5, ,( )=

B 18.5 0 0, ,( )=

C 28.5 0 15, ,( )=

D 18.5 0 5, ,( )=

FAB 5KNi 18.5 18.5–( ) j 0 21.5–( ) k 0 5–( )++

21.52

52

+--------------------------------------------------------------------------------------------

i 0( ) j 4.9–( ) k 1.1–( )++= =

Find the force vectors,

FAC TAC

i 28.5 18.5–( ) j 0 21.5–( ) k 15 5–( )++

102

21.52

102

+ +-----------------------------------------------------------------------------------------------

=

F∑ FAB FAC Pi–+ i 0( ) j 0≠( ) k 0( )+ += =

Sum forces,

∴ i 0.39TAC P–( ) j 4.9– 0.84TAC–( ) k 1.1– 0.39TAC+( )+ +=

∴ i 0.39TAC( ) j 0.84– TAC( ) k 0.39TAC( )++=

1.1– 0.39TAC+∴ 0= TAC∴ 2.82KN=

0.39TAC P–∴ 0= P∴ 1.1KN=

4.1.2 References


4.2 FREE BODY DIAGRAMS (FBD)

• Up to this point we assumed very simple forces acting on very small particles.

• In reality mechanical systems have many parts, and we draw an FBD for each part.

• We should divide forces on free body diagrams into two categories,Internal - these forces act only within a free body, and cancel out, unless we are looking at

a section of a free body.External - these forces act on a free body, and they induce reaction forces. Examples are

gravity, and other free bodies.

• An example of using free body diagrams for a system is given below with a system of masses, ropes, pulleys and anchors.

M1 M2

T1

T2

T3

P1

P2

R1

Step 1: label all of the components in the system. In this case there are two masses, two pulleys, one ring, and three cables (one cable is threaded through two pulleys.

M1

P1P2

Step 2: break the system into parts. The rule of thumb here is that each FBD should have only one rigid body.

M2

R1

M1

P1P2

Step 3: Draw arrow heads on the force vectors, and label each with a variable. NOTE: it is essential that the same force shown on two different FBD’s is equal and opposite in direction.

M2

R1

T1

T1

T1T1 T1

T2

T2

T3

M2gM1g

T1 FP1FP2

4.2.1 Pulleys and Springs

• Pulleys are basically a wheeled roller that a rope can roll over freely,

• A simple example of a pulley used for lifting a mass is given below,

ASIDE: The main advantage of free body diagrams is that we can ignore what is happening in other rigid bodies. Although they don’t make the solution easier, they do make it easier to develop equations, and it cuts the problem into smaller steps.

T1 T2T1 = T2

For a perfect pulley (no mass or fric-tion) the tension on one side of a pulley will be equal to the force on the other side.

The pulley support will have a maxi-mum reaction of 2T.

60°

10 kg

60 kg

FBD Block:

10 kg

T1

Mg 10Kg 9.81 NKg-------

98.1N= =

FBD Man:

60 kg

Fy∑ T1 Mg– 0= =

T1∴ 98.1N=

+

When doing calculations with scalar values, always indicate which direction is positive. It is also very useful to maintain consistency throughout solutions.

++ ++

ASIDE:

T1 98N=

FFRICTION

+ +

30°

FR

Fx∑ FFRICTION– T1 30°cos+ 0= =+

FFRICTION∴ 85N=

• Springs are a very important engineering tool,

FBD Pulley:

T1 98N=

T1 98N=

FRθR

60°

Fx∑ T1 60°sin– FR θRsin+ 0= =

Fy∑ T1 T1 60°( )cos– FR θRcos+– 0= =

+

+

FR∴T1 60°sin

θRsin----------------------

T1 T1 60°cos+

θRcos------------------------------------= =

60°sin1 60°cos+---------------------------∴

θRsin

θRcos--------------- θRtan= =

θRtan∴ 0.8661 0.5+----------------=

θR∴ 30°=

98 60°sin FR 30°sin=

FR∴ 170N=

• A sample problem that uses springs is given below, ([Hibbeler, 1992] prob 3-16, pg. )

M = 10 kg

k = 20 N/m

FBD Mass:

MgFR1

Fy∑ FR1Mg– FR1

10kg 9.81 Nkg------

– 0= = =

FR1∴ 98N=

+

FBD Spring:

FR1

FR2

Fy∑ FR2FR1

– FR298N– 0= = =

+

FR2∴ 98N=

But, the spring will be compressed, as governed by Hookes law,

x

F kx= Hookes Law

x∴FR1

k-------- 98N

20Nm----

---------- 4.9m= = =

FR1

F

1.5 m

6.0 m

k = 500 N/m

k = 500 N/mA rubber cord that is originally 6.0m long is stretched as shown in the dia-gram. We can model the elasticity of the cable as two springs to either side of the pulley. Find the force F for the deformation shown.

First select variables for the cable length,

l0 6m the undeformed cable length= =

l1 the deformed cable length=

l1

2---- 6.0m

2------------

21.5m( )2

+ 3.35m= =

l1 6.70m=

T ∆l2-----

kl1 l0–

2--------------

k 6.70m 6m–2

---------------------------- 500N

m---- 175N= = = =

Next, find the tension in the cord as a result of the change in length,

Next, try a FBD of the pulley,

F

T

T

Fx∑ F– 1.5l1

2----

---------

T 1.5l1

2----

---------

T+ + 0= =+

F∴ 6Tl1------ 6 175N( )

6.70m--------------------- 160N= = =

NOTE: this is equivalent to

4.2.2 Summary

• equilibrium of forces and moments• free body diagrams (FBD’s)

pulleysspringsanchorscablesmassesrings

4.3 PRACTICE PROBLEMS

1. Determine the reactions at B, C and D.

2. Four masses are suspended by cables that are supported by pulleys. The frictionless pulleys are mounted on a flat ceiling. Each of the pulleys is a distance of 2m from the others. Determine the height of the center mass.

200N

60deg 60deg

A

B C

D

100lb

100lb

100lb200lb

h?

2m

2m

2m

(ans. 1.03m)

3. The rod has a mass of 20lb., and the mass of joints A & B are both 5 lb. What is the tension ‘T’ in the rope?

(ans. 18.2lb)

4 For the mass pulley system on the left, a) draw the force triangle on the right, with all angles and magnitudes indicated, then b) find the mass.

T30°

10°

5ft.

A

B

M

10 Kg

3.0 ft

2.0 ft

θA

B

C

a)

5. Given the three masses below, connected by a cable through three pulleys, determine the final resting height (h2) for the centre mass. Assume the pulleys are very small.

4.3.1 References


5. MOMENTS

• a force is an important type of mechanical effect, but it doesn’t explain bending. For this we use moments.

Find M = ______________

b)

M=10kgM=10kg M=10kg

4.0m

1.0m1.5m

h2

h2= 2.4m

• The classic example is the see-saw,

up to now we have only dealt with forces through central points (virtual particles). But, in the case of the pirate on the gang-plank, there would be great concern about the plank bending.

MSMALL Fg102m( ) 10Kg 9.81 N

Kg-------

2m( ) 196Nm= = =

MMEDIUM Fg201m( ) 20Kg 9.81 N

Kg-------

1m( ) 196Nm= = =

To solve the problem using proper notation,

M∑ 0=+

MSMALL∴ MMEDIUM– 0=

196∴ 196– 0=

This shows that the system is static, andthe children will balance.

1 m 2 m

10 kg

20 kg

M Fd=

The two children sit on the teeter-tot-ter, because they have different masses, they must sit different dis-tances from the centre of rotation, or face catastrophic impact. In mathe-matical terms the moments on either side of the centre must balance.

First, recall the basic equation for a moment.

MMEDIUM

MSMALL

• Consider the case of a vise. The bottom member must resist the moment caused when the jaws are tightened.

egr20900.jpg

• If we look at a normal step ladder we can see that both halves can be analysed using moments about the top platform.

egr20903.jpg

• Certain varieties of tools amplify forces by using moment arms.egr20907.jpg

5.1 CALCULATING SCALAR AND VECTOR MOMENTS

• Next lets consider a case where the forces are not so simple,

10 kg

20 kg

To carry on, also consider that the forces must also balance,

FMEDIUM

FSMALL

FR

Fy∑ FMEDIUM– FR FSMALL–+ 0= =+

FR∴ 20 10+( )Kg 30Kg 9.81 NKg-------

294N= = =

2.0 m

M

40°

F = 10 N

Given the force applied to the end of the moment arm, find the induced moment.

Solution 1: Find the force component normal to the moment arm,

M Fd F 40°cos–( ) 2m( ) 15Nm–= = =

Solution 2: find the distance normal to the force

M Fd F–( ) 2.0m 90° 40°–( )sin( ) 15Nm–= = =

2.0 m

M

40°

F = 10 N

d

ASIDE: When dealing with moments in planar problems we are free to choose which direction (clockwise or counter-clockwise) is positive. Also, when we sum moments we are doing it about some centre of rotation. Consider the exam-ple below,

AB

F1

F2

MB∑+

clockwise is positive

it is the sum of momentsabout point BA

BF1

F2

BF1

F1 creates a counter-clockwisemoment and therefore should havea negative value

B

F2

B

F2

F2 creates a counter-clockwisemoment and therefore shouldhave a negative value.

BB In this case the force has been

F2y

F2x

+ve

-ve

broken into two components, thesigns on the moments created byeach component are opposite.

Solution 3: Use vectors and a cross product

F 10N 40°sin–( )i 10N 40°cos( )j 0N( )k+ +=

F∴ 6.43i– 7.66j 0k+ +( )N=

d 2i 0j 0k+ +( )m=

M d F×i j k

2m 0m 0m

6.43N– 7.66N 0N

= =

NOTE: note that the cross prod-uct here is for the right hand rule coordinates. If the left handed coordinate system is used F and d should be reversed.

M∴ 0m0N 0m 7.66N( )–( )i 2m0N 0m 6.43N–( )–( )– j +=2m 7.66N( ) 0m 6.43N–( )–( )k 15.3k mN( )=

NOTE: there are two things to note about the solution. First, it is a vector. Here there is only a z component because this vector points out of the page, and a rotation about this vector would rotate on the plane of the page. Second, this result is positive, because the positive sense is defined by the vector system. In this right handed system find the positive rotation by pointing your right hand thumb towards the positive axis (the ‘k’ means that the vector is about the z-axis here), and curl your fingers, that is the positive direction.

ASIDE: The cross (or vector) product of two vectors will yield a new vector per-pendicular to both vectors, with a magnitude that is a product of the two magni-tudes.

V1

V2

V1 V2×

V1 V2× y1z2 z1y2–( )i z1x2 x1z2–( )j x1y2 y1x2–( )k+ +=

V1 V2×i j k

x1 y1 z1

x2 y2 z2

=

V1 V2× x1i y1j z1k+ +( ) x2i y2j z2k+ +( )×=

ASIDE: The positive orientation of angles and moments about an axis can be determined by pointing the thumb of the right hand along the axis of rotation. The fingers curl in the positive direction.

x

y

z

+z

x

y

+y

z

x

+

ASIDE: The cross product is distributive, but not associative. This allows us to col-lect terms in a cross product operation, but we cannot change the order of the cross product.

r1 F× r2 F×+ r1 r2+( ) F×= DISTRIBUTIVE

r F× F r×≠ NOT ASSOCIATIVE

• Each of these three methods will be required at some time or another, each should be considered fundamental, any that is not learned will make certain problems difficult/impossible to solve.

• Consider the bent pipe in the practice problem below, ([Hibbeler, 1992], prob 4-30, pg. )

x

y

z

A

B

C

D

24 in.

15 in.

6 in.

F1 4i 10j 8k–+( )lb·=

F2 9i 4j– 10k–( )lb·=

Given the pipe CBAD, and the two forces applied to the end, find the moment that is twisting section AB about its axis.

Lets find the moment about point A using vector methods,

dAD 0i 0j 6k–+( )in=

MA dAD F1 F2+( )×=

MA∴0in

0in

6in–

4lb 9lb+

10lb 4lb–

8lb– 10lb–

×0in

0in

6in–

13lb

6lb

18lb–

×= =

MA∴i j k

0in 0in 6in–

13lb 6lb 18lb–

6in–( ) 6lb( )–( )i 6in–( ) 13lb( )( )j+[ ] in lb•( )= =

MA∴ 36i 78j–( )in lb⋅=

The axis unit vector of AB can be used to project the moment using a dot product

uAB 1( )i 0( )j 0( )k+ +=

MAB MA uAB•=

MAB∴ 36i 78j–( )in lb⋅ 1( )i 0( )j 0( )k+ +( )• 36in lb⋅= =

ASIDE: later you will learn how to calculate how much this pipe would twist when exposed to the moments calculated above. But, before these calcula-tions can be done, the basic statics calculations above (or equivalent) must be done. A basic question to ask your self is will this beam fail?

• Finally consider an application of moment calculations as seen below. Here we would need to balance the maximum load allowed against the hydralic cylinders acting on the lifter. This can be done using a sum of moments.

egr20922.jpg

When we want to do a cross product, followed by a dot product (called the mixed tripple product), we can do both steps in one operation by finding the determinant of the following. An example of a problem that would use this shortcut is when a moment is found about one point on a pipe, and then the moment component twisting the pipe is found using the dot product.

d F×( ) u•ux uy uz

dx dy dz

Fx Fy Fz

=

NOTE: When finding the moment about a point we need to usethe vector (cross) product, and the result is a vector. When we want to find the moment about a line, we use the tripple product to get a scalar value.

As an alternative to the vector methods, lets try solving this problem using scalars,

MAB∑ F2y6in( ) F1y

6in( )+ 4lb–( ) 6in( ) 10lb( ) 6in( )+ 36lb in⋅= = =

NOTE: in this case the solution was almost trivial. This is because the distances all happened to be split up into cartesian coordinates. In some cases the scalar solution will become very messy, and undoubtedly lead to mistakes.

5.1.1 References


5.2 FORCE COUPLES TO MAKE CENTERLESS MOMENTS

• Sometimes we are faced with moments that are in awkward positions, We can move these by replacing them with forces and moments in new positions. (Force couples also allow us to rotate the effects of moments)

• Keep in mind that couples cause rotation only, without any translation.

• Consider the basic force couple,

• A force couple acts about an axis. In the previous example the axis was out of the page. The axis can be moved to any position on the rigid body, as long as the direction is maintained. This axis of the couple will be perpendicular to the plane of the forces.

• A sample problem is given using a rope pulling a wedge ([Hibbeler, 1992], prob 4-81, pg. )

d

F

F

Consider a case where we have a moment (M), but we want to replace it with forces, or we have a force that we want to move, and we are willing to introduce a new moment. We can convert a moment to a pair of forces as pictured below.

MO Fd F d2---

F d2---

+= =

Now, also consider that force couple can be moved freely about the part, and by impli-cation, so can the moment.

+M

d

F

F

M

d

F

F

A

M Fd F d A+( ) F A( )–= =

OO

OO O+

ASIDE: Moment problems can be solved without resorting to force couples, but these are powerful concepts when dealing with mechanical systems. In effect they allow us to deal with a moment that is independent of a centre of rotation.

3”

4”

6”

P

-P

F

-F10 lb

10 lb

30°

30°

3”

Find the forces F and P to keep the block from moving.

5.2.1 Moving Forces and Equivalent Force Moments

I will use the vector approach to solve the problem, this problem could also be solved with more effort using scalars.

To begin I will find the moments caused by forces P and F. These are easy to find because they lie on the planes. (we should also define axis)

xy

z

MF F– 6in( )( )j 6– Fj( )in= =

MP P 6in( )–( )k 6– Pk( )in= =

Next, to find the moment caused by the ropes, we must find the normal to the plane that the moment acts upon. If this is then multiplied by the scalar moment, we will have the moment vector.

uNR

0i 3j 4k+ +

32

42

+---------------------------- 0.6j 0.8k+= =

MR 10lb 3in( ) 30lb in⋅= =

MR MR uNR30lb in⋅( ) 0.6j 0.8k+( ) 18j 24k+( )lb in⋅= = =

Finally, we sum the moments, and solve for the static condition,

M∑ MF MP MR+ +=

M∑∴ 6– Fj( )in 6Pk( )in– 18j 24k+( )lb in⋅+ 0= =

6– Fj 6Pk– 18j 24k+( )lb+ 0=∴

6– F 18lb+( )j 6P– 24+( )k+ 0=∴

6– F 18lb+ 0=∴

F∴ 18lb6

----------- 3lb= =

6P 24lb– 0=∴

P∴ 24lb–6

-------------- 4lb= =

• If a force is moved along the line of action, there are no other modifications needed, (this is called transmissibility)

• If a force is not moved along its axis (line of action), then we must create a resultant (make believe) force couple (and it’s resulting moment).

• A sample of a problems that is simplified by moving forces is seen below, ([Hibbeler, 1992], prob 4-102, pg. )

F

F

As long as a force is moved along a line of action, it will appear to be the same force (in a rigid body)

F

These three systems of forces are equivalent, but they show how to move forces about an object. Obviously these forces do not move in reality, but this is a calculation trick that will make solving some problems much easier.

FF

F

F

M

M r F×=r

FR

A

3’

30° 48 lb

5’ 2’

60 lb

18 lb

7’

C

Given this beam with applied forces,a) replace all of the known forces with an equivalent force and

moment at C.b) find the resulting force FR using the results in a)

part a)for the 18 lb force:

FCx18lb–=+

MC 7ft 18lb–( ) 126lb– ft⋅= =+

for the 60 lb force:

FCy60lb–=+

MC 2ft 60lb–( ) 120lb– ft⋅= =+for the 48 lb force:

FCy48lb 30°cos– 42lb–= =+

MC 48lb 30° 5ft 2ft+( ) 48lb 30° 7ft( )cos–sin– 459ft– lb⋅= =+

FCx48lb 30°sin– 24lb–= =+

for the total at C:

FCy60lb– 42lb– 102lb–= =+

MC 126lb– ft⋅( ) 120lb– ft⋅( ) 459ft– lb⋅( )+ + 705lb ft⋅–= =+

FCx18lb– 24lb– 42lb–= =+

part b):

Using the results from the last section, and the rather simple force at the end of the beam, we can sum the moments about C to find the force FR.

M∑ 705lb ft FR 2ft 5ft 3ft+ +( )+⋅– 0= = ASIDE: the sum of the moments about the pinned joint C will total zero because a pinned joint has no resistance to moments.

FR∴ 70.5lb=

+


1. The pole OA is 5m long and is held firm at base O. It is pulled by two cables AC and AB that each have 1KN of tension. Find the magnitude of the moment at O caused by the two cables.

2. A 500 lb cylindrical tank, 8ft in diameter, is to be raised over a 2ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing the corner of the obstruction at A is rough so that the cylinder slips at C but not at A, find the required tension in the cable.

NOTE: Remeber that once we have found a force couple we are free to move it over a rigid body or structure as needed. In calculations to come later this will make some calculations simpler. - In other words a couple is added into a moment equation without considering its position, and it is not considered in a sum of forces.

30°60°

(-3, 4, 5) m

2m

A

B

C

O

45°

60°

x

y

z

60°

3. A force of 1200N acts on a bracket as shown. Determine the moment Ma of the force about A.

4. A 30lb force acts on the end of the 3ft lever as shown. Determine the moment of the force about O.

8ft

2ft

A

C

1200N

30°

A

C

120mm

140mm

40mm

180mm

5. Considering moments, convert the cases given on the left to those requested on the right.

O

3ft

50°

20°

30lbA

M = 10 lb ft

x

y

a)A Moment

x

y

3

4

2.0 ft

1.0 ft

A Force

F _________=

M = 10 lb ft

x

y

b)A Moment

x

y

3

4

d = __________

d

A Distance

F 5.0 lb=

6. Given the two forces F1 and F2 acting on point A and the force of gravity acting on the centre of the beam Fg, write the equation for the sum of the moments about C.

7.The frame below has a ball joint at A, and the other end of the beam at C is smooth and slides freely along the y-axis. A mass of 10kg is added midway along the beam AC. Find the reaction forces at A and C.

y

z

c)A 3D Force A Moment Magnitude

MO ___________=

O F = { 5.0i - 4.0j - 6.0k } MN

A = (-20m, 50m, 30m)

x

x

y

zA

B C

F2F1 = {2.0i + 3.0j + 4.0k }lb

2.0m

Fg = -200k Kg

8. Find the moment at A given the two forces at B and C.

9. A box is shown in the figure below. Find the moment that the tension cable causes about the

4m

5m

x

y

z

65°

70°

A

B

Ccan slide freelyalong edge

10kg

Fax = 2.5NFay = 98.1NFaz = -2NFcx = -2.5NFcy = 0Fcz = 2N

A B

C

45°

F1 20N=

F2 30N=

5m

2m

MA 79.3Nm=

origin using a cross product. Give the results using components and a magnitude.

10. Find the tension in the cable DE.

11. A force P acts on a corner of a frame (at A) and creates a moment about the origin (O). We know that the force P is in the y-z plane, the moment about the y-axis is -20Nm, and the moment about the z axis is -40 Nm. Find the magnitude of the moment about the x-axis.

x

y

z 6’

3’

5’

(10, 0, -7) ft.

T=1kip

MO 7716ftlb= MOx3309ftlb–= MOy

5122ftlb= MOz4729ftlb–=

A

B

CD

E5000 Nm

10 KN

3m

3m 3m

TDE 4.17KN=

ANS.

x

y

z

P

α

5 cm

3 cm30°

O

A

12. Given the force at C and couple at B, find the magnitude and direction of the reaction forces at the supports A and E below.

Find the vector moment about O,

MO rA O⁄ P× i Mx( ) j 20–( ) k 40–( )+ +{ } Nm= =

Next, find the moment arm (or point A), and the force P,

A Ax 0.05 0.03, ,( )=

30°tan0.03m

Ax---------------= Ax∴ 0.052m=

rA O⁄ 0.052 0.05 0.03, ,( )m=

P 0 P αsin– P αcos, ,( )=

Do the cross product to get equations,

MO

i j k

0.052 0.05 0.03

0 P αsin– P αcos

i Mx( ) j 20–( ) k 40–( )+ += =

∴ i 0.05P αcos 0.03P αsin+( ) j 0.052P αcos–( ) k 0.052P αsin–( )+ +=

0.05P αcos 0.03P αsin+ Mx=

Write the equations and solve,

0.052P αcos– 20–=

0.052P αsin– 40–=

0.052P αsin–0.052P αcos–

---------------------------------∴ 40–20–

---------= α∴ 63.4°= P∴ 860N=

Mx 0.05P αcos 0.03P αsin+ 42Nm= =

5.2.3 References

Beer, F.P., Johnson, E.R., Statics & Mechanics of Materials, McGraw-Hill, 1992.


6. MECHANISMS

6.1 REACTIONS AND SUPPORTS

• These symbols are very important to engineers. Their equivalent is the symbols for resistors, transistors, etc in electrical engineering. They are a way to talk about systems in general terms before investing time and resources designing and building the system.

• A subset of these symbols are,

2 yd.

2 yd.

3 yd. 3 yd. 2 yd.

500 lb.

10000 lb.yd.

A

B

C D

E

ANS.FAx = -532 lb.FAy = 1564 lb.FE = 1190 lb <-27°

SIMPLE LINK: This type of link is commonly used, and only has pinned joints at two ends. These members don’t trans-mit moments, and all forces are tension or compression along an axis between the pins.

F

F

F

CABLE: a cable (like a string) can be made of natural/artifi-cial fibres, metals (usually stranded), etc. These mem-bers can only be used in ten-sion - you can’t push with a rope.

e.g., a cable does not provide force when compressed

The cable cannot supplycompressive forces neededto hold the bar in position

F

F

• Other joints include- spring- rod on a smooth surface- rod on a rough surface- pin supports- beams cemented into walls (fixed support)- ball (3D)- ball socket- universal joint- hinge- collar and shaft

ROLLER: Force is only normal to the rolling plane. If the applied force is away from the surface, the roller exerts no force.

F

F

SLIDER: this is the same as the roller, except the force can be in both directions.

F F

******* NOTE: Add more on support reactions

6.2 EQUILLIBRIUM OF FORCES AND MOMENTS

• It is often necessary to solve problems using both sums of moments and forces. Indeed, both of these sums must be equal to zero for the body to be static.

• An example of a problem that has both forces and moments is given below, ([Hibbeler, 1992], prob 5-4, pg. )

Supports Reactions Unknowns

Roller

FrictionlessSurface

Rocker

1

M∑ 0= F∑ 0=AND

3

45

A B

C

D

E

2.0 m 2.0 m 1.5 mM = 80 kg

For the beam suspended by a cable, with a mass at the end, find the forces in the cable, and at point A.

NOTE: the slope trian-gle is sometimes used in place of angles, but note that it can be used to give trig. values directly.

stat0000.wm

Before starting any calculations, we should develop a free body diagram. This problem is best solved by scalars because of the simple planar forces at right angles.

A B

CE

2 m 2 m 1.5 mM = 80 kg

FAx

FAy TT

Next, we can sum forces and moments to get three equations. We know that there are two unknowns at point A, and the cable tension is the second unknown. Therefore we will have three equations and three unknowns.

We can start by summing the moments about point A. Because this point has two of the three unknown forces go through it, the sum of moments equation will only have one unknown (and can be solved immediately).

MA∑ T–( ) 2m( ) 45---– T

2m 2m+( )+=

80Kg 9.81 NKg-------

2m 2m 1.5m+ +( )+ 0=

+

T 5.2m( )∴ 4316N m⋅=

T∴ 830N=

Fx∑ FAx

35---–

T+ 0= =+ FAx∴ 498N=

Fy∑ FAyT 4

5---

T 80Kg 9.81 NKg-------

–+ + 0= =+

FAy∴ 709– N=

FAy∴ 830N– 4

5--- 830N( )– 785N+=

ASIDE: please note that -ve sign here indicates that I assumed the wrong direction for the force on the FBD.

• Consider the pictures of both the hoists below, they are much like the examples that follow. The red dampers apply a force that is almost static to slow the lowering of the upper arms.


• Lets consider another complex system to solve. In this case free body diagrams are beginning to become essential. ([Hibbeler, 1992], prob 5-44, pg. )

8 ft

12 ft

14 ft

10°

A

B

C

D

M = 800 lb

When passengers must leave a large ship, lifeboats are lowered over the side. A large hydraulic cylinder tips the arm. Given the diagram to the right, and the maximum load in the boat,

a) find the force exerted by the hydraulic piston CB to support the assembly.

b) find the forces at pin A.

stat0001.wm

1. First we must construct a FBD. In this case it is essential to reduce the quantity of detail. The arm seems to be the principle structure here (it touches all of the parts) so it will be the first FBD.

8 ft

12 ft

10°A

BD

M = 800 lb

FCYL

FAx

FAy

14 ft

FBD Arm ABD:

2. If we find the sum of the moments about point A, then we can simplify the first step to one equation and one unknown.

MA∑ 800lb( ) 6ft( ) FCYL 10°cos–( ) 14ft( ) FCYL 10°sin( ) 12ft( )+ + 0= =+

FCYL∴ 6 800( )ft lb⋅10°cos( )14ft 10°sin( )12ft–

---------------------------------------------------------------------- 410lb= =

3. Finally, we can sum up the forces to find both components of the force at A.

Fx∑ FAxFCYL 80°cos+ 0= =+

Fy∑ 800lb– FCYL 80° FAy+sin+ 0= =

+

etc.....

• Now, lets try a problem using vectors, ([Hibbeler, 1992], prob 5-83, pg. )

8 ft

6 ft

4 ft

12 ft

x

y

z

A

B

C

D

The angled bar slides on a square col-lar. A force is pulling the bar in a downwards direction, but this is resisted by a rope between points C and B. Find the tension in the rope, and the reactions at A.

FD 20i 40j– 75k–( )lb·=

1. first, draw a FBD of the main arm. This is chosen because it is the main member. Then, define the force and moment vectors. The diagram is drawn in 2D, but it is still a full 3D set of vectors.

FA

MA

FCB

FA FAxi FAy

j+=

Because the collar at A can only translate along the z-axis (no rotation),

MA MAxi MAy

j MAzk+ +=

FCB FCBC B–C B–---------------- FCB

0 12–( )i 8 4–( )j 6 0–( )k+ +

122

42

62

+ +-------------------------------------------------------------------------

= =

Next find the rope force vector using the endpoints of the rope.

FCB∴ 0.86 FCB–( )i 0.29 FCB( )j 0.43 FCB( )k+ +=

FD 20i 40j– 75k–( )lb·=

2. Next we can find the sum of the moments about A. To do this in vector form we will need the displacement vectors between the centre of rotation, and the point, but since the point of rotation is the origin, and the dimensions are simple, there is no calcula-tions required.

MA∑ MA dAB FCB× dAD FD×+ + 0= =

MAxi MAy

j MAzk 12i( ) 20i 40j– 75k–( )×+ + +∴

12i 4j+( ) 0.86 FCB–( )i 0.29 FCB( )j 0.43 FCB( )k+ +( )×+ 0=

MAxi MAy

j MAzk

i j k

12 0 0

20 40– 75–

FCB

i j k

12 4

0.86– 0.29 0.43

+ + + +∴ 0=

MAx4 0.43( ) FCB+( )i MAy

12 75–( )– 12 0.43( ) FCB–( )j+∴

MAz12 40–( ) 12 0.29( ) FCB 4 0.86–( ) FCB–+ +( )+ 0=

3. find the tension in the rope by summing the forces. In particular, we can see that there are only two z-components of forces that will apply, therefore, sum forces in the z-axis.

Fz∑ 75lb–( ) 0.43 FCB( )+ 0= = FCB∴ 174lb=

MAx4ft 0.43( ) FCB+ 0= MAx

∴ 299ft– lb⋅=

MAy12 75–( )– 12 0.43( ) FCB– 0= MAy

∴ 2ft– lb⋅=

MAz12 40–( ) 12 0.29( ) FCB 4 0.86–( ) FCB–+ + 0= MAz

∴ 724ft– lb⋅=

6.3 SPECIAL CASES

• There are a number of standard load cases that we will encounter. One of the simplest is the two force member discussed before.

egr20940.jpg

• Three force members can be solved quickly using force triangles, We can also keep in mind that the forces must act concurrently through a common point on the rigid body, or else all three must be parallel.

4. Using previously discussed techniques to find the reaction forces at A. etc.......

ASIDE: At this point you should expect to find many problems that will require that you find a few equations and then solve for an answer. Trying to avoid this will make many problems unsolvable and frustrating. Also keep in mind that the sum of forces and moments in 3D yields a total of 6 equations, allowing the solution of up to 6 unknowns.

F1

F2

F3

egr20941.jpg

6.3.1 References


6.4 STATICALLY INDETERMINATE

• There are a number of problems that cannot be solved with statics methods. In fact, all problems are inexact, but luckily we can solve some using statics.

• These types of problems often occur when the number of forces known is equal to the number of useful governing equations. This is known as completely constrained.

• When there are more equations than unknowns the problem may be overconstrained.

• When there are fewer equations than unknown forces, the problem may be underconstrained.

• The basic types of cases are,

• be cautious with this class of problems. You may actually be able to calculate a solution, but the answer will not be correct realistically, and mathematically speaking.

6.4.1 Summary

• basic moments and calculations- using normal components- using perpendicular components- using vectors and cross products

• a review of cross products, and the positive direction for angles/moments• force couples and moments were shown to be able to freely move about a rigid body.• a force can be moved away from its axis if a moment is also added.• mechanical schematic symbols were discussed with applications to statics.• indeterminate problems

1. The defined problem does not have balanced forces, or improper supports, resulting in a dynamics problem.

2. There are redundant constraints, and the problem solution depends on materials proper-ties. (generally violates the rigid body assumption)

F M

F

M

F

F F

M M

F

F


1. Determine the reactions at A and E.

6.4.3 References


7. TRUSSES AND FRAMES

• We will look at a variety of analysis techniques for trusses and frames. These techniques endeavor to find the internal forces in trusses, and external forces for frames.

7.1 WHAT ARE TRUSSES?

• Trusses are at the heart of many engineering projects. We can see one of these in the bridge across the grand river.

egr20908.jpg

200mm

200mm150mm

200mm30deg

30degA

B C

D

E

500N

ANS.FAx = 37NFAy = 433NFE = 287N

• Basically, a truss is a collection of beams joined together to carry simple and complex loads.

• We can see trusses use in cranes,


• Trusses are typically made from beams that are joined with gusset plates.

e.g., a house roof truss- snow load- roof mass- wind load

FF

F

- bearing wall

egr20923.jpg

• Other times obvious pin joints are used. Consider the pin joint on the end of the tension member below,

egr20911.jpg

• With these types of problems the tension or compression of the beams/members should be clearly indicated. Materials and structures will not fail at the same load when in tension (neck-ing then fracture), than in compression (buckling).

• We can see a tension member in a bridge with turnbuckles for tensioning,egr20912.jpg

• The basic assumptions used in most truss and frame problems are,1. the joints have pinned ends, so the forces exerted by the beam has a direction that is

along the line between pins.2. Forces are exerted at pins, but no moments.

• Types of trusses are shown below,

Beams are quite often joined with gusset plates. These can be seen where there are exposed steel structures, such as subway tun-nels, bridges, etc.

But, for all practical purposes, we treat these joints as if they are pinned joints.

egr20929.jpg

PRATT

HOWE

FINK

PRATT

HOWE

WARREN

BALTIMORE

PRATT

STADIUM

cantilever sectionof a truss

• Picture of other types of trusses can be seen below,


• Many of the methods in this section can also be extended to the analysis of trusses in 3D.egr20935.jpg

7.1.1 References



7.2 STABILITY OF TRUSSES

• Trusses are composed of beams and pin joints. Generally, if the frame is constructed of only tri-angles (internally) it will be stable. (not collapse when pushed the wrong way)

• Quite often aditional members are added to frames to stifen it (make it stable), but not to carry loads. We can see such members in a bridge,

egr20913.jpg

• We can predict stability using a basic topological relationship where we look for equality. The equation shown below requires that the truss be composed of simple pinned beams, and when drawn to scale on paper the beams must not cross or touch except at the pinned joints

• We can also verify stability by making nodes that are held in place either by supports, or triangu-lar structures.

• As an example,

m 2n 3–= where,m = the number of beamsn = the number of joints

A

B

C

D

E

F

G

H

First, we will try the topological test, m = 14n = 8

14 2 8( ) 3–≠ 13=therefore this is not a simple truss (there isan extra beam).

Second, we can walk through the truss looking for stable substructures,

pick a triangle touching a support to start (fill it in to mark it as sta-ble

if another triangle touches it with two points, fill it in.

All of the joints in the struc-ture are touching the shaded area, therefore the structure is stable

stat0002.wm

7.2.1 References


7.3 THE METHOD OF JOINTS

• The basic steps when solving problems with the method of joints is,1. Draw an FBD for each joint as if it is a particle.2. Solve for the applied forces with the “joint particle” in equilibrium. The simplest joints

are often the best to start with.

• We can see an example of a two force member below being used to support a canopy,egr20909.jpg

• This method is best shown using a sample problem, ([Hibbeler, 1992], prob 6-7, pg. )

ASIDE: Two force members can greatly simplify problems. (As is done in the method of joints and section problems). In these problems there is typically a beam with a force at either end. These forces must be along the same line, either towards or away from each other. As a result these forces are described as tension or compres-sion along the line between the two pins.

θA θB

FA FB

Consider the case where θB is any angle other than 0° or 180°. This would result in a tangential component, the sum of the moments about A would not be zero, there-fore it would go into motion. Thus for a static frame θB = θB and FB = FB.

A

B C

D

EF

3 KN

8 KN 4 KN 10 KN

1.5 m

2 m2 m

Find the force in each beam, indicating clearly tension or compression, and find the reaction forces at the supports.

ASIDE: with this type of problem you will learn to find easy places to start. The best place to start is often the supports, or joints that the smallest number of forces. When reviewing consider that by observation, beams AB, BC and FE can all be solved immediately.

NOTE: These things come with practice, more problems must be solved to gain expe-rience. And, experience will dramatically cut solution time.

1. In this case the support forces have to be found, and they are easy to identify, so we can determine those first. Please recall, the pinned joint at A will have x and y com-ponents, while the roller at E will only have a y component.

MA∑ 3KN 1.5m( ) 4KN 2m( ) 10KN 2m 2m+( ) FEy2m 2m+( )–+ + 0= =+

FEy∴ 4.5KN 8KN 40KN+ +

4------------------------------------------------------- 13.1KN= =

+

Fx∑ FAx3KN+ 0= =+

FAx∴ 3KN–=

Fy∑ FAy8KN– 4KN– 10KN– FE+ 0= = FAy

∴ 8.9KN=

FBD of entire truss:

A

B C D

E

3KN

8KN 4KN 10KN

2 m2 m

1.5 m

FEyFAy

FAx

2. We can now look at some of the joints and, by inspection, determine the forces in some of the members.In this case I am trying to eliminate many of the obvious forces. The most common of these is a zero force member, and full force members, where joints have members meeting at 90° angles. In this case the joints will be explained, but in most cases just stating the reason is sufficient.

8 KN

3 KN TBC

TBA

Fx∑ 3KN TBC+ 0= = TBC∴ 3KN–=+

Fy∑ 8KN– TBA– 0= = TBA∴ 8KN–=+

NOTE: all of the forces meet at right angles. For the forces in line what goes in one side, comes out the other. In other words, the force from the left is opposed by a force from the right, and the force from the top is opposed by a force from the bottom.

ASIDE: also note the convention for describing the forces. You will first notice that I am using a T instead of an F. This notation means that the beam is assumed to be in tension. A C could have also been used, indicating compres-sion. In very blunt terms, you can use any method you want, but consistency is of paramount importance, you are best off to pick one method and reli-giously adhere to it - it is the shortcuts and changes in method that will cause mistakes. Another point to notice is that TBC=TCB=-CCB=-CBC, but if FBC, or FCB is being used, the force direction must be defined someplace.

NOTE: because I am using T (tensions) for each beam, the force arrows are always away from the joints because a beam in tension is pulling the joint. When the result of calculations is negative, the beam is actually in compres-sion.

By the same reasoning as above,

TDE∴ FE– 13.1KN–= =

TEF∴ 0=

NOTE: EF is a zero force member, this is obvious if we use the case above (joint B), and consider that the force from the right is zero, therefore the force to the left will be zero.

B

egr20934.jpg

NOTE: zero force members might seem to have no purpose at first glance, but when building large structures these members are sometimes added to “stiffen” long beams and prevent buckling. (Zero force members can be seen in the structure below)

3. So far we have a few obvious forces in the upper left and lower right corners, it can help to keep track of the forces in your mind, or on paper by checking off the forces that are known. The next action is to select the simplest joint to solve. Looking at the figure below (this figure is normally not drawn, but is used here for illustration) with known forces having thicker lines, I observe that there are three unknowns at joints C and F, and two unknowns at joints A and D. So, I arbitrarily select A to solve next.

A

B C D

E

3KN

8KN 4KN 10KN

FEyFAy

FAx

F

ASIDE: known forces to this point have darker lines

FBD A:

FAx

FAy

TAF

TAB TAC

3

4

5

A

Fx∑ FAxTAF

45---TAC+ + 0= =+

TAF∴ 3KN 45---TAC–=

Fy∑ TAB35---TAC FAy

+ + 0= =+

8– KN 35---TAC 8.9KN+ +∴ 0=

TAC∴ 1.5– KN=

TAF∴ 4.2KN=

4. At this point each of the remaining joints has two unknowns, so I arbitrarily select D as the next joint to solve for.

FBD D:

TDE

TCD

TFD3

4

5

D

Fy∑ 10KN– TDE– 35---TFD– 0= =

+

10KN– 13.1KN 35---TFD–+∴ 0=

Fx∑ TCD– 45---TFD– 0= =

+

TFD∴ 5.2KN=

TCD∴ 4.2KN=

10 KN

5. There is only one member (CF) still unknown. Therefore we can solve for either joint C or F. F has only 4 forces on the beam, as opposed to the five on joint C. Therefore I will solve for joint F.

FBD F:

TFE

TAF

TFD

3

4

5

F

Fy∑ TCF35---TFD+ 0= =+

TCF∴ 3.1KN=

TCF


1. Use the method of joints to find the force in each member of the truss below.

2. Determine the force in each member of the truss shown. Indicate tension or compression.

6. Finally, it is always a good idea (in this course is mandatory) to summarize the results so that when you start the next stages of design work, it is much easier to find the essential numbers.

Link/Support

ABBCCDDEEFFAACCFDFFAxFAyFEy

Force (Ten./Comp.)

8 KN (C)3 KN (C)4.2 KN (C)13.1 KN (C)04.2 KN (T)1.5 KN (C)3.1 KN (C)5.2 KN (T)3 KN (left)8.9 KN (up)13.1 KN (up)

A B

CD

1000lb

2000lb8ft

4ft


4. Find the forces in each of the beams of the structure below (indicate tension or compression).

12 ft 12 ft

8 ft

6 ft12 ft6 ft

2000 lb 1000 lb

16 ft

12 ft

800 lb

800 lb

12 ft

Assume all joints are pinned.

5. For the frame below, find the tension/compression in each member, under the loading condi-tions given, and assuming all joints are pinned. (Note: you must clearly indicate tension or compression when dealing with this type of problem)

1m 1m

1m

2m

10KN

5KN

A B

C D E

F

ans.

ABACBCBECDCFDEDFEF

5KN(C)10KN(C)7.07KN(T)5KN(C)07.07KN(C)007.07KN(C)

3’ 3’ 3’

6’

6’

500lb 500lb

A

B

CD

E

F

ans.

ABBCCDDEEFAEBEBD

774lb(T)515lb(C)125lb(C)1289lb(C)515lb(C)401lb(C)625lb(C)1563lb(T)

(ans. EF=515lb(C), BC=515(C), CD=125(C), AB=774(T), AE=401(C), BD=1563(T), BE=625(C), DE=1289(C)

6. Given the frame below, find the tension/compression in each member.

7. Find the forces in each beam of the bridge shown. (As usual, clearly indicate tension or com-pression)

0.5N 0.5N

A

B

C

D

E

F

I

H

L

J

K M

G

FIX*******

8. Find the tension/compression in each member of the frame below, (HINT: look for 6 zero force members and use symmetry)

200KN

5m

5m

5m

10m 10m 10m 10m

A

B

C

D

E

F

G

H I

J K L

AB= -141KN

AJ= 100KN

BC= -141KN

BJ= 0

CD= -112KN

CH= 0

CJ= 50KN

DE= -112KN

DH= -60KN

DI= -60KN

EF= -141KN

EI= 0

EL= 50KN

FG= -141KN

FL= 0

GL= 100KN

HJ= -60KN

IL= -60KN

JK= 133KN

KL= 133KN

1ft

1ft

1ft

1ft

2ft 2ft

A

B

C

D

E

F

G

100lb

AB = 71 lb [C]AC = 0BC = 0BD = 71 lb [T]BE = 100 lb [C]CD = 0DE = 71 lb [T]DF = 0EF = 0EG = 71 lb [C]FG = 0

ANS.

7.3.2 References



7.4 THE METHOD OF SECTIONS

• Basically: cut out a part of a truss, and then treat it as if it is a rigid body. When done wisely, this allows simplified solutions. The alternative is using the method of joints to find all (or many) of the forces in the frame.

• Keep in mind that while moments are very popular with the method of sections, it can also be used for forces as well.

• Consider an example where we want to find forces in a structure, ([Hibbeler, 1992], prob 6-24, pg. )

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

4 m

4 m

4 m

4 m

In the truss shown find the forces in EF, BE and CB.

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

1. first, we must decide where to cut the section. A good rule of thumb to follow when deciding which section to cut is that the section line can cut through any joint, or member, but it can cut through at most one unknown member. In this case member BE could not be cut out without cutting through 2 members, but EF can be cut if the section line passes through one joint. The result of this sectioning is a FBD shown to the right of the page.

A

A

B

C

D

10 KN

10 KN

5 KN

E

5 KN

TEF

MB∑ 5KN 4m 4m+( ) 10KN 4m( ) 5KN 4m( ) TEF 4m( )+ + + 0= =+

TEF∴ 25KN–=

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

2. Continuing with the same idea, if I cut from joint C, through BE and EF, there will only be one unknown. Again, the diagram to the left is included only for illustra-tion purposes, normally it is omitted.

B

B

C

D

10 KN

5 KN

E

5 KN

TEF

MC∑ 5KN 4m( ) 10KN 0m( ) 5KN 4m( ) TEF 4m( ) 1

2-------T

BE 4m( )+ + + + 0= =

+

TBE∴ 21.2KN=

TBE

2

11


1. Determine the tensions/compressions in members EF, JK and HJ for the bridge truss shown below. The method of sections is recommended.

3. To find the force in CB we can cut the section as shown,

C

D

10 KN

5 KN

E

5 KN

TEF

Fy∑ 5KN– TEF– 1

2-------TBE– TCB– 0= =

TCB∴ 5KN=

TBE

2

11TCB

+

5KN–∴ 25KN–( )– 1

2------- 21.2KN( )– TCB=

4m 4m 4m 4m 4m 4m

2m

2mA

BC

D

E F

GHJ

K

2kN 3kN 5kN 4kN 3kN

2. Find the forces in FG and DF (indicate tension or compression). Assume all joints are pinned.

3. In the frame pictured below, find the forces in DE, EF, DF and CB.

1m 1m 1m 1m 1m 1m 1m 1m

4m

2.5 kip

A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

ans. AH=1.77kip(C), FG=0)

15°

10°

30°

95”

105”

1000lb2000lb

150”

A

B

C

D

E

F

G

H

I ans.

ED=1970lb (T)EF=1532lb (C)DF=0CB=401lb(T)

4. Find the forces in members CD, CF, CG. (You will benefit most by using the method of sec-tions to solve this problem)

(ans. CD= 11.25KN(C), CF= 3.21(T), CG= 6.8(C))

5. In the frame below members DL and EL can support up to 1kip of tension or compression, find the maximum load P that may be applied. (Hint: you could mix the method of joints and sections)

6. Find the forces in JM and KM.

2kN 4kN 4kN 5kN 3kN5m5m5m5m

2m

3m

2m

A B C D E

F

G

H

J

1m1m1m1m1m1m

2m

A B C D E F G

P

H I J K L M N

7. Find the tension/compression in members CE, DE.

1m 1m

.5m

1m

1m

1.5m 1.5m20KN 100KN 20KN

A

B C

D

E

F

G H

I J

K

L N

M P

1m

JM= 83KN

KM= 14KN

1m

0.75m

1m 1m

1m 1m

A

B

C

D

E

F

G

1KNANS.

CE = .667 KN [T]DE = -.4 KN [C]

7.4.2 References


7.5 METHOD OF MEMBERS

• Sometimes we must deal with structures that do not have simple two-force beams, in this case we must use the method of members.

egr20906.jpg

• These structures are commonly called frames, referring to the fact that they have at least one member that has more than two forces. We can see some examples of these members in the picture below,

egr20931.jpg

• In basic terms we are just making good use of free body diagrams, and quite often solving para-metric equations. Consider how we could isolate the free body diagrams in the figures below.

egr20930.jpg

• If we were to assume that beams in a truss have a mass, then we would have to use the method of members to solve the problem.

• A sample problem is given to illustrate the method, ([Hibbeler, 1992], prob. 6-60, pg. )

9”

9”

18”

A

B

C

24”

P30°

30°

30°

70 lb.(comp.)

The air pump shown is used to inflate bicycle tires. In effect a force P is applied by standing on the top. This in turn causes the lever to push on the cylinder which is compressed. The cylinder must be compressed with at least 70 lb. before the pump will work. What is the minimum force P?

D

1. First, we must select a suitable part (member) of the assembly to start with, and draw an FBD of it. Looking at the question, and the structure, force P touches the lever, and so does the cylinder, so this seems to be the reasonable part to start with.

9”

9”

18”

P30°

30°

30°

FCYL 70lb·=

θ

FBx

FBy

FBD BAD:

2. Find the angle of application of the cylinder force. This just involves a bit of basic trigonometry.

θ 9in( ) 60°sin24in( ) 9in( ) 60°cos–

----------------------------------------------------- atan 21.8°= =

60°

24”

9”θ

θ

• Now, lets consider a practice problem,


3. Next, the only unknown left of the drawing is P, and the reaction forces at B, so to reduce the problem to one variable, I will take the sum of the moments about B.

MB∑ P 9in 9in+( ) 60°cos 18in( ) 30°cos+[ ]=

70lb θcos( ) 9in 60°sin( )[ ]– 70lb θsin( ) 9in 60°cos( )[ ]– 0=+

P∴ 70lb θcos( ) 9 60°sin( ) 70lb θsin( ) 9 60°cos( )+9 9+( ) 60°cos 18( ) 30°cos+

------------------------------------------------------------------------------------------------------------------ 25lb= =

1 m

6 m3 m

2 m

4 m

1 m 2 m 1 m

10 KN/m

A

B

C

D

E

F

For the structure to the right, find the reaction forces in all of the joints. You can use any methods (or combination) that you find suitable

1. Determine the reactions at A and B.

2. In the figure below, determine the force exerted by pin B. The method of members will be most useful.

3. What are the reactions on each of the members (clearly indicate the results on FBDs)?

400 N

600 mm

250 mm

250 mm

375mm

A

B

D

C

R0.15m

50kg

D

0.5m

0.5m

0.5m

45°

45°

A

B

C

x

y

4. Calculate the x- and y-components of the force at D which member AD exerts on member DE. The deflection of the spring in the equilibrium state shown is 2.5 inches. The mass of the

A

B

C

D

E

100N

1m 1m 1m1m

1m 0.5m

0.25m

(ans. Ax=150N, Ay=150N, Bx=0, By=50N, Cx=150N, Cy=0, Dx=0, Dy=150N)

members and friction are negligible.

(ans. 70.7N in both)

5. The three member frame below is exposed to a load of L=60lb. The beams that the frame is made from weight 8lb/ft. Find the reaction at the base, and find the reaction at each joint and in each member.

4’ 4’ 8’

8’

2’ 2’

K=100lb/in.

750lb

A

B C D

E

F

G

H I

6. For the structure below, find the reaction forces in pins B and E.

3’

3’

4’

5’60lb

A

B

C

D

E

3’ 3’

x

y

ans.

Cx=164lbBy=351lbCx=164lbCy=224lbDx=164lbDy=413lbFEx=0FEy=261lbME=655lb ft

1 m

6 m3 m

2 m 4 mA

B

C

D

E

F

40KN

7. Find the forces at all of the pins, and at the supports. (clearly indicate direction using FBDs)

8. Find the forces acting on DCBA at pin D,

1m

.5m

2m 1m 2m 1m

50kg

A B C

D

E

F

G

H

A B C G

D F

E

D

C

E

F

G

9. We work for a maker of hand tools, and today we are evaluating a new design for vice grip pli-ers. In the configuration pictured we want a gripping force of 50 lb. between the jaws. How much force P must be applied at the handles to achieve this?

6 in

4 in

2 in

1 in

1 in

3 in 2 in

5 kip

A

B

C

D

E

F

G

ANS.35.4 kip <135°

P

P

50lb.

0.5” 5”

0.5”

2”

2.5”

1.5”

ANS. P = 12.9 lb.

7.5.2 References


7.6 SUMMARY

• A simple way to differentiate between the different methods is,

• You are encouraged to mix and match methods in any way that will simplify a solution. For example, in a couple of cases a problem that is being solved by the method of sections could easily use the method of joints in a couple of places.

8. DRY STATIC FRICTION

• Friction is a force that exists between any two object in contact. This can sometimes work against the engineer, other times it can be of great advantage.

8.1 THE BASIC PHYSICS OF FRICTION

• This natural phenomenon explains the resistance of one object to slide across another when they have common surfaces in contact.

Method

Method of Joints

Method of Sections

Method of Members

Use

Suitable for solving entire structures when we have trusses of beams with two pinned joints.

Allows forces and reactions to be determined in trusses (as above) without solving the entire truss.

For more complicated mechanical systems. These struc-tures will have members that may have more than 2 forces or moments on the beam.

• It is primarily the result of surface roughness, material properties, and if the object is moving.

• The graph of applied load versus friction helps illustrate the nature of friction. Notice that while the force is static, the force increases linearly up to the limit. After the object begins moving the force can be approximated with a constant value, using the dynamic coefficient of friction. Note that dynamic friction is shown to be lower that the maximum static friction.

If static on a dry surface we can approximate the maximum friction force using the equation,

FS µSN=N = Mg

FR

FS

µS a coefficient of friction for the two surfaces=

NOTE: coefficients of static friction are experi-mentally determined, and will typically be found using lookup tables. So me approximate sample values are, [Hibbeler, 1992, pg.347]

metal on ice 0.03-0.05wood on wood 0.3-0.7leather on wood 0.2-0.5leather on metal 0.3-0.6aluminum on aluminum 1.1-1.7

ASIDE: this equation is more correctly written.

FS µSN≤

• The basic assumptions that we will use are,1. the maximum friction force is proportional to the normal force2. the maximum friction force is not proportional to the area of contact3. the static friction force is always higher than the dynamic friction force4. the kinetic friction force is independant of velocity

• A couple of the major applications for friction calculations is the determination if an object will slip or tip. The following problem shows a typical application, ([Hibbeler, 1992], prob 8-8, pg. )

F

N

FN

FS

F

FS

F=µsN

FS=µsN

static dynamic

µkN

rubbing slowly

high speed rubbing

2 ft

4 ft

4 ft

C

FG

θ

The car pictured is parked on an extreme slope, and we want to determine at what angle the car will tip over, or begin to slip if the coeffi-cient of friction is 0.4, and the mass of the car is 4000 lb.

1. First, for tipping, we want to find the angle that would place the centre of mass beyond one of the wheels. In other words, if the gravity vector does not pass between the wheels, then the car will tip over.

2 ft

4 ft

4 ft

C

FG

θ

θ 4ft

4ft( )22ft( )2

+---------------------------------------

asin 63°= =

• Consider the simple tip/slip problem below,

wm_demos/stat0005.wm

• The general approach to slip-tip problem is,1. Find the center of gravity for the object.2. Determine which corner the object is most likely to tip (as if the corner is a pin joint).

Sum the moments about the corner. If the sum of moments is equal to zero to block is about to tip. If not equal to zero look at the resulting moment to see if it will cause motion about the corner.

3. Find the component of the gravity and any other non-friction forces acting perpendiculr to the surface of contact. Find the components of applied forces acting parallel to the plane of contact.

4. Compare the actual parallel compoent to the maximum friction force possible. The the resultant is larger than the maximum the block will slip.

2. Next for slipping, we want to find the angle at which the component of the gravity force pulling the car downhill overcomes the maximum friction force.

FS µSFR 0.4FR= =FS

FR

θ

FGFS

FR

FG

FR FG θcos=

FS FG θsin=

FG∴ θsin 0.4 FG θcos( )=

θsinθcos

------------∴ 0.4 θtan= =

θ∴ 22°=

3. Based on the two numbers we can see that the lesser angle is for slipping, therefore the car will slip (at 21.8°) before it will tip (at 63.4°).

NOTE: the slip angle (or angle of fric-tion) can be found using,θ µSatan=

*************** More friction examples

8.1.1 Practice Problems1. We conduct an experiment using the 10 kg. block below on a slope that is being slowly tilted.

The block tips over at 20°, and then stops moving, but then it starts to slip at 40°. What is the height ‘h’, and the coefficient of friction?

8.1.2 References



8.2 APPLICATIONS OF FRICTION

• When dealing with these problems the direction of friction forces must be assigned with care. If the directions are selected backwards, the solutions will be incorrect.

• Before assigning friction forces the impending motion should be analyzed. To do this think of the possible cases that might cause the bodies to start moving - do not assume that all friction

w = 50cm

h

θ

Block

Slope

Tilting

ANS.coeff. = 0.84h = 137 cm

surfaces must go into motion. At times this approach may mean that multiple solutions will have to be done to solve a problem.

• The general methods to be followed in these problems involves,1. Examine the problem to determine impending motion for each individual object, and

the overall system. There may be one or more possible cases, each will require a separate solution.

2. Based upon the assumed motion at the points of contact, drawn on friction forces that oppose the motion. Also draw on normal forces.

3. Solve the problem using normal statics (but avoid using sums of moments for friction forces when they don’t act on a clear point).

4. Examine the solution (and compare to others) for anomalies such as normal forces that separate friction surfaces. This will help detemine problems, and to eliminate unreasonable solutions.

8.2.1 Wedges

• Wedges are a useful engineering tool, and the approach used for wedges also finds its way into other engineering applications.

• A good rule to stick to is that when a wedge is in use, the forces on the faces will both be in the same direction. That is either towards, or away from the point of the wedge.

• When solving friction problems we look for friction that is about to let go and start slipping. Keep in mind that not all surfaces will slip, this should be verified after the solution. For all surfaces that slip the friction force will be at the maximum value.

• The example below shows how to deal with a multiple wedge problem. ([Hibbeler, 1992], prob 8-55, pg. )

P

FS1

FS2

N1

N2

P FS1

FS2

N1

N2Pushing the wedge in Pulling the wedge out

µAB 0.1=15°

15°

3000 lb

P

A

B

C

D

The two wedges are stacked as shown, and a load is applied. What is the minimum force P required to pull the bottom wedge out?

µAC 0.2=

µBD 0.2=

FSBD

1. These types of problems involve a large number of forces, and should not be attempted without FBDs. So, the first step is to draw FBDs of one of the wedges. The first wedge to be considered is B. When the bottom wedge slips out, B will push against the wall, and slide down, but A will try to pull it away from the wall.

FBD B:

15°

3000 lb

B FRBD

FSAB

FRAB

Fx∑ FSAB15°cos– FRAB

15°sin FRBD–+ 0= =+

µABFRAB15°cos–∴ FRAB

15°sin+ FRBD=

Fy∑ 3000lb– FSBDFSAB

15°( ) FRAB15°cos+sin+ + 0= =+

µBD µABFRAB15°cos– FRAB

15°sin+( ) FSAB15°( ) FRAB

15°cos+sin+∴ 3000lb=

FRAB∴ 3000lb

µABµBD 15°cos– µBD 15°sin+ µAB 15° 15°cos+sin+------------------------------------------------------------------------------------------------------------------------------------ 2929lb= =

FSAB∴ 293lb=


8.2.1.1 - Practice Problems

1. A vertical force, F, of 500N acts at one end of a bracket while a wedge is pushed against the other end, as depicted in the diagram below. Given that the weight of the wedge and the bracket can be neglected and that the coefficient of static friction for the contacting surfaces of

FSAC

2. Now that we know the force that wedge B applies to A, we are ready to begin finding the forces on wedge A. We can begin with a FBD.

FBD A:

15°

PA

FRAC

FSAB

FRAB

Fx∑ P– FSACFSAB

15°cos FRAB15°( )sin–+ + 0= =+

Fy∑ FSAB15°sin– FRAB

15°cos– FRAC+ 0= =+

FRAC∴ 2905lb=

P∴ 106lb=

FSAC∴ 581lb=

the wedge is 0.2, determine the horizontal force P required to push the wedge.

2. The angled bar below has two forces applied that are tending to push it in a counterclockwise direction. These forces are resisted by a wedge that is kept in place by friction (the coefficient of friction is 0.20). determine the force P that is required to pull the block out.


3. We are designing a firing mechanism for a new gun that uses two identical rails that are pressed together to accelerate a projectile. In the figure below we see the two rails at an angle before firing begins. When firing begins, the force ‘F’ will be applied, overcoming the coefficient of friction of 0.05. The length of the projectile is negligible, but it has a wedge shape that matches the rails before firing. What is the initial force ‘F’ that must be applied to the rails before the shot begins to move?

300mm

40mm

40mm

500mm

15°P

A

BC

D

500N

ANS. 216N

3

45

A

B

C

6m 6m

3m

1m

8kN

15kN

P

1m15°

(ans. 6.55KN)

4. What force P will have to be applied to the wedge A to force the blocks B and C apart? You can assume that the coefficient of friction is 0.4 at all points of contact.

8.2.1.2 - References

F

F

projectile motion

A

B

3”

1” 0.9”

6”

projectile

1”

P

75° 75°

A

B C

D

1 MN1 MN

ANS. P = 0.854 MN


8.2.2 Belt Friction

• Belts are a common tool for transmission of forces, motions and velocities.

• If we have a flat belt, it primarily depends on friction to hold it in place.

• The basic rules of static friction still apply for local friction between the belt and the drum, but over the length of the belt the effective normal force changes.

• If we consider one element of the belt we can see an element of friction and a differential of ten-sion.

∆θ2

-------

∆θ2

------- ∆θ2

-------

∆θ2

-------

T ∆T+T∆FS∆FN

∆FS µS∆FN=

Fx∑ T∆θ2

-------cos – T

∆θ2

-------cos ∆T

∆θ2

-------cos ∆FS–+ + 0= =

∆T∆θ2

-------cos∴ ∆ FS=

+

Fy∑ T∆θ2

-------sin– T∆θ2

-------sin– ∆T∆θ2

-------sin– ∆FN+ 0= =+

∆FN∴ 2T∆θ2

-------sin ∆T∆θ2

-------sin+=

∆FS µS∆FN=

Next combine the equations to get a single expression,

First, find the equations for forces acting on the belt element,

∆T∆θ2

-------cos µS 2T∆θ2

-------sin ∆T∆θ2

-------sin+ =∴

Now find the differential form of the equation,

∆T∆θ2

-------cos µS2T∆θ2

-------sin µS∆T∆θ2

-------sin+=∆θ 0→lim

dT∴ µ S2Tdθ2

------ µSdTdθ2

------+ µSdTdθ2

------= =

0

dTT

------ µSdθ=

• An example of belt friction is given below (8.123, pg. 40, Beer and Johnston). In this problem the upper drum is moving slowly (this means the belt sticks with static friction), the lower drum allows the belt to slide (the belt slides with dynamic friction). We need to find the force W that will balance the 150lb load on the other side.

1T---

TdT1

T2

∫ µS θdθ1

θ2

∫=

Tln[ ] T1

T2∴ µ Sθ[ ] θ1

θ2=

T2ln T1ln–∴ µ Sθ2 µSθ1–=

T2

T1-----

ln∴ µ S θ2 θ1–( )=

T2

T1-----∴ e

µS θ2 θ1–( )=

• V-belts use the same principle as flat belts, except the friction is increased by the angle of the sides.

θA

θB

θ

8”

Both drums have a radius of 2”.

150lb W

T1T2

µS 0.3=

µk 0.25=

θ 24---

asin 30° 0.524rad= = =

First we can find the various angles,

θA 0.524rad=

θB π 2 0.524( )+ 4.19rad= =

Next we can develop the equations for relative tensions in cables,

T1

W----- e

µkθA=

T2

T1----- e

µsθB=

150lbT2

-------------- eµkθA=

Finally, we combine the equations into a single expression,

W150lb-------------- W

T1-----

T2

T1-----

T2

150lb--------------

eµkθA( )

1–e

µsθB( ) eµkθA( )

1–= =

W150lb-------------- e

µsθB 2µkθA–( )=∴

W 410lb=∴


1. A belt is wrapped about drums A and B, both have a radius of 3”. How much force F will have to be applied to lift a weight of 1 lb. if the drums are not moving, and the coefficient of static fric-tion is 0.2?

α

T2

T1----- e

µSθα2---sin

-----------

=

BELT

PULLEY

1 lb

A

B

F

ANS.2.57lb.

9. FORCES

9.1 INTRODUCTION

• WHAT? - We look at mechanical structures, and determine the distribution of forces and moments.

• WHY? - a fundamental subject for every form of Mechanical Engineering (and every other branch of engineering that has ever existed.

[picture]

• The difference between statics and dynamics, in brief,Statics - does nothing, just sits thereDynamics - moving things

[working model file]

• Consider some of the applications of statics design techniques,[picture][picture][picture]

e.g., an electric transmission tower

When the weight of the transmission line (a force) is applied, how much force does each part (beam) of the tower carry. How much support is needed on the ground

If we do the analysis we can then determine how large the beams must be, but this will be taught in Mechanics of Materials.

media/egr20935.jpg

media/egr20921.jpg

media/egr20932.jpg

media/egr20926.jpg

media/EX1.WM

9.2 SOME BASIC CONCEPTS

• Mass and force - A mass can exert forces through gravity and other effects. Forces can also be exerted by other phenomenon, such as magnetism.

• We can emphasize the relationship between mass as an absolute and gravity as a local. The effects of gravity are dealt with as forces in most statics problems.

• Force has magnitude and direction. Therefore it is well suited to vectors.

• many forces can also operate on the same object, we can replace these with equivalent forces, called RESULTANTS.

g 9.81 NKg-------

=

F

MassM = 10kg

(This gravity vector is the averagefor ground level on earth)

F gM=

F 9.81 NKg-------

10 Kg( ) 9.81 10( ) N Kg×Kg

----------------- 98.1N= = =

Consider, the same mass on the moon,

g 1 NKg-------

???=

F 1 NKg-------

10 Kg( ) 1 10( ) N Kg×Kg

----------------- 10N= = =

*Note: the same mass is exerting a different force. Please recall from basic physics that the mass is a grouping of matter. The force is the attraction it feels when it approaches another body. Also beware, when using the Imperial system of units, gravity is often measured in slugs, as a result you must be VERY careful with units.

FGM1M2

r2

-------------------- M1g= = g∴GM2

r2

------------=

g 9.78 NKg-------

= (At the equator)

g 9.83 NKg-------

= (At the north pole)

• We have both action and reaction forces as well. As we apply ACTION forces, there are forces that will resist, these are called REACTIONS.

• Some approximations,- we are pretending the forces are applied at points, but in reality a force must be distrib-

uted,

- we generally assume there are no deflections. This is known as the RIGID BODY assumption.

- we often use PARTICLE approximations that assume bodies have no size. This simpli-fies calculations significantly.

- TRANSMISSIBILITY - a force can be moved along a line of action.- Parallelogram law - a method for adding two forces to get a resultant vector.- Forces that are the result of mechanical contact are called surface forces, and are trans-

mitted by pressure on the surface of an object.- Body forces are the result of Physics (read magic) and will pull on all of the mass of the

object, such as gravity, electrostatic, inertia, magnetic, etc.

9.3 VECTOR VS. SCALAR QUANTITIES

• definitions,

FP

Fg

each part of the thread on the screw will trans-mit a bit of the force, and each will be a dif-ferent amount (a very hard problem that is dealt with by finite element methods).

F

• Parallelogram Law - Recall that vectors can be added or subtracted. This is a variation of the tri-angle law. In both cases we are putting vectors head to tail. These methods favor drafting solu-tions to problems that are not really necessary with calculators, but they are still very useful for understanding.

9.4 MATH REVIEW

VECTORS - a magnitude and a direction

SCALARS - a simple quantity (no direction)

e.g. gravity is a vector

e.g. mass is 10 kg

9.81 NKg-------

10 kg

A

B

R=A+B A

B

R=A+B

A

B

R=A-B

-B

• Aside: recall the following terms when discussing vectors and algebra,- head/tail OR terminus/origin

- line of action

- fixed/sliding/free vectors

- colinear

- coplanar

- concurrent

- resultant

commutative/associative

- sine law (law of sines)

- cosine law (law of cosines)

- basic sin, cos, tan relationships - in triangle

- orthogonal - cartesian

- pythagorean formula

- unit vectors

right/left handed coordinates - coordinate axes and positive rotations

- cross products

- dot products

- identity matrix

- matrix multiplication, addition, subtraction, division with matrices/scalars

- matrix determinants

matrix inversion

- matrix transpose

- equation solutions in matrix form: gauss-jordan row reduction; cramer’s rule

- parametric equations

- plug-and-chug solutions: trial and error, Newton-Raphson

- inequalities: greater than, less than, etc.

- differential calculus

integration single and double, and selecting elements rectangular, triangular, cylindrical

- developing linear equations

9.5 RECTANGULAR FORM OF VECTORS

• Vectors can be added to get resultant forces in vector (rectangular component) form.

READ

2.6-2.7

PROBLEMS

SUGGESTED

67, 70

REQUIRED

64(MC)

9.6 POLAR FORM OF VECTORS

• We can also represent the same forces as scalar magnitudes, and direction,

e.g.

x

y

oil tanker

2 tug boats

F1 = (5000N, 5000N)

FR

F2 = (14000N, 0N)

14000N 5000N

5000NFR

FR F1 F2+ 5000N 5000N,( ) 14000N 0N,( )+ 19000N 5000N,( )= = =

rectangularnotation

F Fx Fy,( )= FR F1 F2+ F1xF2x

+( ) F1yF2y

+( ),[ ]= =

e.g.

x

y

oil tanker

FR 19000N( )25000N( )2

+ 386 106N

2× 19646N= = =

θFR

5000N19000N-------------------

atan 14.7°= =

θFR

θF1

F1

F2

FR

F1 5000N( )25000N( )2

+ 50 106N

2× 7071N= = =

θF1

5000N5000N----------------

atan 45°= =

F2 14000N( )20N( )2

+ 14000N= =

θFR

0N14000N-------------------

atan 0°= =

19646N 14.7°∠polar notation

7071N 45°∠

14000N 0°∠

To add the vectors in polar form we can,

1) Convert to rectangular form

9.6.1 Cartesian Vector Notation

• This is like the rectangular form, except that i, j & k are used as placeholders.

FR

F2

F1

180°-45°=135°

F2 14000N=

F1 7071N=

FR2

F12

F22

2 F1 F2 135°cos–+=

FR∴ 70712

140002

2 7071( ) 14000( ) 135°cos–+ 19646N= =

Given,

We can find the magnitude of FR, using the cosine law

θFR

We can also find the angle of FR, using the sine law,

F1

θFRsin----------------

FR

135°sin-------------------=

θFR∴

F1 135°sin

FR-----------------------------

asin 7071 135°sin14000

-------------------------------- asin 14.7°= = =

19646N 14.7°∠

2) Use trigonometry

x

y

oil tanker

2 tug boats

F1 = (5000N, 5000N)

FR

F2 = (14000N, 0N)

we can also write the vectors in the form,

F1 5000i 5000j+( )N=

F2 14000i( )N=

FR F1 F2+ 5000Ni 5000Nj 14000Ni+ += =

FR∴ 19000Ni 5000Nj+ 19000i 5000j+( )N= =cartesian vector notation

i unit vector along x axis=

j unit vector along y axis=

k unit vector along z axis=( )

F Fxi Fyj+= where Fx, Fy are scalars

FR F1 F2+ i F1xF2x

+( ) j F1yF2y

+( )+= =

Note: The i, j & k values will be underlined to minimize the potential for confusion. Also, notice that I am keeping the units in the calculations, as if they are variables. I will do this more for illustrative purposes, but this method will reduce the chance for unit based calculation mistakes.

9.6.2 Scalar Notation

• We can represent forces as simple scalar values (not vectors). But, we must still remember that theses are still vectors. For scalar values we need to take care to define their direction for the problem, or for specific values. One common way to do this is to define positive ‘x-y’ axes, and then refer to ‘x’ and ‘y’ components. From this a positive ‘x’ component implies one direction - a negative component implies the opposite direction.

• For example,

• Scalar notation is often made obvious by using ‘x’, and ‘y’, or similar subscripts.

• direction, location, signs, etc. are all defined by convention, and very compact mathematical methods can be used.

• These problems can also be solved using cosine and sine law force additions on force triangles. Considering the last example,

F1

F2

FR

70° 20°

x

y

F1 = 5NF2 = 2N

where,

FRxF1 20°cos F2 180° 70°–( )cos+ 4.0N= =

FRyF1 20°sin F2 180° 70°–( )sin+ 3.6N= =

FRx4.0N=

FRy3.6N=

Scalar notation

• Consider the large pendulum below as an example where a force triangle could be used to find the tensions in the cables.

[picture][picture]

FR

F2=2N

F1=5N

20°

φ

70°

use cosine law (we could also use pythagoras) to find FR,

FR2

F12

F22

2F1F2 90°cos–+=

FR2

52

22

2 5( ) 2( ) 0( )–+=∴ 29=

Next, you can try using the sine law to find the direction angles,

media/egr20928.jpg

media/egr20927.jpg

• NOTE: all of the vectors added are position vectors. If we are to consider rotation vectors, they cannot be simply added. You must consider alternatives such as Euler angles, etc.


9.6.3 Unit Vector Representation

• The unit vector representation follows.

α

β

A

B

C

TA

TB

u

v

Find an equation that relates all of the tensions and angles so that the tug boats A and B to over-come the drag force.

Fd

• The unit vector representation can be developed from basics.

F F λxi λyj+( )=

Here the vector is defined by a magnitude and direction. The magnitude is a scalar and thedirection is a unit vector.

λxi λyj+ =

λx2 λy

2+ =

Fx F θxcos=

F Fxi Fyj+=

Fy F θycos=

F F θxcos( )i F θycos( )j+=

F F θxcos i θycos j+( )=

F F λxi λyj+( )=

λx θxcos=

λy θycos=

λx2 λy

2+ 1 θxcos( )2 θycos( )2

+= =

More important in 3D

READ

SLI 2.1-2.3

PROBLEMS

SUGGESTED

2, 4, 13

REQUIRED

15(WM), 16(MC)

9.6.4 3D Vectors

• We will use right-handed coordinates

x

y

z

z is upy is to the rightx is towards you

i

j

k

** How to remember: on the right hand the thumb points along the positive z-axis, and when you curl your fingers into a fist it should push the positive x-axis 90° into the positive y-axis. Try this with the coordinate axes above. If you need to push the x-axis 270° you have left handed coordinates.

ASIDE: these axes could also be shown in different orientation, but the right hand rule will still apply.

x

y

z

NOTE: 3D axis are typically drawn in a manner similar to the oblique views used in traditional drafting. The axis drawn horizontal and vertical lie on the plane of the page, while the axis on an angle is

- out of the page if it points downward- into the page if it points upwards

ASIDE: Most 3D axes follow drafting conventions.

Isometric Cavalier, etc.

• Consider the following example,

θα

θγ

θβ

x

y

z

F 3i 4j 5k+ +( )N=

F 3N( )24N( )2

5N( )2+ + 7.07N= =

1) Find the magnitude of the vector (using the 3D pythagorean theorem),

2) Find a unit vector representation,

F 7.07N7.07N-------------- 3Ni 4Nj 5Nk+ +( ) 7.07N 3

7.07----------i 4

7.07----------j 5

7.07----------k+ +

= =

F∴ 7.07N 0.42i 0.57j 0.71k+ +( )=

A unit vector of length 0.422

0.572

0.712

+ + 1.0= =

We can find the angles between the axis, and the force, using direction cosines

θαcosFx

F------ 3

7.07---------- 0.42= = = θα∴ 65.2°=

θβcosFy

F------ 4

7.07---------- 0.57= = = θβ∴ 55.2°=

θγcosFz

F------ 5

7.07---------- 0.71= = = θγ∴ 44.8°=

NOTE: We will sometimes use θx, θy, θz as a replacement for θα, θβ, θγ respectively, these are often referred to as α, β, γ in other places.

θα 65.2°= θβ 55.2°= θγ 44.8°=F 7.07N=

(direction) cosine notation

• To emphasize the main relations,

ASIDE:

F F λxi λyj λ zk+ +( )=

1 λx2 λy

2 λz2

+ +=

1 λx2 λy

2 λ z2

+ +=

1Fx

F------

2 Fy

F------

2 Fz

F------

2+ + θαcos( )2 θβcos( )2 θγcos( )2

+ += =

you can use this to checkresults.

1 θαcos( )2 θβcos( )2 θγcos( )2+ +=

F F θαcos i θβcos j θγcos k+ +( )=

F Fx2

Fy2

Fz2

+ +=

θαFx

F------

acos= θβFy

F------

acos= θγFz

F------

acos=

Aside - for the purposes of using computer algebra tools we will use another type of nota-tion called matrix notation. The basic principles are shown below,

A

Ax

Ay

Az

Axi Ayj Azk+ += =

matrix notation

cartesian notation

i1

0

0

= j0

1

0

= k0

0

1

=

aAA------

Ax

A------

Ay

A------

Az

A------

= = unit vector

Note: lower case for unit vector

**************** Solve previous problem with Mathcad *******

• An example to illustrate this technique is, ([Hibbeler, 1992], prob. 2-56, pg. 49)

x

y

z

120°

60°

45°

45°

60°

F1 300N=

F2 500N=

Find the resultant force for the two forces shown in vector projection (F1) and cosine notation (F2).

NOTE: this problem has two different, but common ways of representing vec-tors. F2 is represented using direction cosines, but F1 is represented by a technique known as vector projection. The common way to deal with this representation is to find the force component that lies on the projection plane, and then use that for some of the calculations.

NOTE: Forces represented with vector projection are similar to forces mea-sured with azimuth and elevation. Surveyors commonly use these angles to specify direction.

For F2:F2 F2 60°icos 45°jcos 120°kcos+ +( )=

F2∴ 500N 60°icos 45°jcos 120°kcos+ +( )=

F2∴ 250i 354j 250k–+( )N=

For F1:

F1xyF1 60°cos 150N= =

First, find the component of the force projected onto the x-y plane.

Now, find the components of the force in the x-y plane,

F1xF– 1xy

45°sin 106.1– N= =

F1yF1xy

45°cos 106.1N= =

Finally, find the z-component,

F1zF1 90° 60°–( )cos 259.8N= =

And, put the forces together into a vector form,

F1 106i– 106j 260k+ +( )N=


Finally add the two forces,

FR F1 F2+ 250i 354j 250k– 106i– 106j 260k+ + +( )N= =

FR∴ 144i 460j 10k+ +( )N=

FR 1442

4602

102

+ + 482N= =

And find the magnitude, and angles of the resultant,

θαFRx

FR---------

acos 72.6°= =

θβFRy

FR---------

acos 17.5°= =

θγFRz

FR---------

acos 88.8°= =

NOTE: One significant figure has been lost.

50°

50°

115°

30°

20°

x

y

z

(10,10,10)

F1 10N=

F2 20N=

F3 10lb=

a) i(41.5) + j(37.8) + k(25.3) N

b) 61.5N

c) θx = 47.6°, θy = 52.1°, θz = 65.7°

d) Fxy = 56.1N

Given the system of vectors pictured, a) give the resultant force using cartesian notation b) find the magnitude of the resultant force in metric units. c) Then then using cosine angles, and finally d) projected onto the x-y plane.

******************* Solve using Mathcad ********************

• Consider the case below, where we know positions, and forces, but we want to find the resultant force,

Boeing Balloon Co.

xy

z

(-2, 3, 0)

(-15, 8, 0)

(-4, 13, 0)

(-10, 10, 10)

F1

F2

F3

If we have an inflated hot air balloon that has a buoyancy force keeping it aloft, and three people of the ground resisting it’s rise by holding ropes. The position of the ends of the ropes is shown. Find the resultant force on the balloon. What will happen?

FB

F1 200N=

F2 300N=

F3 500N=

FB 1000N=

we know that,Basket

to start, we should recognize that the ropes are actually force vectors, and the direction of the rope is the direction of the force vectors. And, knowing the endpoints of the ropes, allows us to calculate direction vectors, and from this we can find force vec-tors for each rope.

Lets first label the points,

P1 = (-2, 3, 0)udP2 = (-15, 8, 0)udP3 = (-4, 13, 0)udPB = (-10, 10, 10)ud

Next, find the relative displacements between the people, and the bottom of the balloon,

r1 B⁄ P1 PB– 2– 3 0, ,( )ud 10– 10 10, ,( )ud– 8 7– 10–, ,( )ud= = =

r2 B⁄ P2 PB– 15– 8 0, ,( )ud 10– 10 10, ,( )ud– 5– 2– 10–, ,( )ud= = =

r3 B⁄ P3 PB– 4– 13 0, ,( )ud 10– 10 10, ,( )ud– 6 3 10–, ,( )ud= = =

Now, use the vectors to find the forces, by multiplying the force magnitude by

F1 F1 r1 B⁄ F1P1 PB–

P1PB------------------ 200N

8i 7j– 10k–( )

82

72

102

+ +------------------------------------

110i 96j– 137k–( )N= = = =

F2 300N5– i 2j– 10k–( )

52

22

102

+ +---------------------------------------

132– i 53j– 264k–( )N= =

F3 500N6i 3j 10k–+( )

62

32

102

+ +------------------------------------

249i 125j 415k–+( )N= =

We can also write out the equation for the buoyancy force,

FB 1000k( )N=

the direction unit vector,

NOTE: the distance units are missing in this prob-lem, so ‘ud’ is used. These cancel out

********************** Mathcad example *********************

• As a practical example of where 3D vectors might be required, consider the power line pole. It uses a tension cable anchored in the ground to resist the forces exerted by the power lines.

[picture]

9.6.5 Dot (Scalar) Product

• The definition of the dot product is,

Finally, we want to sum the forces, and find the magnitude,

FR F1 F2 F3 FB+ + +=

FR∴ i 110 132– 249+( ) j 96– 53– 125+( ) k 137– 264– 415– 1000+( )+ +=

FR∴ 227i 24j– 184k+( )N=

FR 2272

242

1842

+ + 293N= =

It is obvious from the results that the balloon will move up, and out of the page as drawn because the z (or k) component is positive. The other two components suggest the balloon will fly forward and the left.

READ

SLI 2.4-2.5

PROBLEMS

SUGGESTED

48, 51, 55

REQUIRED

F1 F2⋅ F1 F2 θcos= (a scalar)

• Evaluating the dot product.

• We can use a dot product to find the angle between two vectors

• Projecting vectors.

• Consider the example below where we find the component of one vector that lies in the direction of the other vector.

i i⋅ 1=

i j⋅ 0=

i k⋅ 0=

j i⋅ 0=

j j⋅ 1=

j k⋅ 0=

k i⋅ 0=

k j⋅ 0=

k k⋅ 1=

F1 F2⋅ F1xF2x

( ) F1yF2y

( ) F1zF2z

( )+ +=

F1 2i 4j+=

F2 5i 3j+=

x

y

θcosF1 F2•F1 F2------------------=

θ∴ 2( ) 5( ) 4( ) 3( )+

22

42

+ 52

32

+-------------------------------------------

acos=

θ∴ 224.47( ) 6( )

----------------------- acos 32.5°= =

θ

θB

A

FA B⁄

FA B⁄ FA θcos=

Note the similarity to the dot product,

FA FB⋅ FA FB θcos=

if,

FB 1=

Therefore we can use the dot product if we have a unit vector for the comparison vector.

• The use for the dot product will become obvious in later sections.

F1 3i– 4j 5k+ +( )N=

V 1j 1k+=

x

y

z

We want to find the component of force F1 that projects onto the vector V. To do this we first convert V to a unit vector, if we do not, the component we find will be multiplied by the magnitude of V.

λVVV------

1j 1k+

12

12

+--------------------- 0.707j 0.707k+= = =

F1VλV F1• 0.707j 0.707k+( ) 3i– 4j 5k+ +( )N•= =

F1V∴ 0( ) 3–( ) 0.707( ) 4( ) 0.707( ) 5( )+ + 6N= = V

F1

F1V

F1vF1 θcos=

θ

ASIDE: UNIT VECTORS AND DOT PRODUCTS

Unit vectors are useful when breaking up vector magnitudes and direction. As an exam-ple consider the vector, and the displaced x-y axes shown below as x’-y’.

x

y

x’y’

45°

60°

F 10N=

If we want to find the x and y components of F relative to the x-y axis we can use the dot product.

λx 1i 0j+= (unit vector for the x-axis)

Fx λx F• 1i 0j+( ) 10 60°cos( )i 10 60°sin( )j+[ ]•= =

∴ 1( ) 10 60°cos( ) 0( ) 10 60°sin( )+ 10N 60°cos= =

This result is obvious, but consider the other obvious case where we want to project a vector onto itself,

λFFF------

10 60°icos 10 60°jsin+

10--------------------------------------------------------- 60°icos 60°jsin+= = =

Incorrect - Not using a unit vector

FF F F•=

10 60°cos( )i 10 60°sin( )j+( ) 10 60°cos( )i 10 30°cos( )j+( )•=10 60°cos( ) 10 60°cos( ) 10 60°sin( ) 10 30°cos( )+=

100 60°cos( )230cos °( )2

+( ) 100= =

Using a unit vector

FF F λF•=

10 60°cos( )i 10 60°sin( )j+( ) 60°cos( )i 30°cos( )j+( )•=

10 60°cos( ) 60°cos( ) 10 60°sin( ) 30°cos( )+=

10 60°cos( )230°cos( )2

+( ) 10= = Correct

Now consider the case where we find the component of F in the x’ direction. Again, this can be done using the dot product to project F onto a unit vector.

ux' 45°icos 45°jsin+=

Fx' F λx'• 10 60°cos( )i 10 60°sin( )j+( ) 45°cos( )i 45°sin( )j+( )•= =

10 60°cos( ) 45°cos( ) 10 60°sin( ) 45°sin( )+=

10 60° 45°coscos 60° 45°sinsin+( ) 10 60° 45°–( )cos( )= =

Here we see a few cases where the dot product has been applied to find the vector pro-jected onto a unit vector. Now finally consider the more general case,


V1

V2

θ1

θ2

x

y

V2V1

V2V1V2 θ2 θ1–( )cos=

Next, we can manipulate this expression into the dot product form,

First, by inspection, we can see that the component of V2 (projected) in the direction of V1 will be,

V2 θ1 θ2coscos θ1 θ2sinsin+( )=

V2 θ1icos θ1jsin+( ) θ2icos θ2jsin+( )•[ ]=

V2V1

V1---------

V2

V2---------• V2

V1 V2•V1 V2------------------

V1 V2•V1

------------------ V2 λV1•= = = =

Or more generally,

V2V1V2 θ2 θ1–( )cos V2

V1 V2•V1 V2------------------= =

V2∴ θ 2 θ1–( )cos V2V1 V2•V1 V2------------------=

θ2 θ1–( )cos∴V1 V2•V1 V2------------------=

*Note that the dot product also works in 3D, and similar proofs are used.

********************** Solve Using Mathcad ************************

x

y

z

F 25i 30j– 10k+{ } N=

F F1 F2+=

F1

F2

F

4”

3”

5”

A force F acts at the end of a pipe. Determine the magnitudes of the components that act perpendicular to, and along the axis of the end of the pipe. (the pipe lies in the y-z plane)

9.6.6 Summary

• the basics of statics as a topic were covered

• engineering units and calculations

• representations covered in this section were,- scalar values- vector values

rectangularpolarcartesiandirection cosinesvector projectiondirection vectors

• The dot product was shown as a way to project one vector onto another, or final angles between them.

9.6.7 Practice Problems0.

READ

SLI 2.8

PROBLEMS

SUGGESTED

78, 84

REQUIRED

93(MC) - use fig 2-92

1. Four forces act on bolt A as shown. Determine the resultant of the forces on the bolt.

2. Find the tension in cable A and B if the tension in cable C is 100N.

Cartesian Vectors Polar Vectors ResultantA=Ax i + Ay j A=|A| at Thetao R=A+B

Ax Ay Bx By |A| ThetaA |B| ThetaB Rx Ry |R| ThetaR

1.00 1.00 2.00 2.00

1.00 2.00 1.00 -2.00

-3.00 7.00 -10.00 4.00

15.00 25.00 -23.00 40.00

-2.00 3.00 -2.00 3.00

7.00 90.00 2.83 135.00

12.37 -165.96 6.40 -51.34

0.80 -6.00 6.32 -161.57

279.51 153.43 173.36 137.10

-34.00 54.33 23.00 -51.55

0.30 12.00 -2.00 7.28

12.00 5.00 17.49 30.96

F2=80N

F1=150N

F4=100N

F3=110N

30°

15°

20°

A

x

y

ANS. Fx=199N, Fy=14N

A

B

C

M=3kg

F=100N

45°

ANS. TA=100N,TB=71N

3. A disabled automobile is pulled by two ropes as shown below. If the resultant of the two forces must be 300lb, parallel to the forward roll of the car, find (a) the tension in each of the ropes, knowing that α = 30°, (b) the value of α such that the tension in rope 2 is minimum.

4. Convert between the representations given on the left, and the results requested on the right.

T1

T2

α

20°A

ANS.a) TA=b) 70°

x

y10dN

45°

Polar Rectangular

( x , y ) = ( __________ , __________ )

a)

ANS.(7.1dN,7.1dN)

x

yPolarCartesian

M = { -10i - 20j } lb.

b)

θ =

M =

ANS. 22.3lb, 243°

y

zDirection Cosines Cartesian

F = i(________)

c)

+j(________)

+k(________)

x

θy 70°=

θz 60°=F 5.0m=

ANS. (4.0m,1.7m,2.5m)

5. An F-117A stealth fighter is supposed top be flying N20°E, but a strong wind from East to West is pushing it off course. If the plane is pointed N20°E, but is actually moving N23°E, and its 22,000 lb engine is at full thrust, a) what force is the wind exerting on the plane? b) What is the answer in newtons?

y

ProjectedCartesian

L = {20i + 5j + 10k}Pa

d)

x

θxy 30°=

θz _________=L 22.9Pa=

z

θxy _________=

L __________=

θz 24°=

3°

20°

N (North)

E (East)a)

b)

Fw= 1137lb

Fw= 5070N

9.6.8 References


Soustas-Little, R.W. and Inman, D.J., Engineering Mechanics Statics, Prentice-Hall, 1997.

10. EQUILLIBRIUM

10.1 INTRODUCTION

• Put simply equilibrium describes the condition where all forces are balanced (no acceleration). Static equilibrium describes the state where all forces are balanced, and the object is not in motion.

• For an object to be in equilibrium, all the forces and moments must be balanced for,- each particle in a rigid body- each rigid body- each object made up of rigid bodies

10.2 THE BASIC EQUATIONS OF STATICS

• There are two basic balances that must exist in any statics problems.

• if the sum of the forces is not zero, then the system will undergo translation, and the problem

Equilibrium of Forces

Equilibrium of Moments (more later)

F∑ 0=

M∑ 0=

STATICS

Acceleration in Direction of Unbalanced Forces

Rotational Accelerations

F∑ 0≠

M∑ 0≠

DYNAMICS

cannot be solved with statics methods.

• if the sum of moments is not zero, then the system will undergo rotation, and the problem cannot be solved with statics methods.

• At least one of these two equations will appear in every statics problem.

• A simple example,

• Let’s consider a force balance problem, ([Hibbeler, 1992]prob 3-40, pg. )

Fx∑ F1xF2x

F3x+ + 0= =

Looking at the three ropes pullingthe ring, each rope exerts a forcewith components in the x and ydirections.

F1

Fy∑ F1yF2y

F3y+ + 0= =

F2

F3

REVIEW: what type of force representation is this?

A

B

C

1.5 ft.

1.5 ft.

3 ft.

2.5 ft.4.5 ft.

3 ft.y

z

20 lb.

x

D

a 20 lb. bucket is suspended with three ropes that are joined at a point (D). each rope is connected to an anchored hook. Hook A is on the x-z plane, B is on the x-axis, and C is on the y-z plane. Find the ten-sions in each of the ropes in Newtons.

First, find the position vectors of the points in the problem,

D 1.5 1.5 0, ,( )ft=

A 4.5 0 3, ,( )ft=

B 1.5 0 0, ,( )ft=

C 0 2.5 3, ,( )ft=

Next, find the displacement vectors of the ropes,

DC C D– 0 1.5–( )i 2.5 1.5–( )j 3 0–( )k+ + 1.5i– j 3k+ +( )ft= = =

DC 1.52

12

32

+ + 3.5ft·= =

DB B D– 1.5 1.5–( )i 0 1.5–( )j 0 0–( )k+ + 1.5j–( )ft= = =

DB 1.52

1.5ft·= =

DA A D– 4.5 1.5–( )i 0 1.5–( )j 3 0–( )k+ + 3i 1.5j– 3k+( )ft= = =

DA 32

1.52

32

+ + 4.5ft·= =

DFDB

FDC

FDA

20 lb.

FBD of Particle D Free Body Diagrams (FBD) are the essential start-ing place for mechanics problems. They define all of the forces, and direction. To draw one of these for a particle, we draw all forces applied - both action and reaction. WARNING: Do not skip this step, even when using Mathcad - paint can be used to create these rough, but complete figures.

Now, using the direction vectors, let’s find force vectors,

FDC FCDCDC-----------

1.5 FC–

3.5--------------------

iFC

3.5---------

j3 FC

3.5-------------

k+ += =

FDB FBDBDB-----------

0 FB

1.5------------

i1.5– FB

1.5--------------------

j0 FB

1.5------------

k+ += =

FDB∴ FB–( )j=

FDA FADADA-----------

3 FA

4.5------------

i1.5 FA–

4.5--------------------

j3 FA

4.5------------

k+ += =

FDA∴ 0.667 FA( )i 0.333– FA( )j 0.667 FA( )k+ +=

Finally, the mass of the bucket is,

FBUCKET 20lb–( )k 20lb 4.448N1lb

----------------- –

k 88.96N–( )k= = =

Now, having all of the force vectors, we can write out the general equation,(keep in mind because we are using vectors, we don’t need to define thepositive direction),

F∑ 0= Fx∑ Fy∑ Fz∑ 0= = =

Fx∑ 0.428 FC– 0.667 FA+ 0= =

Fy∑ 0.286 FC FB– 0.333– FA+ 0= =

Fz∑ 0.857 FC 0.667 FA 88.96N–+ 0= =

Basically there arethree equations, andthree unknowns, sowe can do a simpleparametric solution.

An algebraic solution of the three equations above leads to the results,

FA 44N=

FB 5N=

FC 69N=

FDC∴ 0.428– FC( )i 0.286 FC( )j 0.857 FC( )k+ +=


1. Two cables (AB and AC) and a force (P) act on the top of a flag pole (AD). Find the magnitude of force P required to keep the flag pole standing? Assume that cable AB is under 5 KN of ten-sion.

PROBLEM SOLVING PHILOSOPHY- NUMBER OF UNKNOWNS AND EQUATIONS:

An important constraint when solving problems is that there must be as many (or more) equations as there are unknowns. If there are more unknowns than equations, the problem is unsolvable at that point, so you should number all equations with the intention of using then later, and having to refer to them. Keep in mind that while it is nice to get numbers right away, in many prob-lems you will have to solve equations in parametric or matrix forms - trying to avoid these problems will only make life complicated.

READ

SLI 2.9-2.10

PROBLEMS

SUGGESTED

105, 110

REQUIRED

108(WM), 113(MC)

x

y

z

PA

B

C=(28.5,0,15)D

5

80°

50°O

ANS

Given,

αOA 50°=OAz 5= γOA 80°=

Find the missing direction cosine angle, and magnitudes,

1 αOAcos( )2 βOAcos( )2 γOAcos( )2+ += βOA 41.7°=∴

OA γOAcos 5= OA 28.8=∴

OAx OA αcos 18.5= =

OAy OA βcos 21.5= =

In summary, the points are,

A 18.5 21.5 5, ,( )=

B 18.5 0 0, ,( )=

C 28.5 0 15, ,( )=

D 18.5 0 5, ,( )=

FAB 5KNi 18.5 18.5–( ) j 0 21.5–( ) k 0 5–( )++

21.52

52

+--------------------------------------------------------------------------------------------

i 0( ) j 4.9–( ) k 1.1–( )++= =

Find the force vectors,

FAC TAC

i 28.5 18.5–( ) j 0 21.5–( ) k 15 5–( )++

102

21.52

102

+ +-----------------------------------------------------------------------------------------------

=

F∑ FAB FAC Pi–+ i 0( ) j 0≠( ) k 0( )+ += =

Sum forces,

∴ i 0.39TAC P–( ) j 4.9– 0.84TAC–( ) k 1.1– 0.39TAC+( )+ +=

∴ i 0.39TAC( ) j 0.84– TAC( ) k 0.39TAC( )++=

1.1– 0.39TAC+∴ 0= TAC∴ 2.82KN=

0.39TAC P–∴ 0= P∴ 1.1KN=

2. The 500 lb pendulum below is hung from beam ABC with three cables (BD, DC and DE). The cables are joined at D with a ring. What is the tension in each of the cables? (25%)

3. F1 is a projected vector with a magnitude of 15KN, the magnitude of F2 is 10KN. F2 is repre-sented with direction cosines, and lies in the negative ‘x’ direction. Find the resultant of these vec-tors in cartesian vector notation. (50%)

A B C

D

E

F

3 ft.4 ft. 3 ft.

4 ft.

7 ft.

4 ft.

6 in.

500 lb

10.2.2 References


10.3 FREE BODY DIAGRAMS (FBD)

• Up to this point we assumed very simple forces acting on very small particles.

• In reality mechanical systems have many parts, and we draw an FBD for each part.

• We should divide forces on free body diagrams into two categories,Internal - these forces act only within a free body, and cancel out, unless we are looking at

a section of a free body.External - these forces act on a free body, and they induce reaction forces. Examples are

gravity, and other free bodies.

x

y

z

F1

F2

γ 65°=

β 70°=

θxz 40°=

θy 30°=

• An example of using free body diagrams for a system is given below with a system of masses, ropes, pulleys and anchors.

M1 M2

T1

T2

T3

P1

P2

R1

Step 1: label all of the components in the system. In this case there are two masses, two pulleys, one ring, and three cables (one cable is threaded through two pulleys.

M1

P1P2

Step 2: break the system into parts. The rule of thumb here is that each FBD should have only one rigid body.

M2

R1

10.3.1 Pulleys and Springs

• Pulleys are basically a wheeled roller that a rope can roll over freely,

M1

P1P2

Step 3: Draw arrow heads on the force vectors, and label each with a variable. NOTE: it is essential that the same force shown on two different FBD’s is equal and opposite in direction.

M2

R1

T1

T1

T1T1 T1

T2

T2

T3

M2gM1g

T1 FP1FP2

ASIDE: The main advantage of free body diagrams is that we can ignore what is happening TO (and IN) OTHER rigid bodies. Although they don’t make the solu-tion easier, they do make it easier to develop equations, and it cuts the problem into smaller steps.

• A simple example of a pulley used for lifting a mass is given below,

T1 T2T1 = T2

For a perfect pulley (no mass or fric-tion) the tension on one side of a pulley will be equal to the force on the other side.

The pulley support will have a maxi-mum reaction of 2T.

60°

10 kg

60 kg

FBD Block:

10 kg

T1

Mg 10Kg 9.81 NKg-------

98.1N= =

FBD Man:

60 kg

Fy∑ T1 Mg– 0= =

T1∴ 98.1N=

+

When doing calculations with scalar values, always indicate which direction is positive. It is also very useful to maintain consistency throughout solutions.

++ ++

ASIDE:

T1 98N=

FFRICTION

+ +

30°

FR

Fx∑ FFRICTION– T1 30°cos+ 0= =+

FFRICTION∴ 85N=

• Springs are a very important engineering tool,

FBD Pulley:

T1 98N=

T1 98N=

FRθR

60°

Fx∑ T1 60°sin– FR θRsin+ 0= =

Fy∑ T1 T1 60°( )cos– FR θRcos+– 0= =

+

+

FR∴T1 60°sin

θRsin----------------------

T1 T1 60°cos+

θRcos------------------------------------= =

60°sin1 60°cos+---------------------------∴

θRsin

θRcos--------------- θRtan= =

θRtan∴ 0.8661 0.5+----------------=

θR∴ 30°=

98 60°sin FR 30°sin=

FR∴ 170N=

• A sample problem that uses springs is given below, ([Hibbeler, 1992] prob 3-16, pg. )

M = 10 kg

k = 20 N/m

FBD Mass:

MgFR1

Fy∑ FR1Mg– FR1

10kg 9.81 Nkg------

– 0= = =

FR1∴ 98N=

+

FBD Spring:

FR1

FR2

Fy∑ FR2FR1

– FR298N– 0= = =

+

FR2∴ 98N=

But, the spring will be compressed, as governed by Hookes law,

x

F kx= Hookes Law

x∴FR1

k-------- 98N

20Nm----

---------- 4.9m= = =

FR1

k spring constant (N/m) or (lb/in)=

• Consider the problem below,

F

1.5 m

6.0 m

k = 500 N/m

k = 500 N/mA rubber cord that is originally 6.0m long is stretched as shown in the dia-gram. We can model the elasticity of the cable as two springs to either side of the pulley. Find the force F for the deformation shown.

First select variables for the cable length,

l0 6m the undeformed cable length= =

l1 the deformed cable length=

l1

2---- 6.0m

2------------

21.5m( )2

+ 3.35m= =

l1 6.70m=

T ∆l2-----

kl1 l0–

2--------------

k 6.70m 6m–2

---------------------------- 500N

m---- 175N= = = =

Next, find the tension in the cord as a result of the change in length,

Next, try a FBD of the pulley,

F

T

T

Fx∑ F– 1.5l1

2----

---------

T 1.5l1

2----

---------

T+ + 0= =+

F∴ 6Tl1------ 6 175N( )

6.70m--------------------- 160N= = =

NOTE: this is equivalent to

************************ Solve using Mathcad **********************

M=10kgM=10kg M=10kg

4.0m

1.0m1.5m

h2

h2= 2.4m

Given the three masses below, connected by a cable through three pulleys, determine the final resting height (h2) for the centre mass. Assume the pulleys are very small and static.

10.3.2 Summary

• equilibrium of forces and moments• free body diagrams (FBD’s)

pulleysspringsanchorscablesmassesrings


1. Determine the reactions at B, C and D.

2. Four masses are suspended by cables that are supported by pulleys. The frictionless pulleys are mounted on a flat ceiling. Each of the pulleys is a distance of 2m from the others. Determine the height of the center mass.

READ

SLI 2.11

PROBLEMS

SUGGESTED

114, 115

REQUIRED

115(WM)

200N

60deg 60deg

A

B C

D

Fd = 200NFb = Fc = 231N

(ans. 1.03m)

3 For the mass pulley system on the left, a) draw the force triangle on the right, with all angles and magnitudes indicated, then b) find the mass if the tension in BC is 70N.

100lb

100lb

100lb200lb

h?

2m

2m

2m

M

10 Kg

3.0 ft

2.0 ft

θA

B

C

a)

Find M = ______________

b)

10.4.1 References



11. STRESS

11.1 INTRODUCTION

• Stress is the force per unit area. This is because the geometry of the parts are not fixed, and we must account for the size of the cross section that is resisting a force or moment.

• So far we have dealt with tensions and compressions in members somewhat simply. If we con-sider a simple member as shown below, we use the force divided by the area to give us the force per unit area.

• The normal units for stress are given below.

P

P

Aσave

PA---=

where,

σave average normal stress (psi, ksi, Pa)=

A cross sectional area=P applied load=

NOTE: This assumes that the material deforms uniformly across the section and each section of material pulls the same amount.

Note: The area A is perpendicular to the force Pfor normal stress.

• The material at the section of interest acts like numerous supports, each exerting internal forces. As a result the stress is statically indeterminate, but a reasonable estimate is made by assuming uniform distribution. In the internal forces are not constant over the cross section of the part.

• For two dimensional objects the internal forces of interest are shear force, normal force and bending moment.

• Using stress we can calculate deformations (strain) and failure loads. This is done by referring to tables of values for particular materials.

PA--- lb

in2

------- psi= = 103psi ksi=

PA--- N

m2

------ Pa= = 106Pa MPa=

109Pa GPa=

Shear Stress

NormalStress

bendingmoment

bendingmoment

READ

GT 1.1-1.2

PROBLEMS

SUGGESTED

2-1, 2-2, 2-5

REQUIRED

11.2 TYPES OF STRESS

• Consider an example of using a pair of scissors (“shears”) to cut a piece of metal,

• Consider the simple example given below,

P

P

A

τAVEPA---=

where,

τAVE the average shearing stress (psi, ksi, PA)=

P The applied load=

A The area of the shear=

SHEAR STRESS

NOTE: like axial stress, shear stress is also not uniform in distribution.

NOTE: for shear stress the area A is parallel to the force ‘P’. Remember that normal stress was perpendicular.

• By convention we will always define tension as a positive stress, and compression as a negative stress.

P

P

Given the 1” diameter bar to the left, what is the inter-nal (normal) stress if the load P is 100KN? If the bar is made of gray cast iron, what is the difference between the ultimate tensile strength and the induced stress?

σ 0> positive stress is tension

σ 0< negative stress is compression

• Single Shear - These calculations are commonly used when examining rivets and bolts. Con-sider the example below with the bolt in single shear.

UNC-8-1”

P1

P1

P2

P2

dia. = 1” A∴ πd2

4---------=

σP2

A------= τ

P1

A------=

• Double Shear - Now consider a bolt that is exposed to double shear. Below we can see a section view of a pin through two members.

• Bearing Stress - We may also consider how the force on a pin effects the surrounding material. In the figure below we are finding the bearing stress. In this case the bearing stress will be a normal stress.

• The bearing stress as shown only needs to be considered when the member is in tension.

PP

PP/2

P/2

τave

P2---

A--------- P

2A-------= =

τave

τave

PA1 + A2

σ PA1 A2+------------------=

11.3 STRESS ANALYSIS

• Let’s consider an example problem [1.15, pg. 16, Beer and Johnston]. For this problem we will examine all links for maximum stresses.

30°

45°

1.2 kip 1.2 kip

A

B

C

0.5”0.5”

0.5”

1.6”

1.6”

* all pins are 0.6” diameter

Free B ody D iagram P in a t B

First, find the tension/compression in each member.

TAB

TBC

60°

45°

1.2+1.2kip

B

FBD B:

Fx∑ TAB 60° TBC 45°sin–sin– 0= =

TAB∴ 0.816TBC–=

+

Fy∑ TAB 60° TBC 45° 2.4kip–cos–cos 0= =+

TBC∴ 2.15kip–=

TAB∴ 1.76kip=

Now consider the pin A,

τA

TAB

2---------

A--------------- 1.76kip

2 0.283in2( )

----------------------------- 3.11ksi= = =

τA τATAB

TAB

2---------

TAB

2---------

Next consider the member AB at the pins A & B,

AABAAABB

1.6in 0.6in–( )0.5in 0.5in2

= = =

σABAσABB

TAB

AABA

----------- 1.76kip

0.5in2

------------------ 3.52ksi= = = =

Now consider the stress in the centre of AB,

AAB 1.6in( )0.5in 0.8in2

= =

σAB

TAB

AAB--------- 1.76kip

0.8in2

------------------ 2.2ksi= = =

Consider the pin at B now, we can do this from left to right,

A 0.283in2

=

τB1τB2 τB2

τB1

1.2kip

τB1A τB1

1.2kipA

--------------- 1.2kip

0.283in2

--------------------- 4.24ksi= = =

1.2kip

τB2A

TCB

2---------

TCB

2---------

1.2kip

τB2A

45°

Fx∑TCB

2--------- 45°sin– 0.76kip= =

Fy∑ 1.2kip–TCB

2--------- 45°cos– 0.44kip= =

τB2

0.762

0.442

+0.283

------------------------------------ 0.88ksi= =

Next consider the shear on pin C,

TCB

TCB/2 TCB/2

τ

TCB

2---------

A---------------

2.152

----------–

0.283------------------- 3.80ksi= = =

Now, lets consider member BC. At pin C the stress will be,

ACBc1.6in 0.6in–( ) 0.5in( ) 0.5in

2= =

σBCc

TCB

ACBc

----------- 2.15kip–

0.5in2

---------------------- 3.30ksi–= = =

Likewise for the support at B,

ACBB2 1.6in 0.6in–( ) 0.5in( ) 1.0in

2= =

σBCB

TCB

ACBBc

------------- 2.15kip–

1.0in2

---------------------- 2.15ksi–= = =

Finally, the stress in the member BC will be,

ACB 1.6in 0.5in( ) 0.8in2

= =

σBC

TCB

ACB--------- 2.15kip–

0.8in2

---------------------- 2.69ksi–= = =

ASIDE: If doing a real design we could then use the maximum stress values to select materials, or to select components to redesign.

Note: Because the member is in com-pression, the reac-tions at the supports will not be a prob-lem.

11.4 STRAIN CAUSED BY AXIAL LOADS

• Under simple axial loads metals and other materials act like simple springs. This behavior is known as elastic

• If the material becomes permanently deformed this is known as plastic deformation.

• Consider a generic load-deflection curve,

• This curve is produced by forcing the deformation, then measuring the holding force using a special tensile tester. The specimen pictured above is fastened into a hydraulic or screw based machine. The specimen is then stretched and the applied load measured. The curves can be directly drawn from using load and deflection.

• Secondly, the graph above is often changed as shown to make it independent of geometry.

P necking

rupture

linearelasticregion

plasticregion

PULTIMATE

δ

∆δ

∆P

P

P

A

L

(Hookes law)

∆δ

K∆P∆δ-------=

Test Specimen

• Note that Young’s modulus is a spring constant for a unit volume of material. To get it into a tra-ditional spring constant we must multiply it by area, and divide by length.

11.5 STRESS STRAIN CURVES

• For low carbon steel,

σULTIMATE

∆ ∈

∆σ

∈

σ where,

σ PA--- stress= =

∈ δL--- strain= =

E∆σ

∆ ∈------------- Youngs modulus( )= =

σ E∈=

δ LELσE

------- LPEA-------= = =

or

or

σU 60=

σ ksi( )

σy 37=

∈

elastic yield strain hardening necking

rupture

0.0012 0.02 0.2 0.25

• While necking, the cross section at one point decreases, thus increasing the stress. In turn this continues rapidly until fracture.

• The strain hardening of some materials occurs as they are stretched, the Ultimate Tensile Strength increases, but Young’s modulus remains the same.

• Each material will have it’s own stress-strain curve and these are determined experimentally, and found in abundance in handbooks.

• If a material is brittle, it does not deform much and simply breaks. Or simply the ultimate and rupture strengths are the same.

• Ductile materials deform quite a bit before the ultimate stress, necking typically occurs before rupture.

• The area under the stress strain curve indicates toughness. A larger area under a stress strain curve will make a material that must be extensively deformed before it will fail.

• Creep is an effect that can lead to permanently elongated specimens,

σ

∈

σ

∈

11.6 ANALYSIS OF MEMBERS

• Consider the simple example below,

σ

∈

1

23

STRAIN HARDENING

READ

GT 1.3-1.5

PROBLEMS

SUGGESTED

3-4, 3-6, 4-1, 5-6

REQUIRED

P = 10,000 lb

P

1” dia. annealed copper rod

6”

1) If the rod is 6” before the load is applied, what is the new length?

2) What load P would result in plastic deformation?

3) What load would result in rupture?

4) How much would it grow before permanently deforming?

• As another example, let’s consider problem 2.26 in [Beer and Johnston]

READ

GT 2.1-2.3

PROBLEMS

SUGGESTED

2-1, 2-13, 3-8

REQUIRED

3-10(MC)

A

B

C

D

P

10” 15”

8”

8”

E 29 106psi×=

Links AB & CD are 1/4” by 1”

Given,

Find the load required for a tip deflection of 0.01”

δ 0.01in=

F ree B o d y D ia g ra m

P

11.7 GENERALIZED STRESS

A 0.25in2

=

MB∑ TCD 10in( )– P 25in( )+ 0= =+

δ1

8inTCD

29 106psi×( ) 0.25in

2( )---------------------------------------------------------=

(1)

MC∑ TAB 10in( )– P 15in( )+ 0= =+

δ2

TAB8in

0.25in2

29 106psi⋅( )

--------------------------------------------------=

(2)

(3)

(4)

At this point there are 4 equations and 5 unknowns!!!

10” 15”

δ2 δ1

δ 0.01in=

δ δ2+

25in--------------

δ1 δ2+

10in-----------------=

P∴ 1066lb=

Using similar triangles,

Now with 5 equations and unknowns we can find P.

• We can consider the more general case of the 3D element,

• If we assume the element is square, then we can replace the stresses with forces,

• Finally, for simplicity, we can look at the element from a single side,

σy

τyx

τyz

x

y

z

σx

τxy

τxz

σz

τzx

τzy

Here we can see 3 of the 6 faces of the square element. The normal stress, and the two shear forces define the stress on each face. All of the normal and shear forces must balance for the element.

∆Aσy

∆Aτyx

∆Aτyz

∆Aσx

∆Aτxy

∆Aτxz

∆Aσz

∆Aτzx

∆Aτzy

• For statics, the sum of the forces and moments must total zero. As a result we would see a shear stress (τxy) induce another shear that is perpendicular (τyx).

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

x

y

• We also consider the case where we rotate the virtual element,

P

P

+

∆Aτyx

∆Aτyx

∆Aτxy

∆Aτxy

NOTE: in the stress element shown above the directions of the stresses will be used as the positive convention for the other stress problems. As a memory tool, consider that the upper and right-most stresses are in the direction of the positive x and y axis.

τxy τyx= τyz τzy= τzx τxz=

ASIDE: When we cover moments and force couples, this will become obvious.

• The value of these relationships show how stress is spread throughout an object, and will become valuable as stress is examined in greater detail.

11.8 REFERENCES


11.9 STRESS ON OBLIQUE PLANES

• Consider the arbitrary definition of a force, and how any force can be replaced with components. If we take a force causing an axial stress on a beam, we could replace it with two components that give the same resultant. The only complication being that as we shift the plane of applica-

P P

45°

σ1

σ1σ2

σ2

τ1

τ1 τ2

τ2

Here we can show that,

τ1 τ2P

2A-------= =

σ1 σ2P

2A-------= =

* This will reappear later

F F

tion, the effective section area changes.

σA σA

PPA

-- IS EQUAL TO --

Pθ τ'A'

σ'A'

A'A

θcos------------=

Fy∑ τ'A' θ σ'A' θsin–cos 0= =+

τ' θcos( ) Aθcos

------------ σ' θsin( ) A

θcos------------

– 0=∴

τ'∴ σ ' θsinθcos

------------ σ' θtan= =

Fx∑ P– τ'A' θ σ'A' θcos+sin+ 0= =+

P θcosA

----------------∴ τ ' θ σ' θcos+sin=

P θcosA

----------------∴ τ ' θτ' θcos

2( )θsin

-----------------------+sin=

P θcosA

----------------∴ τ ' θsin( )2 θcos( )2+

θsin--------------------------------------------

τ'θsin

-----------= =

PA--- θ θsincos∴ τ ' θ θσsincos= =

σ'∴ τ' θcosθsin

---------------- θ θσsincos( ) θcosθsin

------------ σ θcos( )2

= = =

P P

τ'∴ θ θσsincos=


30 deg

2 kip

2 kip

4 in.

4 in.

The two pieces of wood below are to be glued together and will carry a tensile load of 2 kip. What are the shear and axial stresses between the glued faces?

READ

GT 2.6

PROBLEMS

SUGGESTED

6-10

REQUIRED

6-18(MC)

11.10 SHEAR STRAIN

• While Young’s modulus is intended for the stress to strain relationship for Axial or normal strain, we will use the Shear modulus G for shear strain.

• First consider the general square cubic element below,

• Now, Assume a shear stress in the positive x direction that induces the deformation as shown below.

σy

τyx

τyz

x

y

z

σx

τxy

τxz

σz

τzx

τzy

• NOTE: these calculations are based upon the assumption of small angles of deflection.

• How a material strains in shear is related to normal strain by Poisson’s Ratio,

• Try the simple problem below,

τyxx

y

zτxy

π2--- γxy+

π2--- γxy+

π2--- γxy–

π2--- γxy–

γxy

2------

γxy

2------

τxy Gγxy= τyz Gγyz= τzx Gγzx=

Where,

τxy τyz τzx, , shear stress in z, x and y planes respectively=

G shear module (or modulus of rigidity)=

γxy γyz γzx, , shear strain in z, x and y planes respectively=

GE

2 1 ν+( )--------------------=

• Consider the example shown below from Beer and Johnson, 1992, pg. 89 #2.86,

100lb.

The block to the left is under a shear load. Find the shear strain if it is made of 2014 aluminum. How much does the top move to the right if the base is fixed? What is the poisson’s ratio for the material?

6”

3”

P

A B

a

a

b

c

Two blocks of rubber (B), connect a plate A to outer brackets. The rubber modulus of rigidity is G=1.75 ksi. Given c=4”, P=10 kip, find a and b if the stress in the rubber is not to exceed 200 psi, and the deflec-tion should be at least 3/16”.

τ 200 lb

in2

------- ≤

A cb 4b in( )= =

τ

P2---

A---------=

200∴ lb

in2

-------

10000 lb( )2

-------------------------

4b in( )------------------------------≥ b∴ 6.25 in( )=

First lets deal with the maximum stress,

Next, lets deal with the deflection resulting from strain,

δ 316------ in( ) aγ a τ

G----

a

200 lb

in2

-------

1750 lb

in2

-------

-------------------------

0.114a= = = = =

a∴ 1.64 in( )=

P

P/2

P/2

11.11 POISSON’S RATIO

• An isotropic material has the same properties in all directions. (materials such as fiber glass do not and are called anisotropic materials).

• When we apply a stress to an isotropic material in one direction, we induce stress in the perpen-dicular direction. The resulting ratio between perpendicular stresses, and strains, is called Pois-son’s Ratio

• In physical terms - as we stretch a bar, it becomes a bit thinner. Stretching of silly putty is a good visual example.


READ

GT 1.6GT 2.4

PROBLEMS

SUGGESTED

6-2, 6-10

REQUIRED

8-4(MC)

ν lateral∈

axial∈---------------------------=

where,

ν Poissons Ratio=

lateral∈ strain perpendicular to applied load=axial∈ strain in direction of applied load=

P = 10,000 lb

P


6”

What would the diameter of the bar become after the load P has been applied?

11.12 GENERALIZED HOOKES LAW

• When we load a material in a single direction the effect of Poisson’s ratio is naturally included in Young’s Modulus. But, when there are multiple loads in multiple directions, we must uses Poisson’s ratio to determine how they intersect.


σy

σy

x

y

z

σx

σz

σx

σz

x∈σx

E-----

νσy

E---------–

νσz

E---------–=

y∈σy

E-----

νσx

E---------–

νσz

E---------–=

z∈σz

E-----

νσx

E---------–

νσy

E---------–=

*Note: use positive stressfor tension

11.13 REFERENCES


The cubical gage block to the right is loaded on two faces with 1 kip of compression. The modulus of elas-ticity for the material is 1 Mpsi, and the poisson’s ratio is 0.3. What are the new outside dimensions?

1 kip

1 kip

0.2500 in.

0.2500 in.

0.2500 in.

12. MOMENTS

12.1 INTRODUCTION

• A force is an important type of mechanical effect, but it doesn’t explain bending. For this we use moments.

up to now we have only dealt with forces through central points (virtual particles). But, in the case of the pirate on the gang-plank, there would be great concern about the plank bending. Where and when will the plank break?

• The classic example is the see-saw,

WW

Introduction to Moments

Free Body Diagram

Equilibrium Condition:

Einstein like Thought Experiment

WW

What Happens??????

Why??? Did our equilibrium conditions change?

MSMALL Fg102m( ) 10Kg 9.81 N

Kg-------

2m( ) 196Nm= = =

MMEDIUM Fg201m( ) 20Kg 9.81 N

Kg-------

1m( ) 196Nm= = =

To solve the problem using proper notation,

M∑ 0=+

MSMALL∴ MMEDIUM– 0=

196∴ 196– 0=

This shows that the system is static, andthe children will balance.

1 m 2 m

10 kg

20 kg

M Fd=

The two children sit on the teeter-tot-ter, because they have different masses, they must sit different dis-tances from the centre of rotation, or face catastrophic impact. In mathe-matical terms the moments on either side of the centre must balance.

First, recall the basic equation for a moment.

MMEDIUM

MSMALL

• Considering moments, convert the cases given on the left to those requested on the right.

10 kg

20 kg

To carry on, also consider that the forces must also balance,

FMEDIUM

FSMALL

FR

Fy∑ FMEDIUM– FR FSMALL–+ 0= =+

FR∴ 20 10+( )Kg 30Kg 9.81 NKg-------

294N= = =

M = 10 lb ft

x

y

a)A Moment

x

y

3

4

2.0 ft

1.0 ft

A Force

F _________=

• Consider the case of a vise. The bottom member must resist the moment caused when the jaws are tightened.

[picture]

• If we look at a normal step ladder we can see that both halves can be analyzed using moments about the top platform.

[picture]

• Certain varieties of tools amplify forces by using moment arms.[picture]

• We will use the terms moment (M) and torque (T). Numerically the two terms are identical, and are calculated the same way. Typically Moment is used when describing a bending, and Torque is used when describing a twisting moment on a shaft.

12.2 CALCULATING SCALAR MOMENTS

• Next lets consider a case where the forces are not so simple,

M = 10 lb ft

x

y

b)A Moment

x

y

3

4

d = __________

d

A Distance

F 5.0 lb=

2.0 m

M

40°

F = 10 N

Given the force applied to the end of the moment arm, find the induced moment.

media/egr20900.jpg

media/egr20903.jpg

media/egr20907.jpg

Solution 1: Find the force component normal to the moment arm,

M Fd F 40°cos–( ) 2m( ) 15Nm–= = =

Solution 2: find the distance normal to the force

M Fd F–( ) 2.0m 90° 40°–( )sin( ) 15Nm–= = =

2.0 m

M

40°

F = 10 N

d

ASIDE: When dealing with moments in planar problems we are free to choose which direction (clockwise or counter-clockwise) is positive. Also, when we sum moments we are doing it about some centre of rotation. Consider the exam-ple below,

AB

F1

F2

MB∑+

clockwise is positive

it is the sum of momentsabout point BA

BF1

F2

BF1

F1 creates a counter-clockwisemoment and therefore should havea negative value

B

F2

B

F2

F2 creates a counter-clockwisemoment and therefore shouldhave a negative value.

BB In this case the force has been

F2y

F2x

+ve

-ve

broken into two components, thesigns on the moments created byeach component are opposite.

12.3 CALCULATING VECTOR MOMENTS

• Consider the problem from before.

ASIDE: The cross (or vector) product of two vectors will yield a new vector per-pendicular to both vectors, with a magnitude that is a product of the two magni-tudes.

V1

V2

V1 V2×

V1 V2× y1z2 z1y2–( )i z1x2 x1z2–( )j x1y2 y1x2–( )k+ +=

V1 V2×i j k

x1 y1 z1

x2 y2 z2

=

V1 V2× x1i y1j z1k+ +( ) x2i y2j z2k+ +( )×=

We can also find a unit vector normal ‘n’ to the vectors ‘V1’ and ‘V2’ using a cross product, divided by the magnitude.

λn

V1 V2×V1 V2×---------------------=

Solution 3: Use vectors and a cross product

F 10N 40°sin–( )i 10N 40°cos( )j 0N( )k+ +=

F∴ 6.43i– 7.66j 0k+ +( )N=

d 2i 0j 0k+ +( )m=

M d F×i j k

2m 0m 0m

6.43N– 7.66N 0N

= =

NOTE: note that the cross prod-uct here is for the right hand rule coordinates. If the left handed coordinate system is used F and d should be reversed.

M∴ 0m0N 0m 7.66N( )–( )i 2m0N 0m 6.43N–( )–( )– j +=2m 7.66N( ) 0m 6.43N–( )–( )k 15.3k mN( )=

NOTE: there are two things to note about the solution. First, it is a vector. Here there is only a z component because this vector points out of the page, and a rotation about this vector would rotate on the plane of the page. Second, this result is positive, because the positive sense is defined by the vector system. In this right handed system find the positive rotation by pointing your right hand thumb towards the positive axis (the ‘k’ means that the vector is about the z-axis here), and curl your fingers, that is the positive direction.

NOTE: Varignon’s theorem permits the cross product to be used. Basically put his theory allows a force to be broken into components. The moment caused by each component is calculated, and added to get the total moment about a point or line. The basic form of the cross product will yield this result.

• Try completing the following calculation.

• Solve the following planar example.

ASIDE: The positive orientation of angles and moments about an axis can be determined by pointing the thumb of the right hand along the axis of rotation. The fingers curl in the positive direction.

x

y

z

+z

x

y

+y

z

x

+

ASIDE: The cross product is distributive, but not associative. This allows us to col-lect terms in a cross product operation, but we cannot change the order of the cross product.

r1 F× r2 F×+ r1 r2+( ) F×= DISTRIBUTIVE

r F× F r×( )–=

NOT ASSOCIATIVEr F× F r×≠but

M _____ _____–( )i _____ _____–( )j _____ _____–( )k+ +=

M r F×i j k

rx ry rz

Fx Fy Fz

= =

• M is a moment vector that is not in the plane of the paper. Given the value below, find M. Keep in mind the right hand rule must be followed for this calculation to be correct.

O

d =2m

40degP = 10N

P _____( )i _____( )j _____( )k+ +=

r _____( )i _____( )j _____( )k+ +=

r P× ___________( )i __________( )j ___________( )k+ +=

r

Fx

y

z

O

F 17i 15j– 2k+=

r 2i 10j 3k+ +=

• Considering moments, convert the cases given on the left to those requested on the right.

12.4 MOMENTS ABOUT AN AXIS

• Remember: Moment of a force about an axis or point is a measure of the tendency of the forces to rotate the body about the axis or point. Moments are vector quantities and have both magni-tude and direction. Since real bodies rotate about axes not points we need to calculate moments about an axis.

y

z

A 3D Force A Moment Magnitude

MO ___________=

O F = { 5.0i - 4.0j - 6.0k } MN

A = (-20m, 50m, 30m)

x

• For the previous figure find the moment about line OB with the following values.

r

Fx

y

z

O

Note: the moment used must have been calculated about a point on the line.

MOB MO λOB•=

BNote: the dot product must be done with a unit vector, in this case AB.

F 17i 15j– 2k+=

r 2i 10j 3k+ +=

B 5i 0j 5k+ +=

Is the result of this calculation a scalar or vector?

How does this fit with our knowledge of moments?

What does a negative answer mean?

• Consider the bent pipe in the practice problem below, ([Hibbeler, 1992], prob 4-30, pg. )

When we want to do a cross product, followed by a dot product (called the mixed triple product), we can do both steps in one operation by finding the determinant of the following. An example of a problem that would use this shortcut is when a moment is found about one point on a pipe, and then the moment component twisting the pipe is found using the dot product.

d F×( ) u•λx λy λz

dx dy dz

Fx Fy Fz

=

NOTE: When finding the moment about a point we need to use the vector (cross) product, and the result is a vector. When we want to find the moment about a line, we use the triple product to get a scalar value.

x

y

z

A

B

C

D

24 in.

15 in.

6 in.

F1 4i 10j 8k–+( )lb·=

F2 9i 4j– 10k–( )lb·=

Given the pipe CBAD, and the two forces applied to the end, find the moment that is twisting section AB about its axis.

Lets find the moment about point A using vector methods,

dAD 0i 0j 6k–+( )in=

MA dAD F1 F2+( )×=

MA∴0in

0in

6in–

4lb 9lb+

10lb 4lb–

8lb– 10lb–

×0in

0in

6in–

13lb

6lb

18lb–

×= =

MA∴i j k

0in 0in 6in–

13lb 6lb 18lb–

6in–( ) 6lb( )–( )i 6in–( ) 13lb( )( )j+[ ] in lb•( )= =

MA∴ 36i 78j–( )in lb⋅=

The axis unit vector of AB can be used to project the moment using a dot product

uAB 1( )i 0( )j 0( )k+ +=

MAB MA uAB•=

MAB∴ 36i 78j–( )in lb⋅ 1( )i 0( )j 0( )k+ +( )• 36in lb⋅= =

ASIDE: later you will learn how to calculate how much this pipe would twist when exposed to the moments calculated above. But, before these calcula-tions can be done, the basic statics calculations above (or equivalent) must be done. A basic question to ask your self is will this beam fail?

12.5 EQUILLIBRIUM OF MOMENTS

• As with forces, moments also experience equilibrium. If we sum moments ABOUT A SINGLE POINT ON A RIGID BODY, they should total zero for the object the remain static.

• Try the problem below,

As an alternative to the vector methods, lets try solving this problem using scalars,

MAB∑ F2y6in( ) F1y

6in( )+ 4lb–( ) 6in( ) 10lb( ) 6in( )+ 36lb in⋅= = =

NOTE: in this case the solution was almost trivial. This is because the distances all happened to be split up into cartesian coordinates. In some cases the scalar solution will become very messy, and undoubtedly lead to mistakes.

4m

5m

x

y

z

65°

70°

A

B

Ccan slide freelyalong edge

10kg

Fax = NFay = NFaz = NFcx = -NFcy = 0Fcz = -N

The frame below has a ball joint at A, and the other end of the beam at C is smooth and slides freely along the y-axis. A mass of 10kg is added midway along the beam AC. Find the reaction forces at A and C.

• Another category of problems that may be encountered is to be given a moment vector, and a displacement vector, and be expected to find the applied force. This type of problems is com-plicated by the fact that the cross product is not reversible, but the solution turns out to be sim-ple.

[picture]


Fr M×r r•

-------------=

media/egr20922.jpg

x

y

z

AB C

F2F1 = {2.0i + 3.0j + 4.0k }lb

2.0m

Fg = -200j Kg

Given the two forces F1 and F2 acting on point A and the force of gravity acting on the centre of the beam Fg, write the equation for the sum of the moments about C.

0.1m

12.5.1 References


12.6 FORCE COUPLES TO MAKE CENTERLESS MOMENTS

• Sometimes we are faced with moments that are in awkward positions, We can move these by replacing them with forces and moments in new positions. (Force couples also allow us to rotate the effects of moments)

• Keep in mind that couples cause rotation only, without any translation.

• Consider the basic force couple,

READ

SLI 3.1-3.2SLI 3.3SLI 3.4-3.5SLI 3.6

PROBLEMS

SUGGESTED

6, 9

26, 3857

REQUIRED

• A force couple acts about an axis. In the previous example the axis was out of the page. The axis can be moved to any position on the rigid body, as long as the direction is maintained. This axis of the couple will be perpendicular to the plane of the forces.

• A sample problem is given using a rope pulling a wedge ([Hibbeler, 1992], prob 4-81, pg. )

d

F

F

Consider a case where we have a moment couple (C), but we want to replace it with forces, or we have a force that we want to move, and we are willing to introduce a new moment. We can convert a moment to a pair of forces as pictured below.

MCO Fd F d2---

F d2---

+= =

Now, also consider that force couple can be moved freely about the part, and by impli-cation, so can the moment.

+C

d

F

F

C

d

F

F

A

C Fd F d A+( ) F A( )–= =

OO

OO O+

ASIDE: Moment problems can be solved without resorting to force couples, but these are powerful concepts when dealing with rigid bodies. In effect they allow us to deal with a moment that is independent of a centre of rotation.

3”

4”

6”

P

-P

F

-F10 lb

10 lb

30°

30°

3”

Find the forces F and P to keep the block from moving.


I will use the vector approach to solve the problem, this problem could also be solved with more effort using scalars.

To begin I will find the moments caused by forces P and F. These are easy to find because they lie on the planes. (we should also define axis)

xy

z

CF F– 6in( )( )j 6– Fj( )in= =

CP P 6in( )–( )k 6– Pk( )in= =

Next, to find the moment caused by the ropes, we must find the normal to the plane that the moment acts upon. If this is then multiplied by the scalar moment, we will have the moment vector.

uNR

0i 3j 4k+ +

32

42

+---------------------------- 0.6j 0.8k+= =

CR 10lb 3in( ) 30lb in⋅= =

CR CR uNR30lb in⋅( ) 0.6j 0.8k+( ) 18j 24k+( )lb in⋅= = =

Finally, we sum the moments, and solve for the static condition,

M∑ CF CP CR+ +=

M∑∴ 6– Fj( )in 6Pk( )in– 18j 24k+( )lb in⋅+ 0= =

6– Fj 6Pk– 18j 24k+( )lb+ 0=∴

6– F 18lb+( )j 6P– 24+( )k+ 0=∴

6– F 18lb+ 0=∴

F∴ 18lb6

----------- 3lb= =

6P 24lb– 0=∴

P∴ 24lb–6

-------------- 4lb= =

• Consider the stress element that was seen earlier in the stress section

A

B

CD

E5000 Nm

10 KN

3m

3m 3m

TDE 4.17KN=

Find the tension in the cable DE

12.6.1 Moving Forces and Equivalent Force Moments

• If a force is moved along the line of action, there are no other modifications needed, (this is called transmissibility)

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

x

y

READ

SLI 3.7

PROBLEMS

SUGGESTED

64

REQUIRED

68

• If a force is not moved along its axis (line of action), then we must create a resultant (make believe) force couple (and it’s resulting moment).

• A sample of a problems that is simplified by moving forces is seen below, ([Hibbeler, 1992], prob 4-102, pg. )

F

F

As long as a force is moved along a line of action, it will appear to be the same force (in a rigid body)

F

These three systems of forces are equivalent, but they show how to move forces about an object. Obviously these forces do not move in reality, but this is a calculation trick that will make solving some problems much easier.

FF

F

F

C

C r F×=r

FR

A

3’

30° 48 lb

5’ 2’

60 lb

18 lb

7’

C

Given this beam with applied forces,a) replace all of the known forces with an equivalent force and

moment at C.b) find the resulting force FR using the results in a)

part b):

Using the results from the last section, and the rather simple force at the end of the beam, we can sum the moments about C to find the force FR.

M∑ 705lb ft FR 2ft 5ft 3ft+ +( )+⋅– 0= = ASIDE: the sum of the moments about the pinned joint C will total zero because a pinned joint has no resistance to moments.

FR∴ 70.5lb=

+

NOTE: Remember that once we have found a force couple we are free to move it over a rigid body or structure as needed. In calculations to come later this will make some calculations simpler. - In other words a couple is added into a moment equation without considering its position, and it is not considered in a sum of forces.

• Try the sample problem below,

part a)for the 18 lb force:

FCx18lb–=+

CC 7ft 18lb–( ) 126lb– ft⋅= =+

for the 60 lb force:

FCy60lb–=+

CC 2ft 60lb–( ) 120lb– ft⋅= =+for the 48 lb force:

FCy48lb 30°cos– 42lb–= =+

CC 48lb 30° 5ft 2ft+( ) 48lb 30° 7ft( )cos–sin– 459ft– lb⋅= =+

FCx48lb 30°sin– 24lb–= =+

for the total at C:

FCy60lb– 42lb– 102lb–= =+

CC 126lb– ft⋅( ) 120lb– ft⋅( ) 459ft– lb⋅( )+ + 705lb ft⋅–= =+

FCx18lb– 24lb– 42lb–= =+

A B

C

45°

F1 20N=

F2 30N=

5m

2m

MA 79.3Nm=

Find the reaction moment and force at A given the two forces at B and C

READ

SLI 3.8SLI 3.9

PROBLEMS

SUGGESTED

7483

REQUIRED


1. The pole OA is 5m long and is held firm at base O. It is pulled by two cables AC and AB that each have 1KN of tension. Find the magnitude of the moment at O caused by the two cables.

2. A 500 lb cylindrical tank, 8ft in diameter, is to be raised over a 2ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing the corner of the obstruction at A is rough so that the cylinder slips at C but not at A, find the required tension in the cable.

30°60°

(-3, 4, 5) m

2m

A

B

C

O

45°

60°

x

y

z

60°

3. A force of 1200N acts on a bracket as shown. Determine the moment Ma of the force about A.

4. A 30lb force acts on the end of the 3ft lever as shown. Determine the moment of the force about O.

8ft

2ft

A

C

1200N

30°

A

C

120mm

140mm

40mm

180mm

5. A box is shown in the figure below. Find the moment that the tension cable causes about the origin using a cross product. Give the results using components and a magnitude.

6. A force P acts on a corner of a frame (at A) and creates a moment about the origin (O). We know that the force P is in the y-z plane, the moment about the y-axis is -20Nm, and the moment about the z axis is -40 Nm. Find the magnitude of the moment about the x-axis.

O

3ft

50°

20°

30lbA

x

y

z 6’

3’

5’

(10, 0, -7) ft.

T=1kip

MO 7716ftlb= MOx3309ftlb–= MOy

5122ftlb= MOz4729ftlb–=

ANS.

x

y

z

P

α

5 cm

3 cm30°

O

A

7. The rod has a mass of 20lb., and the mass of joints A & B are both 5 lb. What is the tension ‘T’ in the rope?

Find the vector moment about O,

MO rA O⁄ P× i Mx( ) j 20–( ) k 40–( )+ +{ } Nm= =

Next, find the moment arm (or point A), and the force P,

A Ax 0.05 0.03, ,( )=

30°tan0.03m

Ax---------------= Ax∴ 0.052m=

rA O⁄ 0.052 0.05 0.03, ,( )m=

P 0 P αsin– P αcos, ,( )=

Do the cross product to get equations,

MO

i j k

0.052 0.05 0.03

0 P αsin– P αcos

i Mx( ) j 20–( ) k 40–( )+ += =

∴ i 0.05P αcos 0.03P αsin+( ) j 0.052P αcos–( ) k 0.052P αsin–( )+ +=

0.05P αcos 0.03P αsin+ Mx=

Write the equations and solve,

0.052P αcos– 20–=

0.052P αsin– 40–=

0.052P αsin–0.052P αcos–

---------------------------------∴ 40–20–

---------= α∴ 63.4°= P∴ 860N=

Mx 0.05P αcos 0.03P αsin+ 42Nm= =

(ans. 18.2lb)

12.6.3 References




T30°

10°

5ft.

A

B

13. TORSION

13.1 INTRODUCTION

• If we consider a cylindrical shaft with torques applied it will tend to rotate an angle proportional to the torque.

• We can find the torque exerted using simple stress equations for a plane cutting through the cyl-inder,

TT

T=0

TT

T>0

dFT

ρ

T ρ Fd∫ ρλ Ad∫= =

dA

(1)

• We can describe the twist, or angle of deflection, using two angles.

TT

ργφ

L

γ ρc---γMAX= (3)

where,c the outer radius=

ρ a variable radius [0, c]=

γ the strain at radius c=

φ a small twist angle at the end=

L the length of the shaft=

γMAX the strain at outer radius c=

γsin =

φsin =

Small angle assumption:

γ = γmax =OR (2)φ Lc---γMAX=OR

Substituting for & in the above equation yields:

Recall from the previous lectures:

Substituting equation 3 into equation 4 yields

λ Gγ= γ λG----=or (4)

λG---- ρ

c---

λMAX

G-------------= λ = (5)

What are the consequences of this equation?

Where are the largest shear stresses due to torsion?

How do the shear stresses over the diameter of the bar?

• Consider the simple example of the beam below,

T ρλ Ad∫λMAX

c------------- ρ2

Ad∫λMAX

c-------------J= = =

γMAX∴ cTJG------- cφ

L------= =

φ∴ =

Now, if we substitute equation 5 into equation 1,

TλMAX

c-------------J= λMAX =OR (7)

(6)

Note: the polar moment of inertia will be covered in the next section

J ρ2Ad∫= Polar Moment of Inertia

But equation 2 shows that

which leads o the equation:

where:

L= T =J =G =c =φ =

13.2 THE RELATIONSHIP BETWEEN STRESS AND STRAIN IN TOR-SION

• Although we are applying a torque (moment) it creates stresses and strains in the material.

• Consider the surface of a rod in torsion.

1000 lb.ft. 1000 lb.ft. What is the maximum shear stress in the bar? If the bar is made of cold-rolled yellow brass, what is the angle of deflection along the bar?

10”

1” dia.

TT

σθ λ 2θ( )sin=

λθ λ 2θ( )cos=

(1), (2)

Therefore the maximum normal stress will occur at

2θ =

θ =

Substituting back into equation (1) and (2)

σ 45( ) =

λ 45( ) =

σ 135( ) =

λ 135( ) =

What does this tell us about how the shaft will break?????????

Graphically this looks like:

• Consider the example problem below. [Gere and Timoshenko, 3.7-4 Pg. 256]

• Consider the case below,

Given:

Douter 60mm=

Dinner 40mm=

ω 2500rpm=

P 150KW=

Part A. Find maximum Shear Stress

Need to know P Tω=

J Jouter Jinner– π2--- Couter

4Cinner

4–( )= =

(use radians/sec not revolutions/sec)

Therefore TPω----=

λmaxTcJ

------=

What is the minimum shear stress in the shaft and where does it occur?

13.2.1 Summary

• If we draw a stress element we can derive the relationship between the normal and shear stresses.

1000 lb.ft. 1000 lb.ft. If the bar is made of cold-rolled yellow brass, what are the maxi-mum axial and shear stresses and strains? How do these com-pare to given values?

10”

1” dia.

• The basic torsion relationships are given below.

τ

τ

τ

τ

First, consider the element that we have dealt with before. For a bar in torsion, the forces are balanced (note: these are force couples)

If we rotate the element by 45 degrees, the shear stresses are now replaced with normal forces. Not the directions indicate tension and compression as perpendicular.

σt

σc σt

σc

With a simple sum of forces we can show that both cases are equal, and that the stresses are all equal.

σt σc τ= =

While the stresses are equal, the axial is half the shear strain. (This is a result of the pois-son’s ratio).

∈ max

τmax

E----------

ντmax

E--------------+

τmax

E---------- 1 ν+( )

τmax

2G----------

γmax

2----------= = = =


2. We need to design a stepped shaft to fit between a 1” diameter and 1/2” diameter holes. The shaft is to transmit 100W of power at 10 rpm. 3” of the 1” diameter shaft is in torsion, and 2” of the 0.5” diameter shaft is in torsion. If the shaft is made of brass,

TT

γφ

L

ρ

λ TρJ

------= λmaxTcJ

------=

γmaxTCJG-------= φ LT

JG-------=

Jtube Jouter Jinner– π2--- Couter

4Cinner

4–( )= =

Jcircleπ2---C

4=

READ

GT 3.1-3.4

PROBLEMS

SUGGESTED

3-4, 4-7

REQUIRED

3-11(MC)

a) What is the minimum radius for a fillet between the shaft sections?b) What is the angle of deflection?c) What are the maximum stresses and strains?

2. a) Given the pipe below find the torsion about the centerline of member AB (AB lies in the x-z plane). The round members from A to B, and B to C are solid with a diameter of 1/8in. Assume that the material has the properties; E = 50 Mpsi and G = 20Mpsi. (marks 25)

b) If the torsion is found to be 100 lb then what is the angle of twist for AB? (marks 25)

13.4 REFERENCES


1”0.5”

3” 2”

x

y

z

3”

5”

7”A

B

CF = (69.3, -54.4, -94.9)lb.

14. MASS PROPERTIES

14.1 INTRODUCTION

• In addition to basic statics this topic will lead in to estimating strengths of beams, dynamic behavior, mass, etc.

• Important terms,

• A simple way to illustrate the moment arm effect of gravity about a centroid is by considering an exaggerated T shape that is supported by two cables, one at either end.

Centre of Mass - This will indicate where the effective gravitational force should be applied to take into account the moment arm effect.

gravity

Centre of Gravity - If the gravity vector is not uniform over an object the force exerted by a volume of mass will change, this takes that into account. Centre of gravity is commonly used to describe the point of application for the gravitational force.

Centroid - typically where distributed forces, such as gravity are applied (this is the same as the centre of mass, if the mass density and gravity is uniform). This is the easiest property to calculate. This is also called the first moment of area.

1 m

1 m

1 cm

1 cm

total mass 200 kg

As shown, the cross section has two major sections, each of equal area (and mass of 100 kg), One section hangs beneath the left cable, The mass of this section will largely be held up by the lefthand cable, but the long section that stretches from left to right will be supported on both ends with about the same force. Therefore an intuitive estimate of forces would suggest that the left cable would support 150 kg, and the right cable would sup-port 50 kg.

MIDDLE OF MASS - lets consider the effect if the middle of mass were used to apply the force of gravity to the two cables. First, the centre of gravity would be 1cm from the left of the section. This would be the same as lever shown below,

T1 T2

M∑ T1 1cm( ) T2 100cm( )– 0= =+

Fy∑ T1 T2 200Kg–+ 0= =+

centre of mass

T11cm

100cm-----------------

T1+ 200Kg=∴

T1∴ 198Kg= T∴ 2 2Kg=

These numbers are clearly INCORRECT, but in the next example we will see that the numbers are much closer.

INCORRECT!

• A note of interest, when the object is symmetrical the centre of mass and centroids will be the same. Another term of significance is the centre of gravity, but unless we have a part with a significant gravity gradient across it, this will be the same as the centre of mass. Unless dealing with very sensitive equipment, or astronomically significantly distances, the centre of gravity, and centre of mass can be considered equivalent.

14.2 CENTRE OF MASS

• This property is simply the point at which the mass on one side of the point is equal to the mass on the other side of the point.

• This property can be found with a summation of mass, weighted by distance.

CENTROID - The centroid uses area weighted (multiplied) by its distance from the cen-troid. As a result, the centroid is the natural point to use for the application of gravita-tional forces.

T1 T2

M∑ T1 25.5cm( ) T2 74.5cm( )– 0= =+

Fy∑ T1 T2 200Kg–+ 0= =+

centroid

T1∴ 25.5cm74.5cm------------------

T1+ 200Kg=

T1∴ 149Kg= T2∴ 51Kg=

These numbers do match the intuitive derivation of the cable tensions given before.

0.255m

• A simple example can illustrate the calculation of this property. A 2D example is given, but extension to 3D is trivial.

• Consider the example of the T-shape from before,

xximi∑mi∑

------------------= yyimi∑mi∑

------------------= zzimi∑mi∑

------------------=

ASIDE: The centroid/center of gravity is calculated using the first moment of iner-tia. We will also use this later when dealing with beam bending.

ximi∑ x Vd

V∫ x Ad

A∫

x 6m 5Kg( ) 8m 6Kg( )+5Kg 6Kg+

------------------------------------------------------ 7.1m= =

2 m

2 m3 m

3 m

6 kg

5 kg

(6m, 3m)

(8m, 2m)

x

yFind the effective centre

of mass for the two plates shown.

y 3m 5Kg( ) 2m 6Kg( )+5Kg 6Kg+

------------------------------------------------------ 2.5m= =


x

d1

d2 d3

F1

F2 F3

d1 x0.01

2----------–=

F1 100kg=

d2x 0.01–

2-------------------=

F2 100kg x 0.01–100

------------------- =

d3101 x–

2------------------ x+=

F2 100kg 101 x–100

------------------ =

MO∑ d1F1– d2F2– d3F3+ 0= =

O

+

x0.01

2----------–

100kg( )– x 0.01–2

------------------- 100kg x 0.01–

100-------------------

– 101 x–

2------------------ x+

100kg 101 x–100

------------------

+

x0.01

2----------–

– x 0.01–2

------------------- x 0.01–

100-------------------

– 101 x–2

------------------ x+ 101 x–

100------------------

+ 0=

200x˜ 0.01–( )– x 0.01–( ) x 0.01–( )– 101 x– 2x+( ) 101 x–( )+ 0=

200x˜ 0.01+– x2

0.02x– 0.0001+ – x

2– 10201+ 0=

2x2

– 199.98x– 10201+ 0=

x 199.98–( )– 199.98–( )2 4 2–( ) 10201( )–±2 2–( )

------------------------------------------------------------------------------------------------------------ 137 37,–= =

14.3 CENTROIDS

20°

15KN

20mm

30mm

60mm

40mm 50mm

150mm

P

M = 500Kgµ = 0.2

The 500 Kg plate below is to be used in an industrial machine. The basic shape is rectan-gular, but there is an angled arch cut in the bottom. Considering the centroid, and the applied force, determine the magnitude of the force (P) that is required to pull out the wedge if the both sides of the wedge have a coefficient of friction of 0.2.

• This is also known as the geometric centre, or first moment of area.

• The fundamental definition for Centroids is given below, although a more efficient method for problems with simple geometries is discussed later.

• When finding centroids, look for symmetry, the centroids will be in the center of symmetrical sections.

x

x Vd

V∫

Vd

V∫

---------------= y

y Vd

V∫

Vd

V∫

---------------= z

z Vd

V∫

Vd

V∫

---------------=VOLUME

x

x Ad

A∫

Ad

A∫

---------------= y

y Ad

A∫

Ad

A∫

---------------= z

z Ad

A∫

Ad

A∫

---------------=AREA

x

x Ld

L∫

Ld

L∫

--------------= y

y Ld

L∫

Ld

L∫

--------------= z

z Ld

L∫

Ld

L∫

-------------=LINE

• A simple example of application for these relationships is given below,

h

dx

dA= (h) (dx) = y(x) dx

h = y(x)

ASIDE: Recall that to set up an integral we must pick a suitable element (slice). As we can see below a vertical rectangular slice is selected. We find an area for the slice by multiplying height by width. The bound for integration are taken in the differential (eg. dx) direction. We can use the same elements to find the y centroid.

0a

x

x'˜ Ad( )0

a

∫

Ad

0

a

∫------------------

x y x( )( ) xd

0

a

∫

y x( ) xd

0

a

∫---------------------------= =

x'˜

y'˜ h2--- y x( )

2----------= =

y

y'˜ Ad( )0

a

∫

Ad

0

a

∫------------------

y x( )2

---------- y x( )( ) xd

0

a

∫

y x( ) xd

0

a

∫-----------------------------------------= =

• Lets try a more practical problem using the centroid, ([Hibbeler, 1992], prob. 9-16, pg. )

xx 4 x 5–( )–( ) xd

5

7∫

2 2( ) 12--- 2 2×( )+

---------------------------------------------- 16--- 9x x

2–

xd5

7∫ 1

6--- 9

2---x

2 x3

3------–

5

7

5.889m= = = =

x

y

4 m

2 m

5 m 7 m

12 kg

Find the centroid, and centre of mass for the x-axis of the object pictured. Assume that the 12 kg mass is evenly distributed by area.

First integrate for the centroid,

Next, use weighted sums of areas for the composite shape (covered later),

xxsquareAsquare xtriangleAtriangle+

Asquare Atriangle+---------------------------------------------------------------------------------------------------=

x∴6m 2

3---

12Kg 5m 1

3--- 7m 5m–( )+

13---

12Kg +

23---

12Kg 13---

12Kg+

-------------------------------------------------------------------------------------------------------------------------- 5.889m= =

*** Calculations for y are similar.

δ 5 lb

ft2

-------=

4 ft

A

B

The quarter of a circle steel plate to the left is to be supported by two cables. The tension in each cable is to be deter-mined. The density per unit area is also provided to aid in the calculations.

1. In this problem it is obvious that the problem involves mass properties, so the basic properties will be calculated first. These include mass (M) and area (A). Area is used when finding the centroid and mass, and mass is needed to find the total force exerted on the cables.

A πr2

4-------- 12.57ft

2= =

M Aδ 12.57ft2

5 lb

ft2

-------

62.85lb= = =

2. Next, we are prepared to find the centroid, so we must set up the integral,

xx 16 x

2– xd

0

4∫

A--------------------------------------

23---

12---–

16 x2

–

32---

0

4

12.57----------------------------------------------------- 16 16–( )

32---

16 0–( )

32---

–3 12.57( )–

------------------------------------------------------- 1.70ft= = = =

And, by symmetry,

y x 1.70ft= =


TA3. Finally, we are prepared to find the ten-

sions in the support cables. To do this, a free body diagram is drawn first, and since all of the forces are in line, a sim-ple sum of moments is used.

TB

1.70 ft

4 ft

62.85 lb

+ MA∑ 62.85lb 1.70ft( ) TB 4ft( )– 0= =

TB\ 27lb=

Fy∑ TA TB 62.85lb–+ 0= =+

TA\ 36lb=

x

y

2cmd

y x( ) 0.1x2

3x– 100+( )cm x 0 10cm,[ ]∈,=

d= 5.62cm

We want to design two similar supports for a new sculpture. This means that the supports should be positioned to carry the same load. Find the distance of the second support from the first, and indicate the loads to be supported by each if the plate weights 1000kg per square meter.

14.3.1 Finding Centroids Using Composite Shapes

• This method basically uses tables, or obvious centroids for basic shapes in a complex shape. By using a summation of centroids, weighted by areas/volumes divided by total area.

• A simple example for a two dimensional problem is given below, ([Hibbeler, 1992], prob. 9-48, pg. )

READ

SLI 4.1-4.3

PROBLEMS

SUGGESTED

5, 18

REQUIRED

xxiAi∑Ai∑

----------------= yyiAi∑Ai∑

----------------=

• Next, lets consider a problem using the composite method in 3D volumes, ([Hibbeler, 1992], prob 9-68, pg. )


15 mm

150 mm15 mm

150 mm

15 mm 100 mm

A

B

C

centroid

y

yAAyA AByB ACyC+ +

AA AB AC+ +-------------------------------------------------------------=

For the beam cross section shown, find the centroid.

First, it is obvious by symmetry that the x centroid will be in the centre of the beam, but the height will be somewhat more difficult to determine because of the lack of symme-try top to bottom. This method can actually be done in one step, but the beam must be divided into simple parts. I will divide this beam into three rectangular sections.

y∴15 150×( ) 15

2------

150 15×( ) 15 1502

---------+ 15 100×( ) 15 150 15

2------+ +

+ +

15 150×( ) 150 15×( ) 15 100×( )+ +--------------------------------------------------------------------------------------------------------------------------------------------------------------------------=

y∴ 16875 202500 258750+ +2250 2250 1500+ +

--------------------------------------------------------------- 79.9mm= =

C

a

h

z

z

V113---πr

2h=

Given the 3D volume shown, find the centroid. This shape is essentially a cylinder with a cone removed.

First, the only difficulty in finding the centroid is the height, the x and y centroids are at the centre by symmetry. To begin the solution we must look up the volumes and centroids for each element (look in the math section).

For the cone:

z134---h=

V2 πr2

h=

For the cylinder:

z2h2---=

zV1z1 V2z2+

V1 V2+----------------------------------=

Then, calculate the centroid,

z∴

13---πr

2h

34---h

πr2

h h

2---

+

13---πr

2h πr

2h+

------------------------------------------------------------------

h2---

h4---–

113---–

------------ 38---h= = =

object

2m2m

k=100KN/m

1m

1m

0.25m

0.5m

0.25m

0.75m

The object:

δ 5000 kg

m3

------=

θslip 11.3= θactual 8.83= θtip 49=

An object is perched on the far end of a lever and we want to determine if it will tip, or slip off, or remain in place if the coefficient of friction is 0.2. (Note: the 2m dis-tance to the right is to the centroid of the block)

• We will often encounter distributed forces in real problems. For the convenience of calculation we want to simplify these to point forces.

READ

SLI 4.3

PROBLEMS

SUGGESTED

41, 48

REQUIRED

50(MC)

NOTE: The distributed forces are typically given a force per unit length/area, and the total equivalent force applied can be found by integrating the force over length/area. We can also find the point to apply this concentrated force by inte-grating the force weighted by the area. (only force per unit length is shown, but for force per unit area the methods are similar)

x

F x( )

x

FR

FR F x( ) xd

L∫=

x

xF x( ) xd

L∫

F x( ) xd

L∫

-------------------------=


READ

SLI 4.6

PROBLEMS

SUGGESTED

52, 59

REQUIRED

Find the reactions at the supports of the beam below if iron rods are piled as shown. Each 1” dia. rod weighs 30 lbs. The beam that the rods sit on weighs 2000 kg.

80” 80”

30”

Parabolic

• An example of a roughly distributed load can be seen below,[picture]

14.3.2 Pappus and Guldinus - Rotated Sections

• If we take some curve and rotate it about some axis, it will sweep out a shell of some volume. We refer to this as a Volume of Revolution.

• We can calculate the surface area of these shapes using the centroid and the arc length of the curve,

• We can easily calculate the volume of these shapes using the centroid and the area under the curve,

• These methods can be of particular interest when the shapes are difficult to integrate in three dimensions, or when we start with a profile of a shape. One example of this would be a pro-gram to drive a CNC lathe.

• Try the example below,

A 2πLy=

V 2πAy=

media/egr20914.jpg


1. The plate below has a triangular hole. We are asked to select a base width ‘b’ for the triangle so

Find the volume of the toroid to the right if the minor radius is 3”, and the major radius is 5”.

READ

SLI 4.4

PROBLEMS

SUGGESTED

28

REQUIRED

that the reaction at roller B is 200N. The plate material weighs 1Kg per square meter.

2. The 500 Kg plate below is to be used in an industrial machine. The basic shape is rectangular, but there is an arch cut in the bottom. Considering the centroid, and the applied force, determine the magnitude of the force (P) that is required to pull out the wedge if both sides of the wedge have a coefficient of friction of 0.2.

2m

4m

6m

4m

12m

b

AB

20°

15KN

20mm

30mm

60mm

40mm 30mm

120mm

P

M = 500Kgµ = 0.2

3. A thin flat plate is suspended by means of a screw eye and 2 cords, AC and BC, as depicted in the diagram below. Given that the plate has a mass of 30kg per square meter of surface area, determine the reaction force at point O and the tension forces in the cords.

4. Find the x-y centroid of the shape below which is essentially rectangular, except for the shape cut out of the bottom (the function is given).

5. What is the x-y centroid of the shape below if the plate is homogenous?

A B

C

O

1m

0.6m

0.3m 0.3m

g=9.806 m/s2

x

y

1 m

1 m

yx

2

3-----= ANS.

x = 0.469 my = 0.550 m

6. The mass below (a rectangle with a parabola cut out) is suspended by two cables with a diame-ter of 2mm. The cables have an undeflected length of 2m.

a) Find the x centroid for the plate and the force of gravity on the plate if the material weighs 50 kg per square meter.

b) Determine the new lengths of the cables at A and at B. Assume that the cables have E = 20GPa.

c) Based on the elongated cables, what angle does the plate sit at after it is released and deflects?

x

y

0 2” 4” 7” 9”

2”

3.5”

6”

5.5”

ANS.

x = 4.6”y = 2.96”

2m 5m

3m

1m

0.5m

A B

h x( ) 0.05x2

= x 0 5m,[ ]∈

15. FORCES AND MOMENTS ON RIGID BODIES

15.1 INTRODUCTION

• In real systems we have bodies that have both moments and forces applied. All must total zero for a static case.

• The methods for analyzing these systems requires a good free body diagram. This diagram will be drawn using the forces applied to the system. The reaction forces are put on (as variable in most cases) by evaluating the connections.

15.2 REACTIONS AND SUPPORTS

• These symbols are very important to engineers. Their equivalent is the symbols for resistors, transistors, etc in electrical engineering. They are a way to talk about systems in general terms before investing time and resources designing and building the system.

• A subset of these symbols are,

SIMPLE LINK: This type of link is commonly used, and only has pinned joints at two ends. These members don’t trans-mit moments, and all forces are tension or compression along an axis between the pins.

F

F

F

CABLE: a cable (like a string) can be made of natural/artifi-cial fibres, metals (usually stranded), etc. These mem-bers can only be used in ten-sion - you can’t push with a rope.

e.g., a cable does not provide force when compressed

The cable cannot supplycompressive forces neededto hold the bar in position

F

F

ROLLER: Force is only normal to the rolling plane. If the applied force is away from the surface, the roller exerts no force.

F

F

• Other joints include- spring- rod on a smooth surface- rod on a rough surface- pin supports- beams cemented into walls (fixed support)- ball (3D)- ball socket- universal joint- hinge- collar and shaft

SLIDER: this is the same as the roller, except the force can be in both directions.

F F

******* NOTE: Add more on support reactions

15.3 EQUILLIBRIUM OF FORCES AND MOMENTS

• It is often necessary to solve problems using both sums of moments and forces. Indeed, both of these sums must be equal to zero for the body to be static.

• An example of a problem that has both forces and moments is given below, ([Hibbeler, 1992], prob 5-4, pg. )

Supports Reactions Unknowns

Roller

FrictionlessSurface

Rocker

1

M∑ 0= F∑ 0=AND

3

45

A B

C

D

E

2.0 m 2.0 m 1.5 mM = 80 kg

For the beam suspended by a cable, with a mass at the end, find the forces in the cable, and at point A.

NOTE: the slope trian-gle is sometimes used in place of angles, but note that it can be used to give trig. values directly.


Before starting any calculations, we should develop a free body diagram. This problem is best solved by scalars because of the simple planar forces at right angles.

A B

CE

2 m 2 m 1.5 mM = 80 kg

FAx

FAy TT

Next, we can sum forces and moments to get three equations. We know that there are two unknowns at point A, and the cable tension is the second unknown. Therefore we will have three equations and three unknowns.

We can start by summing the moments about point A. Because this point has two of the three unknown forces go through it, the sum of moments equation will only have one unknown (and can be solved immediately).

MA∑ T–( ) 2m( ) 45---– T

2m 2m+( )+=

80Kg 9.81 NKg-------

2m 2m 1.5m+ +( )+ 0=

+

T 5.2m( )∴ 4316N m⋅=

T∴ 830N=

Fx∑ FAx

35---–

T+ 0= =+ FAx∴ 498N=

Fy∑ FAyT 4

5---

T 80Kg 9.81 NKg-------

–+ + 0= =+

FAy∴ 709– N=

FAy∴ 830N– 4

5--- 830N( )– 785N+=

ASIDE: please note that -ve sign here indicates that I assumed the wrong direction for the force on the FBD.

media/stat0000.wm


2 yd.

2 yd.

3 yd. 3 yd. 2 yd.

500 lb.

10000 lb.yd.

A

B

C D

E

ANS.FAx = -532 lb.FAy = 1564 lb.FE = 1190 lb <-27°

Given the force at C and couple at B, find the magnitude and direction of the reaction forces at the supports A and E below.

• Consider the pictures of both the hoists below, they are much like the examples that follow. The red dampers apply a force that is almost static to slow the lowering of the upper arms.

[picture][picture]

• Lets consider another complex system to solve. In this case free body diagrams are beginning to become essential. ([Hibbeler, 1992], prob 5-44, pg. )

8 ft

12 ft

14 ft

10°

A

B

C

D

M = 800 lb

When passengers must leave a large ship, lifeboats are lowered over the side. A large hydraulic cylinder tips the arm. Given the diagram to the right, and the maximum load in the boat,

a) find the force exerted by the hydraulic piston CB to support the assembly.

b) find the forces at pin A.

media/egr20901.jpg

media/egr20902.jpg


• Now, lets try a problem using vectors, ([Hibbeler, 1992], prob 5-83, pg. )

1. First we must construct a FBD. In this case it is essential to reduce the quantity of detail. The arm seems to be the principle structure here (it touches all of the parts) so it will be the first FBD.

8 ft

12 ft

10°A

BD

M = 800 lb

FCYL

FAx

FAy

14 ft

FBD Arm ABD:

2. If we find the sum of the moments about point A, then we can simplify the first step to one equation and one unknown.

MA∑ 800lb( ) 6ft( ) FCYL 10°cos–( ) 14ft( ) FCYL 10°sin( ) 12ft( )+ + 0= =+

FCYL∴ 6 800( )ft lb⋅10°cos( )14ft 10°sin( )12ft–

---------------------------------------------------------------------- 410lb= =

3. Finally, we can sum up the forces to find both components of the force at A.

Fx∑ FAxFCYL 80°cos+ 0= =+

Fy∑ 800lb– FCYL 80° FAy+sin+ 0= =

+

etc.....

media/stat0001.wm

8 ft

6 ft

4 ft

12 ft

x

y

z

A

B

C

D

The angled bar slides on a square col-lar. A force is pulling the bar in a downwards direction, but this is resisted by a rope between points C and B. Find the tension in the rope, and the reactions at A.

FD 20i 40j– 75k–( )lb·=

1. first, draw a FBD of the main arm. This is chosen because it is the main member. Then, define the force and moment vectors. The diagram is drawn in 2D, but it is still a full 3D set of vectors.

FA

MA

FCB

FA FAxi FAy

j+=

Because the collar at A can only translate along the z-axis (no rotation),

MA MAxi MAy

j MAzk+ +=

FCB FCBC B–C B–---------------- FCB

0 12–( )i 8 4–( )j 6 0–( )k+ +

122

42

62

+ +-------------------------------------------------------------------------

= =

Next find the rope force vector using the endpoints of the rope.

FCB∴ 0.86 FCB–( )i 0.29 FCB( )j 0.43 FCB( )k+ +=

FD 20i 40j– 75k–( )lb·=

2. Next we can find the sum of the moments about A. To do this in vector form we will need the displacement vectors between the centre of rotation, and the point, but since the point of rotation is the origin, and the dimensions are simple, there is no calcula-tions required.

MA∑ MA dAB FCB× dAD FD×+ + 0= =

MAxi MAy

j MAzk 12i( ) 20i 40j– 75k–( )×+ + +∴

12i 4j+( ) 0.86 FCB–( )i 0.29 FCB( )j 0.43 FCB( )k+ +( )×+ 0=

MAxi MAy

j MAzk

i j k

12 0 0

20 40– 75–

FCB

i j k

12 4

0.86– 0.29 0.43

+ + + +∴ 0=

MAx4 0.43( ) FCB+( )i MAy

12 75–( )– 12 0.43( ) FCB–( )j+∴

MAz12 40–( ) 12 0.29( ) FCB 4 0.86–( ) FCB–+ +( )+ 0=

3. find the tension in the rope by summing the forces. In particular, we can see that there are only two z-components of forces that will apply, therefore, sum forces in the z-axis.

Fz∑ 75lb–( ) 0.43 FCB( )+ 0= = FCB∴ 174lb=

MAx4ft 0.43( ) FCB+ 0= MAx

∴ 299ft– lb⋅=

MAy12 75–( )– 12 0.43( ) FCB– 0= MAy

∴ 2ft– lb⋅=

MAz12 40–( ) 12 0.29( ) FCB 4 0.86–( ) FCB–+ + 0= MAz

∴ 724ft– lb⋅=

************************* Solution in Mathcad ***************************

15.4 SPECIAL CASES

• There are a number of standard load cases that we will encounter. One of the simplest is the two force member discussed before.

[picture]

• Three force members can be solved quickly using force triangles, We can also keep in mind that the forces must act concurrently through a common point on the rigid body, or else all three must be parallel.

4. Using previously discussed techniques to find the reaction forces at A. etc.......

ASIDE: At this point you should expect to find many problems that will require that you find a few equations and then solve for an answer. Trying to avoid this will make many problems unsolvable and frustrating. Also keep in mind that the sum of forces and moments in 3D yields a total of 6 equations, allowing the solution of up to 6 unknowns.

READ

SLI 5.1-5.4SLI 5.5

PROBLEMS

SUGGESTED

6, 9, 1013, 20

REQUIRED

46(MC)

media/egr20940.jpg

[picture]

15.4.1 References

Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 6th edition, MacMillan Publishing Co., New York, USA, 1992.

15.5 STATICALLY INDETERMINATE

• There are a number of problems that cannot be solved with statics methods. In fact, all problems are inexact, but luckily we can solve some using statics.

• These types of problems often occur when the number of forces known is equal to the number of useful governing equations. This is known as completely constrained.

• When there are more equations than unknowns the problem may be overconstrained.

• When there are fewer equations than unknown forces, the problem may be underconstrained.

• The basic types of cases are,

F1

F2

F3

media/egr20941.jpg

• be cautious with this class of problems. You may actually be able to calculate a solution, but the answer will not be correct realistically, and mathematically speaking.

• Consider problem 2.4-1 from Hibbler

1. The defined problem does not have balanced forces, or improper supports, resulting in a dynamics problem.

2. There are redundant constraints, and the problem solution depends on materials proper-ties. (generally violates the rigid body assumption)

F M

F

M

F

F F

M M

F

F

Given: Aluminum cylinder with Brass Core

L 14.0in=

Dbrass 1.0in=

Dalum 1.6in=

Ebrass 15.0Mpsi=

Ealum 10.5Mpsi=

ε 0.001=Find:

1. Magnitude of P

2. Pmax σbrassmax18Kpsi=if

σalummax12Kpsi=

FBD and Statics

Deflections in terms of the Applied Forces

Relationship between the deflections

• Example #2

P

A B

x

Find x such that bar AB remains horizontal

A1 0.25in2

=

A2 0.6in2

=

L1 2ft=

L2 3.5ft=

• Example #3, problem 2.4-19 from Gere and Timoshenko

Given: Geometry shown below

Find: 1. the tensions in the cables in terms of P2. the vertical displacement of point D

1.5b

2b bP

A

BCD θAC θAB

P

BCD

FBD TAC TAB

15.5.1 Summary

• basic moments and calculations- using normal components- using perpendicular components- using vectors and cross products

Exaggerated Displacement

A

D

δAC δAB

δC δB

Note: The elongations of the cables are not the same as the beam.

READ

SLI 5.6-5.7

PROBLEMS

SUGGESTED

69, 80

REQUIRED

• a review of cross products, and the positive direction for angles/moments

• force couples and moments were shown to be able to freely move about a rigid body.

• a force can be moved away from its axis if a moment is also added.

• mechanical schematic symbols were discussed with applications to statics.

• indeterminate problems


1. Determine the reactions at A and E.

15.5.3 References



200mm

200mm150mm

200mm30deg

30degA

B C

D

E

500N

ANS.FAx = 37NFAy = 433NFE = 287N

16. TRUSSES AND FRAMES

16.1 INTRODUCTION

• We will look at a variety of analysis techniques for trusses and frames. These techniques endeavor to find the internal forces in trusses, and external forces for frames.

16.2 WHAT ARE TRUSSES?

• Trusses are at the heart of many engineering projects. We can see one of these in the bridge across the grand river.

[picture]

• Basically, a truss is a collection of beams joined together to carry simple and complex loads.

• We can see trusses use in cranes,[picture][picture]

• Trusses are typically made from beams that are joined with gusset plates.

e.g., a house roof truss- snow load- roof mass- wind load

FF

F

- bearing wall

media/egr20908.jpg

media/egr20920.jpg

media/egr20910.jpg

[picture]

• Other times obvious pin joints are used. Consider the pin joint on the end of the tension member below,

[picture]

• With these types of problems the tension or compression of the beams/members should be clearly indicated. Materials and structures will not fail at the same load when in tension (neck-ing then fracture), than in compression (buckling).

• We can see a tension member in a bridge with turnbuckles for tensioning,[picture]

• The basic assumptions used in most truss and frame problems are,1. the joints have pinned ends, so the forces exerted by the beam has a direction that is

along the line between pins.2. Forces are exerted at pins, but no moments.

• Types of trusses are shown below,

Beams are quite often joined with gusset plates. These can be seen where there are exposed steel structures, such as subway tun-nels, bridges, etc.

But, for all practical purposes, we treat these joints as if they are pinned joints.

media/egr20923.jpg

media/egr20911.jpg

media/egr20912.jpg

[picture]

PRATT

HOWE

FINK

media/egr20929.jpg

PRATT

HOWE

WARREN

BALTIMORE

PRATT

STADIUM

cantilever sectionof a truss

• Picture of other types of trusses can be seen below,[picture][picture]

• Many of the methods in this section can also be extended to the analysis of trusses in 3D.[picture]

16.2.1 References



16.3 STABILITY OF TRUSSES

• Trusses are composed of beams and pin joints. Generally, if the frame is constructed of only tri-angles (internally) it will be stable. (not collapse when pushed the wrong way)

• Quite often additional members are added to frames to stiffen it (make it stable), but not to carry loads. We can see such members in a bridge,

[picture]

• We can predict stability using a basic topological relationship where we look for equality. The equation shown below requires that the truss be composed of simple pinned beams, and when drawn to scale on paper the beams must not cross or touch except at the pinned joints

• We can also verify stability by making nodes that are held in place either by supports, or triangu-lar structures.

• As an example,

m 2n 3–≥ where,m = the number of beamsn = the number of joints

media/egr20924.jpg

media/egr20925.jpg

media/egr20925.jpg

media/egr20913.jpg

A

B

C

D

E

F

G

H

First, we will try the topological test, m = 14n = 8

14 2 8( ) 3–≠ 13=therefore this is not a simple truss (there isan extra beam).

Second, we can walk through the truss looking for stable substructures,

pick a triangle touching a support to start (fill it in to mark it as sta-ble

if another triangle touches it with two points, fill it in.

All of the joints in the struc-ture are touching the shaded area, therefore the structure is stable


16.3.1 References


16.4 THE METHOD OF JOINTS

• The basic steps when solving problems with the method of joints is,1. Draw an FBD for each joint as if it is a particle.2. Solve for the applied forces with the “joint particle” in equilibrium. The simplest joints

are often the best to start with.

• We can see an example of a two force member below being used to support a canopy,[picture]

• This method is best shown using a sample problem, ([Hibbeler, 1992], prob 6-7, pg. )

ASIDE: Two force members can greatly simplify problems. (As is done in the method of joints and section problems). In these problems there is typically a beam with a force at either end. These forces must be along the same line, either towards or away from each other. As a result these forces are described as tension or compres-sion along the line between the two pins.

θA θB

FA FB

Consider the case where θB is any angle other than 0° or 180°. This would result in a tangential component, the sum of the moments about A would not be zero, there-fore it would go into motion. Thus for a static frame θB = θB and FB = FB.

media/stat0002.wm

media/egr20909.jpg

A

B C

D

EF

3 KN

8 KN 4 KN 10 KN

1.5 m

2 m2 m

Find the force in each beam, indicating clearly tension or compression, and find the reaction forces at the supports.

ASIDE: with this type of problem you will learn to find easy places to start. The best place to start is often the supports, or joints that the smallest number of forces. When reviewing consider that by observation, beams AB, BC and FE can all be solved immediately.

NOTE: These things come with practice, more problems must be solved to gain expe-rience. And, experience will dramatically cut solution time.

1. In this case the support forces have to be found, and they are easy to identify, so we can determine those first. Please recall, the pinned joint at A will have x and y com-ponents, while the roller at E will only have a y component.

MA∑ 3KN 1.5m( ) 4KN 2m( ) 10KN 2m 2m+( ) FEy2m 2m+( )–+ + 0= =+

FEy∴ 4.5KN 8KN 40KN+ +

4------------------------------------------------------- 13.1KN= =

+

Fx∑ FAx3KN+ 0= =+

FAx∴ 3KN–=

Fy∑ FAy8KN– 4KN– 10KN– FE+ 0= = FAy

∴ 8.9KN=

FBD of entire truss:

A

B C D

E

3KN

8KN 4KN 10KN

2 m2 m

1.5 m

FEyFAy

FAx

2. We can now look at some of the joints and, by inspection, determine the forces in some of the members.In this case I am trying to eliminate many of the obvious forces. The most common of these is a zero force member, and full force members, where joints have members meeting at 90° angles. In this case the joints will be explained, but in most cases just stating the reason is sufficient.

8 KN

3 KN TBC

TBA

Fx∑ 3KN TBC+ 0= = TBC∴ 3KN–=+

Fy∑ 8KN– TBA– 0= = TBA∴ 8KN–=+

NOTE: all of the forces meet at right angles. For the forces in line what goes in one side, comes out the other. In other words, the force from the left is opposed by a force from the right, and the force from the top is opposed by a force from the bottom.

ASIDE: also note the convention for describing the forces. You will first notice that I am using a T instead of an F. This notation means that the beam is assumed to be in tension. A C could have also been used, indicating compres-sion. In very blunt terms, you can use any method you want, but consistency is of paramount importance, you are best off to pick one method and reli-giously adhere to it - it is the shortcuts and changes in method that will cause mistakes. Another point to notice is that TBC=TCB=-CCB=-CBC, but if FBC, or FCB is being used, the force direction must be defined someplace.

NOTE: because I am using T (tensions) for each beam, the force arrows are always away from the joints because a beam in tension is pulling the joint. When the result of calculations is negative, the beam is actually in compres-sion.

By the same reasoning as above,

TDE∴ FE– 13.1KN–= =

TEF∴ 0=

NOTE: EF is a zero force member, this is obvious if we use the case above (joint B), and consider that the force from the right is zero, therefore the force to the left will be zero.

B

[picture]

NOTE: zero force members might seem to have no purpose at first glance, but when building large structures these members are sometimes added to “stiffen” long beams and prevent buckling. (Zero force members can be seen in the structure below)

media/egr20934.jpg

3. So far we have a few obvious forces in the upper left and lower right corners, it can help to keep track of the forces in your mind, or on paper by checking off the forces that are known. The next action is to select the simplest joint to solve. Looking at the figure below (this figure is normally not drawn, but is used here for illustration) with known forces having thicker lines, I observe that there are three unknowns at joints C and F, and two unknowns at joints A and D. So, I arbitrarily select A to solve next.

A

B C D

E

3KN

8KN 4KN 10KN

FEyFAy

FAx

F

ASIDE: known forces to this point have darker lines

FBD A:

FAx

FAy

TAF

TAB TAC

3

4

5

A

Fx∑ FAxTAF

45---TAC+ + 0= =+

TAF∴ 3KN 45---TAC–=

Fy∑ TAB35---TAC FAy

+ + 0= =+

8– KN 35---TAC 8.9KN+ +∴ 0=

TAC∴ 1.5– KN=

TAF∴ 4.2KN=

4. At this point each of the remaining joints has two unknowns, so I arbitrarily select D as the next joint to solve for.

FBD D:

TDE

TCD

TFD3

4

5

D

Fy∑ 10KN– TDE– 35---TFD– 0= =

+

10KN– 13.1KN 35---TFD–+∴ 0=

Fx∑ TCD– 45---TFD– 0= =

+

TFD∴ 5.2KN=

TCD∴ 4.2KN=

10 KN

5. There is only one member (CF) still unknown. Therefore we can solve for either joint C or F. F has only 4 forces on the beam, as opposed to the five on joint C. Therefore I will solve for joint F.

FBD F:

TFE

TAF

TFD

3

4

5

F

Fy∑ TCF35---TFD+ 0= =+

TCF∴ 3.1KN=

TCF

• This problem can also be solved using matrices techniques by framing the truss members as lin-ear equations.

********************** Solve in Mathcad *********************


6. Finally, it is always a good idea (in this course is mandatory) to summarize the results so that when you start the next stages of design work, it is much easier to find the essential numbers.

Link/Support

ABBCCDDEEFFAACCFDFFAxFAyFEy

Force (Ten./Comp.)

8 KN (C)3 KN (C)4.2 KN (C)13.1 KN (C)04.2 KN (T)1.5 KN (C)3.1 KN (C)5.2 KN (T)3 KN (left)8.9 KN (up)13.1 KN (up)

A B

CD

1000lb

2000lb8ft

4ft

Use the method of joints to find the force in each member of the truss below.




READ

SLI 6.1-6.5

PROBLEMS

SUGGESTED

7, 16

REQUIRED

8(MC)

12 ft 12 ft

8 ft

6 ft12 ft6 ft

2000 lb 1000 lb

3. Find the forces in each of the beams of the structure below (indicate tension or compression). Assume all joints are pinned.

4. For the frame below, find the tension/compression in each member, under the loading condi-tions given, and assuming all joints are pinned. (Note: you must clearly indicate tension or compression when dealing with this type of problem)

16 ft

12 ft

800 lb

800 lb

12 ft

1m 1m

1m

2m

10KN

5KN

A B

C D E

F

ans.

ABACBCBECDCFDEDFEF

5KN(C)10KN(C)7.07KN(T)5KN(C)07.07KN(C)007.07KN(C)

(ans. EF=515lb(C), BC=515(C), CD=125(C), AB=774(T), AE=401(C), BD=1563(T), BE=625(C), DE=1289(C)

5. Given the frame below, find the tension/compression in each member.

3’ 3’ 3’

6’

6’

500lb 500lb

A

B

CD

E

F

ans.

ABBCCDDEEFAEBEBD

774lb(T)515lb(C)125lb(C)1289lb(C)515lb(C)401lb(C)625lb(C)1563lb(T)

0.5N 0.5N

A

B

I

C

I

E

F

D

GH

20mm

30mm

30mm 20mm10mm

10mm20mm

30mm

6. Find the forces in each beam of the bridge shown. (As usual, clearly indicate tension or com-pression)

7. Find the tension/compression in each member of the frame below, (HINT: look for 6 zero force members and use symmetry)

200KN

5m

5m

5m

10m 10m 10m 10m

A

B

C

D

E

F

G

H I

J K L

AB= -141KN

AJ= 100KN

BC= -141KN

BJ= 0

CD= -112KN

CH= 0

CJ= 50KN

DE= -112KN

DH= -60KN

DI= -60KN

EF= -141KN

EI= 0

EL= 50KN

FG= -141KN

FL= 0

GL= 100KN

HJ= -60KN

IL= -60KN

JK= 133KN

KL= 133KN

1ft

1ft

1ft

1ft

2ft 2ft

A

B

C

D

E

F

G

100lb

AB = 71 lb [C]AC = 0BC = 0BD = 71 lb [T]BE = 100 lb [C]CD = 0DE = 71 lb [T]DF = 0EF = 0EG = 71 lb [C]FG = 0

ANS.

16.4.2 References



16.5 THE METHOD OF SECTIONS

• Basically: cut out a part of a truss, and then treat it as if it is a rigid body. When done wisely, this allows simplified solutions. The alternative is using the method of joints to find all (or many) of the forces in the frame.

• Keep in mind that while moments are very popular with the method of sections, it can also be used for forces as well.

• Consider an example where we want to find forces in a structure, ([Hibbeler, 1992], prob 6-24, pg. )

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

4 m

4 m

4 m

4 m

In the truss shown find the forces in EF, BE and CB.

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

1. first, we must decide where to cut the section. A good rule of thumb to follow when deciding which section to cut is that the section line can cut through any joint, or member, but it can cut through at most one unknown member. In this case member BE could not be cut out without cutting through 2 members, but EF can be cut if the section line passes through one joint. The result of this sectioning is a FBD shown to the right of the page.

A

A

B

C

D

10 KN

10 KN

5 KN

E

5 KN

TEF

MB∑ 5KN 4m 4m+( ) 10KN 4m( ) 5KN 4m( ) TEF 4m( )+ + + 0= =+

TEF∴ 25KN–=

A

B

C

D

E

F

G

5 KN

10 KN

10 KN

5 KN

2. Continuing with the same idea, if I cut from joint C, through BE and EF, there will only be one unknown. Again, the diagram to the left is included only for illustra-tion purposes, normally it is omitted.

B

B

C

D

10 KN

5 KN

E

5 KN

TEF

MC∑ 5KN 4m( ) 10KN 0m( ) 5KN 4m( ) TEF 4m( ) 1

2-------T

BE 4m( )+ + + + 0= =

+

TBE∴ 21.2KN=

TBE

2

11


3. To find the force in CB we can cut the section as shown,

C

D

10 KN

5 KN

E

5 KN

TEF

Fy∑ 5KN– TEF– 1

2-------TBE– TCB– 0= =

TCB∴ 5KN=

TBE

2

11TCB

+

5KN–∴ 25KN–( )– 1

2------- 21.2KN( )– TCB=

1m1m1m1m1m1m

2m

A B C D E F G

P

H I J K L M N

In the frame below members DL and EL can support up to 1kip of tension or compression, find the maximum load P that may be applied. (Hint: you could mix the method of joints and sections)


1. Determine the tensions/compressions in members EF, JK and HJ for the bridge truss shown below. The method of sections is recommended.

READ

SLI 6.6

PROBLEMS

SUGGESTED

32, 48

REQUIRED

45(MC)

4m 4m 4m 4m 4m 4m

2m

2mA

BC

D

E F

GHJ

K

2kN 3kN 5kN 4kN 3kN

2. Find the forces in FG and DF (indicate tension or compression). Assume all joints are pinned.

3. In the frame pictured below, find the forces in DE, EF, DF and CB.

1m 1m 1m 1m 1m 1m 1m 1m

4m

2.5 kip

A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

ans. AH=1.77kip(C), FG=0)

15°

10°

30°

95”

105”

1000lb2000lb

150”

A

B

C

D

E

F

G

H

I ans.

ED=1970lb (T)EF=1532lb (C)DF=0CB=401lb(T)

4. Find the forces in members CD, CF, CG. (You will benefit most by using the method of sec-tions to solve this problem)

(ans. CD= 11.25KN(C), CF= 3.21(T), CG= 6.8(C))

5. Find the forces in JM and KM.

6. Find the tension/compression in members CE, DE.

2kN 4kN 4kN 5kN 3kN5m5m5m5m

2m

3m

2m

A B C D E

F

G

H

J

1m 1m

.5m

1m

1m

1.5m 1.5m20KN 100KN 20KN

A

B C

D

E

F

G H

I J

K

L N

M P

1m

JM= 83KN

KM= 14KN

16.6 ADDITIONAL TOPICS

16.6.1 Space Trusses

• If a truss is not planar, but still has two force members (note ball joints at ends and no internal forces) we can use the methods of joints, or sections to determine forces.

16.6.2 Compound Trusses

• Compound Trusses are made of smaller trusses.

• We can do a complete analysis of these simpler trusses and combine the solutions for larger problems.

1m

0.75m

1m 1m

1m 1m

A

B

C

D

E

F

G

1KNANS.

CE = .667 KN [T]DE = -.4 KN [C]

16.6.3 References



17. STRESS FAILURE

• We need to do an analysis of our designs for suitability. This involves looking at static forces in the design, and then analyzing the members to see if they will carry the prescribed loads.

• Although static values are usually quite accurate, the failure load for a member is hard to predict accurately, and impossible to predict exactly. This is because we are subject to the effects of geometry, material properties, unexpected loads, etc.

17.1 FACTOR OF SAFETY

• When we do basic design work we typically use a process of the basic format,

• In particular the design process often involves a factor of safety that allows for,

START: select aneeded design

determinefunctional elements

pick membersgeometry, etc.

select valuesand properties

analyze forsuitability

decide on

DONE: Approvedesign

deficiencies

major changes e.g.replace cable with beam

small changese.g. diameter

no deficiencies

e.g. a support cable

e.g. maximum tension

e.g 1/2” steel cable

e.g. find stress and compareto ultimate strength

e.g. use factorof safety

variations in material properties- variations in expectations load- the non-uniform distribution of stress- cyclic loading- etc

• The factor of safety is applied as,

• The values vary for various systems. The values given below are reasonable for static systems.

• For example let us consider a two member mechanism 1.44 from [Beer and Johnston]

F·S· σULTIMATE

σESTIMATED-----------------------------

PULTIMATE

PESTIMATED------------------------------= =

where,FS the factor of safety=

σULT ultimate shear stress before failure=

PULT ultimate load before failure=

σEST the working, normal, stress using assumed conditions=

PEST the working, normal load using assumed conditions=

FS < 1 for a system that should failFS = 1 for a critical system (likely to fail)FS = 2 for a reasonable designFS = 5 for a safe designFS >10 is overdesigned

*NOTE: many values are given in design tables

P = 8kips

A

B C

D

30°

12”

6”

w

given,FS = 3σULTIMATE 60ksi=

find the required width of BC.

ABC thickness( ) w( ) 14---in

w= =

σBC

σBCULTIMATE

FS--------------------------

TBC

ABC---------= =

TBC∴ 60ksi3

------------- w

4----

=

MD∑ TBC 12in 30°cos( )– P 18in 30°sin( )+ 0= =+

w4----

20ksi( )12in 30°cos∴ 8kip 18in 30°sin( )=

w∴ 4( ) 8kip( ) 18in 30°sin( )20ksi( ) 12in 30°cos( )

---------------------------------------------------------- 1.39in= =


1. The mechanism below is to be analyzed. Before answering any of the questions, read the notes, and study the diagram.

- The cable is 1/8 inch in diameter, and made of 6061-T6.- The upper bracket is made of cold rolled steel.- The pulley is made of nylon - assume it is frictionless.- The beam is made of oak.- The pulley is supported by a brass pin. There are brackets on both sides of the pulley.

READ

GT 1.7

PROBLEMS

SUGGESTED REQUIRED

a) Find the new length of the cable (assume the cable does not stretch on the pulley). Is the factor of safety high enough?

b) Is the factor of safety for stress failure in the beam high enough?c) What is the minimum pin diameter for the pulley?e) How thick should the upper brackets be for a factor of safety of 3?

17.2 REFERENCES


M = 1000Kg

A

A’

Section A-A’

150 120

150

120

all dimensions in mm

1200

600550

400

150

150

300R75

18. STRAIN FAILURE

• Reconsider the simple example below,

P = 10,000 lb

P


6”

If the rod is 6” before the load is applied, what is the new length? What load P would result in plastic deforma-tion? What load would result in rupture? What is the factor of safety?

• When a load is applied and removed in cycles the material may fatigue. This means the ultimate strength is effectively reduced. Design curves that can take this into account are found in hand-books, and look like those shown below,


• An isotropic material has the same properties in all directions. (materials such as fiber glass do not).


# cycles

σ ksi( )

60

50

40

30

20

10

103

104

105

106

107

108

109

1020 Hot Rolled Steel

2024 Aluminum

READ

GT 2.9

PROBLEMS

SUGGESTED REQUIRED

• In physical terms - as we stretch a bar, it becomes a bit thinner.


ν lateral∈

axial∈---------------------------=

where,

ν Poissons Ratio=


P = 10,000 lb

P


6”



• When we load a material in a single direction the effect of Poisson’s ratio is naturally included in Young’s Modulus. But, when there are multiple loads in multiple directions, we must uses Poisson’s ratio to determine how they intersect.


σy

σy

x

y

z

σx

σz

σx

σz

x∈σx

E-----

νσy

E---------–

νσz

E---------–=

y∈σy

E-----

νσx

E---------–

νσz

E---------–=

z∈σz

E-----

νσx

E---------–

νσy

E---------–=

*Note: use positive stressfor tension

18.3 STRESS CONCENTRATIONS

• Saint Venant’s Principle states that regardless of how a force is applied, when we move far enough away, the distribution becomes even. For example, if we are applying forces as point loads, they will have very high stress concentrations near the point of application. But, as we move away from the point of application, the force distribution evens out. Consider the pin in the hole where the pin applies a load of P.

The cubical gage block to the right is loaded on two faces with 1 kip of compression. The modulus of elas-ticity for the material is 1 Mpsi, and the poisson’s ratio is 0.3. What are the new outside dimensions?

1 kip

1 kip

0.2500 in.

0.2500 in.

0.2500 in.

• Stress concentrations are hard to predict, and this must often be done using experiments, Finite Element Analysis (FEA), or other techniques.

• During design we must pay attention to the stress concentrations. At some points the stress will be higher that the average stress.

P

Stress distribution is relatively evenfar away from the point load.

PP

P

• One technique for estimating maximum stress concentrations is to use tables derived experimen-tally.

PP

PD

D d

KσMAX

σAVG-------------=

where,

K stress concentration factor=

σMAX the maximum stress estimate=

σAVG stress calculated with normal methods=

PP

d/2

d/2

r

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0r/d

3.0

2.5

2.0

1.5

K

Maximum stress

• Consider the sample problem below,

PP

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0r/d

2.5

2.0

1.5

1.0

KD

r

d

D/d = 2.0D/d = 1.5D/d = 1.25

D/d = 1.1

Maximum stress

10 KN 10 KN5”3”

R0.5”

Find the maximum stress that will be expected in this bar with a hole in it.

READ

GT 2.10

PROBLEMS

SUGGESTED

10-6, 10-7

REQUIRED

10-9(MC)

18.4 TORSION STRESS CONCENTRATIONS

• We can estimate the maximum stress for two shafts connected with a fillet using the following graph,

• What is the maximum stress in the shaft shown below,

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1r/d

2.5

2.0

1.5

1.0

K

D

r

d

D/d = 2.0D/d = 1.33D/d = 1.2

D/d = 1.1

TT

18.5 REFERENCES


0.25”

1”

The shafts shown are joined with a fillet of radius 1/16”. What is the maximum shear stress if a torque of 100 lb.in. is applied?

19. METHOD OF MEMBERS

19.1 INTRODUCTION

• Sometimes we must deal with structures that do not have simple two-force beams, in this case we must use the method of members.

[picture]

• These structures are commonly called frames, referring to the fact that they have at least one member that has more than two forces. We can see some examples of these members in the picture below,

[picture]

• In basic terms we are just making good use of free body diagrams, and quite often solving para-metric equations. Consider how we could isolate the free body diagrams in the figures below.

[picture]

• If we were to assume that beams in a truss have a mass, then we would have to use the method of members to solve the problem.

• A sample problem is given to illustrate the method, ([Hibbeler, 1992], prob. 6-60, pg. )

media/egr20906.jpg

media/egr20931.jpg

media/egr20930.jg

9”

9”

18”

A

B

C

24”

P30°

30°

30°

70 lb.(comp.)

The air pump shown is used to inflate bicycle tires. In effect a force P is applied by standing on the top. This in turn causes the lever to push on the cylinder which is compressed. The cylinder must be compressed with at least 70 lb. before the pump will work. What is the minimum force P?

D

1. First, we must select a suitable part (member) of the assembly to start with, and draw an FBD of it. Looking at the question, and the structure, force P touches the lever, and so does the cylinder, so this seems to be the reasonable part to start with.

9”

9”

18”

P30°

30°

30°

FCYL 70lb·=

θ

FBx

FBy

FBD BAD:

2. Find the angle of application of the cylinder force. This just involves a bit of basic trigonometry.

θ 9in( ) 60°sin24in( ) 9in( ) 60°cos–

----------------------------------------------------- atan 21.8°= =

60°

24”

9”θ

θ

• Now, lets consider a practice problem,


3. Next, the only unknown left of the drawing is P, and the reaction forces at B, so to reduce the problem to one variable, I will take the sum of the moments about B.

MB∑ P 9in 9in+( ) 60°cos 18in( ) 30°cos+[ ]=

70lb θcos( ) 9in 60°sin( )[ ]– 70lb θsin( ) 9in 60°cos( )[ ]– 0=+

P∴ 70lb θcos( ) 9 60°sin( ) 70lb θsin( ) 9 60°cos( )+9 9+( ) 60°cos 18( ) 30°cos+

------------------------------------------------------------------------------------------------------------------ 25lb= =

1 m

6 m3 m

2 m

4 m

1 m 2 m 1 m

10 KN/m

A

B

C

D

E

F

For the structure to the right, find the reaction forces in all of the joints. You can use any methods (or combination) that you find suitable

400 N

600 mm

250 mm

250 mm

375mm

A

B

D

C

Determine the reactions at A and B.

• Sometimes these problems are made easier with Mathcad,


1. In the figure below, determine the force exerted by pin B. The method of members will be most useful.

1m

.5m

2m 1m 2m 1m

50kg

A B C

D

E

F

G

H

A B C G

D F

E

D

C

E

F

G

Find the forces at all of the pins, and at the supports. (clearly indicate direction using FBDs)

READ

SLI 6.8-6.9

PROBLEMS

SUGGESTED

81, 92

REQUIRED

92(MC)

2. What are the reactions on each of the members (clearly indicate the results on FBDs)?

3. Calculate the x- and y-components of the force at D which member AD exerts on member DE. The deflection of the spring in the equilibrium state shown is 2.5 inches. The mass of the

R0.15m

50kg

D

0.5m

0.5m

0.5m

45°

45°

A

B

C

x

y

A

B

C

D

E

100N

1m 1m 1m1m

1m 0.5m

0.25m

(ans. Ax=150N, Ay=150N, Bx=0, By=50N, Cx=150N, Cy=0, Dx=0, Dy=150N)

members and friction are negligible.

4. The three member frame below is exposed to a load of L=60lb. The beams that the frame is made from weight 8lb/ft. Find the reaction at the base, and find the reaction at each joint and in each member.

4’ 4’ 8’

8’

2’ 2’

K=100lb/in.

750lb

A

B C D

E

F

G

H I (ans. 70.7N in both)

5. For the structure below, find the reaction forces in pins B and E.

3’

3’

4’

5’60lb

A

B

C

D

E

3’ 3’

x

y

ans.

Cx=164lbBy=351lbCx=164lbCy=224lbDx=164lbDy=413lbFEx=0FEy=261lbME=655lb ft

1 m

6 m3 m

2 m 4 mA

B

C

D

E

F

40KN

6. Find the forces acting on DCBA at pin D,

7. We work for a maker of hand tools, and today we are evaluating a new design for vice grip pli-ers. In the configuration pictured we want a gripping force of 50 lb. between the jaws. How much force P must be applied at the handles to achieve this?

6 in

4 in

2 in

1 in

1 in

3 in 2 in

5 kip

A

B

C

D

E

F

G

ANS.35.4 kip <135°

P

P

50lb.

0.5” 5”

0.5”

2”

2.5”

1.5”

ANS. P = 12.9 lb.

19.2 REFERENCES

Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 6th edition, MacMillan Publishing Co., New York, USA, 1992.

19.3 SUMMARY

• A simple way to differentiate between the different methods is,

• You are encouraged to mix and match methods in any way that will simplify a solution. For example, in a couple of cases a problem that is being solved by the method of sections could easily use the method of joints in a couple of places.

19.4 REFERENCES


Method

Method of Joints

Method of Sections

Method of Members

Use

Suitable for solving entire structures when we have trusses of beams with two pinned joints.

Allows forces and reactions to be determined in trusses (as above) without solving the entire truss.

For more complicated mechanical systems. These struc-tures will have members that may have more than 2 forces or moments on the beam.

20. INTERNAL FORCES IN MEMBERS

20.1 INTRODUCTION

• Up to this point beams have been treated as completely rigid, in truth they are not.

• Even a simple beam with only two forces applied can have a variety of forces and moments internally. These include,

Normal force - This is the tension/compression force along the axis of the member.Shear force - This is the force acting across the axis of the member.Bending moment - This moment attempts to bend the beam.

• To start, we need to know the forces and moments acting on the beam, as well as the geometry of the member. The forces on the beam are often found using techniques such as the method of members. In the example below the beam is simply supported.

P

P

Ra Rb

1ft 2ft 2ft 1ft

P=200lbZ=400lb/ft

Ra =

Rb =

Z

• Internal forces and moments will be found by cutting a beam at the point of interest, and using the internal forces as a reaction to external forces. Note that these can be found at any location in the beam as they vary between pins and other supports.

• We can calculate the magnitudes of these forces and draw them on shear force and bending moment diagrams.

If we cut the beam for the section for 0 <= x <= 1ft the FBD is,

x

Ra

Note: a force ‘V’ and a moment ‘M’ are shown at the right side of the FBD to represent the internal shear force and moment that the rest of the solid must exert to remain a sin-gle body. There would also be an normal force if there were horizontal forces.

For 1ft <= x <= 3ft the FBD is,

Ra

P


Ra

P


Ra

P

Z

Z

V

M

V

M

V

M

V

M

• To review the basic procedure is:

• Consider the example below.

V

M

x

x

Note: you can check your answers using the rela-tionship:

dMdx-------- V=

1. Solve for the reactions.2. Cut the beam between each force and moment.3. Show internal force and moments on FBDs and solve.4. Do not replace distributed loads with point loads.5. Move to the next beam section and repeat.

• Consider problem 7.12a) in [Beer and Johnston],

P=300lb/ft

L=8ft

A B C

D

a a

a

P

45°

First, lets try to find the internal forces and moments between points B and D on the beam. First we need to draw a complete FBD of the member, and find the reaction forces.

FDx

c

b

P

FDy

FAx

FAy

F E

FDxFD 45°cos

FD

2-------= =

FDyFD 45°sin

FD

2-------= =

FDxFDy

FD

2-------= =∴

FAxFDx

–= (by insp.)

MA∑ a P( ) a FDx( )– 2a FDy

( )– 0= =+

P∴ 3FDx3FDy

= = FDx∴ FDy

P3---= = FAx

P3---=∴

Fy∑ P– FAyFDy

+ + 0= =+FAy

∴ Pp3---–

2P3

-------= =

Next, lets cut the beam at an arbitrary point, and draw the FBD. We will add in the internal forces at this point as if they are now external loads.

FDx

b

FDy

a

FEx

FEy

ME Fy∑ FEyFDy

+ 0= =

FEy∴ P

3---–=

(At all points)

+

Fx∑ F– ExFDx

+ 0= =

FEx∴ P

3---=

(At all points)

+

ME∑ ME a FDx( )– b FDy

( )– 0= =+ MEaP bP+

3--------------------= b 0 a,[ ]∈

As we can see, the internal force FEy (called shear) is constant, as is the tension/compres-sion FEx. But, we see that the bending moment ME varies up to a maximum value of -0.666aP at the point where load P is applied.

************* Solve with Mathcad *******************************

• We can be a little more formal when describing the applied forces. Consider a small part of a beam as shown below. The external shear and bending moments are indicated. Both the shear force V and the bending moment M are drawn in equilibrium. And by convention, both are shown as positive. This is best remembered by looking at the reactions on the left hand side. If we are to draw the internal forces at some point in the beam they would look like those below.

Now consider the internal forces between points A and B.

P/3

a

P/3

a

FFx

FFy

MF Fy∑ FFyP–

P3---+ 0= =

FFy∴ 2P

3-------=

(At all points)

+

Fx∑ F– Fx

P3---+ 0= =

FFx∴ P

3---=

(At all points)

+

MF∑ MF P c a–( ) aP3---– cP

3---–+ 0= =+ MF P

4a3

------ 2c3------–

= c a 2a,[ ]∈

ASIDE: Consider that these forces and moments calculated will eventually be used to determine how much the member bends, stretches, and how it fails

P

c

RECALL: When we look at these beams we can have only 3 or fewer unknowns for the problem to be statically determinate.

• It can be useful to draw diagrams for both shear and bending moments. These can be done easily by calculating the internal forces for an arbitrary point on the beam.

• As an example lets consider problem 7.38 in [Beer and Johnston]

M

V

M

V

M

V

M

V

F F

FF

50 lb 100 lb 50 lb

10”

8”

6”10”

10”

6”

4”

10

712.2

A B

C

D

E F G

HFHx

T

FHy

MH∑ 50lb 26in( )– 1012.2----------T

4in( )– 712.2----------T

20in( ) 100lb 10in( )– 50lb 6in( )+ +=

T∴ 244lb=

+

Fy∑ 50lb– 712.2----------T 100lb– FHy

50lb–+ + 0= =+ FHy∴ 60lb=

Fx∑ 1012.2----------T FHx

+ 0= =+ FHx∴ 200– lb=

We can draw the beam in free body form,

A E

F

G B

50lb 100lb 50lb

60lb

712.2----------T 140lb=

1012.2----------T 4in( ) 800lb in( )=

FHx8in– 1600lb in( )=

We are now ready to start drawing the shear force, and bending moment diagrams. To do this we will draw a FBD for the beam left of the joint, and continue across to the right (keeping in mind the conventions for positive values). It is commonly not done, but full equations will be developed to illustrate this solution.

A

50 lb

L

V

M

x

x [0 6in ),∈

ML∑ 50lb x( )– M– 0= = M∴ 50xlb–=+

Fy∑ 50lb– V– 0= =+ V∴ 50lb–=

We then look at the next segment of the beam,

50lb

140lb

800lb(in)

L

V

M

x

x 6in 16in,( )∈

ML∑ 50lb x( )– 140lb x 6in–( ) 800lb in( ) M–+ + 0= =

M 40lb in( )– 90lb x( )+=∴

Fy∑ 50lb– 140lb V–+ 0= =

+

+ V∴ 90lb=

• NOTE: don’t section at a point load or couple on a beam. At these points the function is unde-fined. Instead section just before or after.

• You may recognize in the last example that the slope of the bending moment diagram was the corresponding value on the shear force diagram.

This continues across the beam, but notice two details seen so far,

i) the shear force is the slope of the bending moment diagram.ii) applied moments simply cause a jump in the bending moment graph.

Keeping these points in mind, let us proceed to the shear force graph first.

V(lb)

90

50

-10

-50

6” 16” 26” 32” x

NOTE: As a check the values at both ends should go to zero.

x

M(lb in)

1400

1300

500

-300

SHEAR FORCE DIAGRAM

BENDING MOMENT DIAGRAM

• Relationships between loads, shear forces and bending moments.

• Consider a simpler technique for developing shear force and bending moment diagrams. First draw the shear force diagram from left to right. The direction of the forces should be added directly to the diagram. Then using the values on the shear force diagram, draw the bending moment diagram.

dMdx-------- V=

Differential elements subjected to distributed load

Summing the forces in the vertical direction yields:

Summing the moments about point O yields

Application:

Only where distributed or no loads are applied.

Integrating yields:

• Solve the following example using the relationship between shear and bending moments,

********************* Do above problem using Mathcad *****************************

• When considering distributed forces, you can lump the force to find reactions, but you must not lump the forces when drawing the shear force diagrams.

12ft

1ft 1ft 3ft

3ft100lb 200lb/ft

900lb

200lb.ft.

5100lb.ft.

A B C D EF


1. a) Find the forces in the tension members HK and JL below.

b) Draw the FBD and then the shear force and bending moment diagrams for the straight section of the beam from K to L.c) Find the centroid and second moment of inertia about the centroid for the beam from K to L. The section is shown below.

READ

SLI 7.1-7.2SLI 7.3-7.4

PROBLEMS

SUGGESTED

4, 1721, 22

REQUIRED

33(MC)

1m1m 2 m 0.7m0 .7m

10 00 K g

H

K

J

L

M

A

A’

d) What is the maximum compressive and tensile stresses and strains in member KL caused by shear and bending? What is the factor of safety for tensile failure if the beam is made from bronze?e) What is the maximum stress in members HK and JL?

f) Find the tensions and compressions in members EH, DF, DG. Notice that the members ana-lyzed in earlier parts of the problems now apply forces at pins H and J. (20%)

1cm

10cm

2cm

7cm

Section A-A’

7cm

10cm

5cm

Notes:1. This part is symmetrical from top to bottom and left to right. The holes are centered.2. The part is 3mm thick.3. The hole diameters are 1cm, and the fillets at the step in the part have a radius of 1cm.

2. a) Find the forces in the tension members HK and JL below.

b) Draw the FBD and then the shear force and bending moment diagrams for the straight section of the beam from K to L.c) Find the centroid and second moment of inertia about the centroid for the beam from K to L. The section is shown below.

1.5m

1.5m

4m4m4m

1.5m0.5m

A

B

C

D

E

F

G

H J

K L

M

1m1m 2 m 0.7m0 .7m

10 00 K g

H

K

J

L

M

A

A’

d) What is the maximum compressive and tensile stresses and strains in member KL caused by shear and bending? What is the factor of safety for tensile failure if the beam is made from bronze?e) What is the maximum stress in members LN and MP?

f) Find the tensions and compressions in all members. Notice that the members analyzed in earlier parts of the problems now apply forces at pins H and J.

1cm

10cm

2cm

7cm

Section A-A’

7cm

10cm

5cm

Notes:1. This part is symmetrical from top to bottom and left to right. The holes are centered.2. The part is 3mm thick.3. The hole diameters are 1cm, and the fillets at the step in the part have a radius of 1cm.

20.3 REFERENCES


4m

4m8m

A

B H J

K L

M

C

D

21. MOMENTS OF INERTIA

21.1 INTRODUCTION

• The basic techniques used for centroids are similar (not identical)

• For centroids we assume the force of gravity was uniform over the face. When the forces are not uniform, but are a function of the distance from the centroid, then we need to use moments of inertia.

21.2 STRESSES IN BEAMS

I. Last lecture we discussed the forces in a beam, today we will relate those forces to the stresses in the beam. These stresses can be used to predict failure of the beam or to size the beam to prevent failure.

dY

ρ

ds dx≅

ρ radius of curvature=

K1ρ---=

From the equation s=rY we obtain

ds ρdY= k1ρ--- dθ

ds------= = (1)

Since the radius of curvature is so large ds can be approximated by dx yielding

1ρ--- dθ

dx------= (2)

e fy

The dotted line represents a plane where the length of the beam does not changewhen the moment M is applied. This plane is called the neutral axis. y is the distanceabove the neutral axis.

Planes above the neutral axis: _________________________________________

Planes below the neutral axis: _________________________________________

Applying s = rY to the length of ef yields:

ef ρ y–( )dθ= (3)

But from (2) we getdθ 1

ρ---dx=

Substituting this into (3) we get

ef dx yρ---dx–=

But, the original length of the line ef was dx then the strain is

εxchange in lengthoriginal length

---------------------------------------

yρ---dx–

dx------------- y

ρ---– ky–= = = =

Note: the expression is negative because the length of ef shrinks.

σx EεxEyρ

------– Eky–= = =

Using Hooke’s law to relate stress to strain we obtain the expression

(4)

The consequences of this equation are:1. The normal stress varies _________________________________________

2. The maximum normal stress occurs _________________________________

3. The sign of the stress is ___________________________________________

But, how do we know where the neutral axis is located?

• The location of the neutral axis:

21.3 MOMENT CURVATURE IN BEAMS

• Consider the basic equation below.

Consider the cross section of the beam we have been analyzing

M

y

cross section

Because there is no net force acting parallel to the x-axis we can write

σx AdA∫ 0=

substituting in equation (4) gives

Kky AdA∫– 0 Ek y Ad

A∫= = (5)

To satisfy equation (5) we must ensure that

0 y AdA∫=

Tis will only happen if the y axis passes through the centroid of the cross sectionarea. Therefore the neutral axis passes through the centroid of the cross section area.

dM σx– ydA= (6)

Integrating yields:

M σxydA∫–=

(7)

Substituting from equation (4) yields:

M kEy2dA∫ kE y

2dA∫ kEI= = =

(8)

Where the moment of inertia is defined as,

I y2dA∫=

Rearranging equation (7) yields:

k1ρ--- M

EI------= =

This is known as the moment-curvature equation and it lads to a definition of apositive moment

Positive moment producing a positive curvatureRemember the sign convention use with internal forces?!?

Negative moment producing a negative curvature

MM

MM

Combining equation (4) in the form of kσx

Ey------–= with equation (8) yields:

σxMyI

--------–= (9)Flexure formula

where:M = the internal or applied momenty = distance from the neutral axisI = second moment of inertia about the axis

21.4 EVALUATING THE SECOND MOMENT OF INERTIA

• These can be evaluated using integration.


• Consider the next example below.

x

y

Ix y2

AdA∫=

Iy x2

AdA∫=

x

y

What is the centroid of this area?

Ix y2

AdA∫ y

2b yd

A∫ y

2b yd

h2---–

h2---

∫by

3

3--------

h2---–

h2---

h3b

12--------= = = = =b/2

h/2

-b/2

-h/2

What is the value of Iy?

What are the consequences of this equation relative to beam design???

Why are floor joist laid in the vertical direction?

21.5 MOMENTS OF INERTIAS B COMPOSITE AREAS

• We can use sums to find moments of inertias for given objects.

• Consider the fully symmetrical shape below.

Determine the value of Ix.

x

yx y

2=

Note: moments of inertias of common plane areas are presented in Appendix D pg. 868 in the textbook.

• The basic application of the parallel axis theorem is shown below.

If the centroids of all the shapes line up, then the

x

y

Ix IxouterIxinner

–=

moment of inertia is,

What happens if the axes do not line up?

x

y

We can use the parallel axes theorem to move them around.


O

Qx

y

x’

y’

Ix' Ix Ad2

+=

where,

Ix' moment of inertia about an axis that IS NOT through the centr=

Ix moment of inertia about an axis that is through the centroid=

d = distance between the x axes

A = area of the cross section

Note: the axes must be parallel

• Consider the example below

x

y

Suppose a beam had the cross section shown below. Determine the value for I that would be used to calculate the normal stresses due to a positive bending moment.

Thought Questions:

Where would the maximum tensile stress occur?

Where would the maximum compressive strength occur?

Could you apply stress concentration factors to this problem?

21.6 POLAR MOMENT OF INERTIA

• The polar moment of inertia may be simple to calculate with rotational parts, but it can also be found using a simple relationship (note: the polar ‘J’ and the ‘o’ for the origin)

x

y

6 in

1 in

1 in 7 in

Find the Second Moment of Inertia about the x and y axes for the beam section shown below.

• Evaluate the polar moment of inertia ‘J’ for the shape below.

Ix y2

Ad∫=

assume A=dA is the area at one point at y=Kx from the x-axis

Kx

Ix

A----=

Ix Kx2A=

likewise, y is found the same way, and for the polar moment we simple use a

Ky

Iy

A----=

Finally, we relate the radius to the x,y position,

Ko

Jo

A-----=

Ko2

Kx2

Ky2

+=

Jo

A-----

2 Ix

A----

2 Iy

A----

2

+=

radius,

Jo Ix Iy+=

Radius of Gyration

The radius of gyration allows us to lump the area (i.e. mass) as a sin-gle particle, at a given radius, the radius of gyration.

J ρ2Ad∫=

dA =


21.7 REVIEW OF BASIC CALCUATIONS

• When we considered the centroids/center of mass we assumed that the force acting on each chunk was the same (uniform gravity).

• If we have a force that varies over a surface linearly we use the second moment of inertia.

x

y

6 in

1 in

1 in 7 in

Find the Polar Moment of Inertia about the origin for the beam section shown below.

• Consider a simple case of a beam (not quite accurate). The material in the beam acts as small springs. As the beam is bent the material in the centre is almost neutral, but towards the sides it is compressed or stretched.

• If we consider the beam above as a continuous force, instead of distributed springs it changes the

MKS1

KS2

KS3

KS4

KS5

4m

δ1 0.01m=

A

δ5 0.01– m=

F1 KS1δ1 0.01KS1= = + MA1 F1 2m( )– 0.02KS1–= =

F2 KS2δ2 0.005KS2= = + MA2 F2 1m( )– 0.005KS2–= =

F3 KS3δ3 0.0KS3= = + MA3 F3 0m( ) 0= =

F4 KS4δ4 0.005– KS4= = + MA4 F4 1m( ) 0.005KS4–= =

F5 KS5δ5 0.01– KS5= = + MA5 F5 2m( ) 0.02– KS5= =

+ MA∑ M 0.02KS1– 0.005KS2– 0.005KS4– 0.02KS5– 0= =

Assume all spring constants are equal to Ks,

M 0.05KS=

Note: the moment effect increases by the distance squared because of the varying force. In this case the moment was balanced about the neutral axis (centroid) of the beam.

A SPRINGAPPROXIMATION

solution (by analogy you might recognize that with centroids we could lump forces, but with second moments we cannot).

M

4m

δ4 0.01m=

A

δ0 0.01– m=

y

F(y)

F δ( ) K δ( )=

First we can find a relationship for the force exerted by the material, assuming a linearchange in force,

where,

delta = the deflection of the material from neutralK = a spring constant of the material (per unit length)F = the force exerted by the material (per unit length)

F y( ) K y4--- δ4 δ0–( ) δ0+

=

where,

y = the distance from the bottom of the beamK = a spring constant of the material (per unit length)F = the force exerted by the material (per unit length)

Next lets find the effective moment. Integration will be required here,

MA∑ M y( ) F y( )( ) yd∫– 0= =+

M yK y4--- δ4 δ0–( ) δ0+

yd

0

4

∫ K y2

4----- δ4 δ0–( ) yδ0+

yd

0

4

∫= =

M K y3

12------ δ4 δ0–( ) y

2

2-----δ0+

0

4

K 6412------ δ4 δ0–( ) 16

2------δ0+ 0.0267K= = =

• Another example of a linearly varying force is hydrostatic pressure based on water depth. Basi-cally we add the mass of the water directly above to find the pressure on an area.

• The general forms of moments resulting from linearly varying forces are,

• We generally calculate second moments of inertia using,

• as an example we can consider a triangle. The calculations will be to find the second moments of inertia about the x and y axis - not the centroid.

M K y2

Ad∫=

Beams

where,M = moment in pure bending (assuming a deflection of 1m for every 1m of y)K = spring coefficient for the material per unit areay,dA = geometrical values

M γ y2

Ad∫=

Fluids

where,M = moment caused by fluid over varying depthy = depth in waterdA = area element of submerged surfacegamma = specific weight of fluid

Ix y2

Ad∫=

Iy x2

Ad∫=

Jo r2

Ad∫=

where,Ix, Iy = the rectangular moments of inertia for planar problems,

where the value is about a line parallel to the x and y axes respectivelyJo = the polar moment of inertia for a rotational second moment

• After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem.

a

a/2

b

x

y

Ix y2a b y–

b-----------

yd0

b

∫ ay

2

b----- y

3–

yd0

b

∫ ay

3

4b------ y

4

3-----–

0

bab

3

12--------= = = =

Iy x2b

xa2---–

a2---

-----------

xd0

a2---

∫ x2b a x–

a2---

-----------

xda2---

a

∫+ x2b 2x a–

a---------------

xd0

a2---

∫ x2b 2a 2x–

a------------------

xda2---

a

∫+= =

2bx4

4a----------- abx

3

3a-----------–

0

a2---

2abx3

3a--------------- 2bx

4

4a-----------– a

2---

a

+ etc= =

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SUGGESTED

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REQUIRED

• Consider the application of the parallel axis theorem to the triangle seen before, To find the moment of inertia about the y centroid, when all we have is the y moment of inertia about the x axis.

I y2

Ad∫ y y'+( )2

Ad∫= =

where,

y distance from first axis to element=

y distance from first axis to centroid=

y' distance from centroid to element=

I y2

2yy' y'2

+ +( ) Ad∫ y2

Ad0

A

∫ 2y y' Ad0

A

∫ y'2

Ad0

A

∫+ += =

Ad0

A

∫ A=

y' Ad0

A

∫ 0 (becasue the distance is from the centroid)=

y'2

Ad0

A

∫ Iy

=

I y2

A( ) 2y 0( ) Iy

( )+ + y2A I

y+= = I d

2A I

y+=


a

b

x

y

Ixab

3

12--------=

yb3---=

Aab2

------=

Ix y2A I

x+=

ab3

12-------- b

3---

2 ab2

------ I

x+=

Ix

b3---

2 ab2

------ ab

3

12--------–

ab3

36--------= =

x

y

6 in

1 in

1 in 7 in

Find the Second Moment of Inertia about the centroid for the beam section shown below.

READ

SLI 9.5-9.6

PROBLEMS

SUGGESTED

22, 29

REQUIRED

• NOTE: when using the parallel axis theorem, the centroid should always be used as a reference.

• Like centroids, we can calculate moments of inertia for simple areas using weighted sums (a slightly different technique to finding centroids). The basic steps are,

1. For each simple shape find the moment of inertia about some global axis. This may require the use of centroids and areas to move the axis.

2. Add moments of inertia (or subtract for cut out areas.

• As an example, let’s consider a problem (9.26 pg. 64 Beer and Johnston)

x

y

R20mm

40mm

Find the 2nd moment of inertiaabout the x and y axes.

First we can break down the shape into primitive shapes, relative to the x-axis.

1

2

3

4

For area 1 about the x-axis,

Ixbh

3

3-------- 40( ) 20( )3

3------------------------ 107 10

3mm

4×= = =


x

x'

x

20mm Ix'πr

4

8-------- 63 10

3mm

4×= =

Aπr

2

2-------- 628mm

2= =

x x'–4r3π------ 8.5mm= =

20mm

Ix' Ix

x x'–( )2A+= I

x∴ 17.6 10

3mm

4×=

Ix Ix

x'( )2A+ 528 10

3mm

4×= =

Finally we find the combined moments of inertia using symmetry for areas 3 & 4.

Ix 2 107 103× 528 10

3×+( ) 1.27 106× mm

4= =


1

2

1

3

4

3

Next to find the second moment of inertia about the y-axis,

Iy 2πr

4

8-------- bh

3

3--------+

2π 20( )4

8---------------- 40( ) 20( )3

3------------------------+

0.34 106mm

4×= = =

x

y

6 in

1 in

1 in 7 in

Find the Second Moment of Inertia about the x-y axes using the composite bodies method.

READ

SLI 9.7

PROBLEMS

SUGGESTED

37

REQUIRED

43(MC)


21.8 PRODUCT OF INERTIA

• When we want to find the maximum moment of inertia of an area we can start by finding the product of inertia.

• This value can be either positive or negative.

• The value will be zero if either x or y is an axis of symmetry. This will allow us to find the prod-uct of inertia if we know the area and location of the centroid.

• If we want to find the maximum and minimum moment of inertia we can use the derived rela-tionship below. We can also find the axis of these maxima and minima,

• We can also shift the moment of inertia values to other arbitrary axes using the following rela-tionships,

Ixy xy AdA∫=

Or in a simpler form,

Ixy I˜x'y' xyA+=

Imin max,Ix Iy+

2--------------

Ix Iy–

2--------------

2Ixy2

+±=

2θtan2Ixy

Ix Iy–--------------=


Ix'

Ix Iy+

2--------------

Ix Iy–

2-------------- 2θcos Ixy 2θsin–+=

Iy'Ix Iy+

2--------------

Ix Iy–

2--------------– 2θcos Ixy 2θsin+=

Ix'y'Ix Iy–

2-------------- 2θsin Ixy 2θcos+=

• If we plot out the values as we rotate the axis, we generate a plot called Mohr’s force circle.

21.9 REFERENCES


x

y

6 in

1 in

1 in 7 in

Find the Second Moment of Inertia for the beam section shown below, about a line that lies 45° between the positive x and y axes.

22. PURE BENDING

22.1 INTRODUCTION

• Consider a section of beam that has two moments applied, thus inducing bending,

• Now consider the overall geometry of the beam,

M=0

θ

M > 0

M M M M

• Consider a cross section of the beam being bent about the z-axis,

θρρ y–( )

y

L ρθ=

L' ρ y–( )θ=

for the arbitrary height y above or

δ L' L– ρ y–( )θ ρθ– yθ–= = =

x∈ δL--- yθ–

L--------- yθ–

ρθ--------- y

ρ---–= = = =

Now considering the extreme dimension is ±c we can find a maximum strain,

MAX∈ cρ---=

x∈( ) yc-- ∈ MAX–=

And, assuming that stress varies linearly from the maximum at the outer limit to the centers

zero at the neutral axis,

σx

E-----∴ y

c--σMAX

E-------------–=

σx∴yσMAX–

c-------------------=

NOTE: We are assuming that the beam takes on a circular shape.

below the neutral axis (centroid).

Note: curvature is,

κ 1ρ---=

• Note that tip deflection for a beam can be found using,

• For the beam below find the maximum stress and radius of curvature,

y

c

c

neutral axis (centroid)

σdA

Mz yσy Ad∫=

Now consider the moment of the section,

M y–( ) σx( ) Ad∫ y–( )yσMAX–

c-------------------

Ad∫σMAX

c------------- y

2( ) Ad∫σMAX

c-------------I= = = =

σMAX∴ cMI

--------=

MAX∈ cρ---

σMAX

E------------- cM

IE--------= = = ρ∴ EI

M------=

δ L 1 θcos–( )=

A

A

Section A-A100 KNm 100 KNm

4m

0.3m

0.4m

The stainless steel beam above is under pure bending. If the beam has the triangular cross section shown, find the maximum stress, the radius of curvature, and the tip deflection.

22.2 TRANSVERSE SHEAR

• Consider the figure below when a split beam is loaded transversly.

• Consider an element in a beam.

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GT 5.1-5.6

PROBLEMS

SUGGESTED

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REQUIRED

4-6(MC), 5-17(MC)

Split in beamNO LOAD

P

WITH LOAD

To keep the ends and middle together a shear stress must be present.

P

Consider the element to the left. The bending moment gener-ated a distributed load across the face of the element. The bending moment is slightly higher to the left, and therefore the distributed forces are too.

If we draw only part of the beam element (in this case the top half, we can thensum forces.

V

F1 =

F2 =

Equilibrium of the forces in the x direction yields:

F3 F2 F1–=

F3 =

Shear stress equal to F3 divided by the area. Area equal to the width, b, time dx.

Symbolically:

τF3

A------=

A bdx=

Substituting we obtain

τ =

• We can find the shear stress in rectangular beams as shown below.

But, we know that

dMdx-------- V= Q =AND

Finally, we obtain the classic shear formula:

τ =

where:V =

Q = The first moment of the cross sectional area above the level at which the shear stress is being evaluated.

I =

b =

Note: shear stress is in the same direction as the shear force!!!!!!!

Cross Section of the beam

Calculating Q

1. Q =multiply area by the distance between the centroid of the area and the neutralaxis of the beam’s cross section.

2. Evaluate the Integral:

Substituting into the shear force formula yields:

• Consider the example below with the rectangular beam that is 4 inches wide and 8 inches tall. We want to find the shear stress 2 inches from the top.

• Try the problem shown below from Beer and Johnson 1992, Pg. 291 #5-1,

160 lb/in

22.3 REVIEW OF TRANSVERSE LOADING

• When loads are applied to beams they tend to deflect

2.5”

2.5”

2.5”

1.5”

1.5”

1.5”

3.5”

The end of the beam is loaded with a vertical shear of 250lb. This beam is made of a set of three 2x4 studs nailed together. Find the shear force in each nail.

READ

GT 5.8

PROBLEMS

SUGGESTED

8-3

REQUIRED

8-10(MC)

• For a beam under a transverse load the stress is a function of the distance from the centroid,

• We can consider the shear on some axis other than the neutral axis of the beam,

PP

P = 0 P > 0

y

P

x

L

y

CA

B

B’’

σxMy

I--------–

PxyI

---------–= =

M = Px(from before)

22.4 REFERENCES


H∴ PQI

-------- x qx= =

σxdA PxyI

---------dA–=

For an element some distance from the neutral axis the force on the element is,

Now we can integrate to find the shear H for the shaded part of the beam cross section,

Fx∑ 0 H PxyI

--------- Ad∫– H PxI

------ y Ada

b

∫– HPxQ

I-----------–= = = =

a

b

where,H = the total shear force for the section

section

Q = the first moment for the section (the centroid multiplied by the area)q = horizontal shear per unit length for the section (Note: H/x)

PQI

-------- q=

23. DRY STATIC FRICTION

23.1 INTRODUCTION

• Friction is a force that exists between any two object in contact. This can sometimes work against the engineer, other times it can be of great advantage.

23.2 THE BASIC PHYSICS OF FRICTION

• This natural phenomenon explains the resistance of one object to slide across another when they have common surfaces in contact.

• It is primarily the result of surface roughness, material properties, and if the object is moving.

• The graph of applied load versus friction helps illustrate the nature of friction. Notice that while the force is static, the force increases linearly up to the limit. After the object begins moving the force can be approximated with a constant value, using the dynamic coefficient of friction. Note that dynamic friction is shown to be lower that the maximum static friction.

If static on a dry surface we can approximate the maximum friction force using the equation,

FS µSN=N = Mg

FR

FS

µS a coefficient of friction for the two surfaces=

NOTE: coefficients of static friction are experi-mentally determined, and will typically be found using lookup tables. So me approximate sample values are, [Hibbeler, 1992, pg.347]

aluminum on aluminum 1.1-1.7copper on copper 1.2copper on steel 0.50glass on glass 0.90leather on metal 0.3-0.6leather on wood 0.2-0.5metal on ice 0.03-0.05metal on wood 0.20 - 0.60rubber on concrete 0.50-0.90rubber on ice 0.05-0.3steel on steel 0.75teflon on teflon 0.04wood on wood 0.25-0.7

ASIDE: this equation is more correctly written.

FS µSN≤

• The basic assumptions that we will use are,1. the maximum friction force is proportional to the normal force2. the maximum friction force is not proportional to the area of contact3. the static friction force is always higher than the dynamic friction force4. the kinetic friction force is independent of velocity

• A couple of the major applications for friction calculations is the determination if an object will slip or tip. The following problem shows a typical application, ([Hibbeler, 1992], prob 8-8, pg. )

F

N

FN

FS

F

FS

F=µsN

FS=µsN

static dynamic

µkN

rubbing slowly

high speed rubbing

2 ft

4 ft

4 ft

C

FG

θ

The car pictured is parked on an extreme slope, and we want to determine at what angle the car will tip over, or begin to slip if the coeffi-cient of friction is 0.4, and the mass of the car is 4000 lb.

1. First, for tipping, we want to find the angle that would place the centre of mass beyond one of the wheels. In other words, if the gravity vector does not pass between the wheels, then the car will tip over.

2 ft

4 ft

4 ft

C

FG

θ

θ 4ft

4ft( )22ft( )2

+---------------------------------------

asin 63°= =

• Consider the simple tip/slip problem below,[woking model file]

• The general approach to slip-tip problem is,1. Find the center of gravity for the object.2. Determine which corner the object is most likely to tip (as if the corner is a pin joint).

Sum the moments about the corner. If the sum of moments is equal to zero to block is about to tip. If not equal to zero look at the resulting moment to see if it will cause motion about the corner.

3. Find the component of the gravity and any other non-friction forces acting perpendicu-lar to the surface of contact. Find the components of applied forces acting parallel to the plane of contact.

4. Compare the actual parallel component to the maximum friction force possible. The the resultant is larger than the maximum the block will slip.


2. Next for slipping, we want to find the angle at which the component of the gravity force pulling the car downhill overcomes the maximum friction force.

FS µSFR 0.4FR= =FS

FR

θ

FGFS

FR

FG

FR FG θcos=

FS FG θsin=

FG∴ θsin 0.4 FG θcos( )=

θsinθcos

------------∴ 0.4 θtan= =

θ∴ 22°=

3. Based on the two numbers we can see that the lesser angle is for slipping, therefore the car will slip (at 21.8°) before it will tip (at 63.4°).

NOTE: the slip angle (or angle of fric-tion) can be found using,θ µSatan=

******************* Solve using Mathcad ********************

w = 50cm

h

θ

Block

Slope

Tilting

ANS.coeff. = 0.84h = 137 cm

We conduct an experiment using the 10 kg. block below on a slope that is being slowly tilted. The block tips over at 20°, and then stops moving, but then it starts to slip at 40°. What is the height ‘h’, and the coefficient of friction?


23.2.2 References



23.3 APPLICATIONS OF FRICTION

• When dealing with these problems the direction of friction forces must be assigned with care. If the directions are selected backwards, the solutions will be incorrect.

• Before assigning friction forces the impending motion should be analyzed. To do this think of the possible cases that might cause the bodies to start moving - do not assume that all friction surfaces must go into motion. At times this approach may mean that multiple solutions will have to be done to solve a problem.

• The general methods to be followed in these problems involves,1. Examine the problem to determine impending motion for each individual object, and

the overall system. There may be one or more possible cases, each will require a separate solution.

2. Based upon the assumed motion at the points of contact, drawn on friction forces that

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SUGGESTED

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oppose the motion. Also draw on normal forces.3. Solve the problem using normal statics (but avoid using sums of moments for friction

forces when they don’t act on a clear point).4. Examine the solution (and compare to others) for anomalies such as normal forces that

separate friction surfaces. This will help determine problems, and to eliminate unreasonable solutions.

23.3.1 Wedges

• Wedges are a useful engineering tool, and the approach used for wedges also finds its way into other engineering applications.

• A good rule to stick to is that when a wedge is in use, the forces on the faces will both be in the same direction. That is either towards, or away from the point of the wedge.

• When solving friction problems we look for friction that is about to let go and start slipping. Keep in mind that not all surfaces will slip, this should be verified after the solution. For all surfaces that slip the friction force will be at the maximum value.

• The example below shows how to deal with a multiple wedge problem. ([Hibbeler, 1992], prob 8-55, pg. )

P

FS1

FS2

N1

N2

P FS1

FS2

N1

N2Pushing the wedge in Pulling the wedge out

µAB 0.1=15°

15°

3000 lb

P

A

B

C

D

The two wedges are stacked as shown, and a load is applied. What is the minimum force P required to pull the bottom wedge out?

µAC 0.2=

µBD 0.2=

FSBD

1. These types of problems involve a large number of forces, and should not be attempted without FBDs. So, the first step is to draw FBDs of one of the wedges. The first wedge to be considered is B. When the bottom wedge slips out, B will push against the wall, and slide down, but A will try to pull it away from the wall.

FBD B:

15°

3000 lb

B FRBD

FSAB

FRAB

Fx∑ FSAB15°cos– FRAB

15°sin FRBD–+ 0= =+

µABFRAB15°cos–∴ FRAB

15°sin+ FRBD=

Fy∑ 3000lb– FSBDFSAB

15°( ) FRAB15°cos+sin+ + 0= =+

µBD µABFRAB15°cos– FRAB

15°sin+( ) FSAB15°( ) FRAB

15°cos+sin+∴ 3000lb=

FRAB∴ 3000lb

µABµBD 15°cos– µBD 15°sin+ µAB 15° 15°cos+sin+------------------------------------------------------------------------------------------------------------------------------------ 2929lb= =

FSAB∴ 293lb=



FSAC

2. Now that we know the force that wedge B applies to A, we are ready to begin finding the forces on wedge A. We can begin with a FBD.

FBD A:

15°

PA

FRAC

FSAB

FRAB

Fx∑ P– FSACFSAB

15°cos FRAB15°( )sin–+ + 0= =+

Fy∑ FSAB15°sin– FRAB

15°cos– FRAC+ 0= =+

FRAC∴ 2905lb=

P∴ 106lb=

FSAC∴ 581lb=

media/stat0006.wm

****************** Solve using Mathcad *******************

300mm

40mm

40mm

500mm

15°P

A

BC

D

500N

ANS. 216N

A vertical force, F, of 500N acts at one end of a bracket while a wedge is pushed against the other end, as depicted in the diagram below. Given that the weight of the wedge and the bracket can be neglected and that the coefficient of static friction for the contacting sur-faces of the wedge is 0.2, determine the horizontal force P required to push the wedge.


1. The angled bar below has two forces applied that are tending to push it in a counterclockwise direction. These forces are resisted by a wedge that is kept in place by friction (the coefficient of friction is 0.20). determine the force P that is required to pull the block out.


2. We are designing a firing mechanism for a new gun that uses two identical rails that are pressed together to accelerate a projectile. In the figure below we see the two rails at an angle before firing begins. When firing begins, the force ‘F’ will be applied, overcoming the coefficient of friction of 0.05. The length of the projectile is negligible, but it has a wedge shape that matches the rails before firing. What is the initial force ‘F’ that must be applied to the rails before the shot begins to move?

READ

SLI 8.3

PROBLEMS

SUGGESTED

73, 84

REQUIRED

3

45

A

B

C

6m 6m

3m

1m

8kN

15kN

P

1m15°

(ans. 6.55KN)

media/stat0007.wm

3. What force P will have to be applied to the wedge A to force the blocks B and C apart? You can assume that the coefficient of friction is 0.4 at all points of contact.

4. Find the force required to pull out the wedge below.

F

F

projectile motion

A

B

3”

1” 0.9”

6”

projectile

1”

P

75° 75°

A

B C

D

1 MN1 MN

ANS. P = 0.854 MN

5. Find the force required to pull out the wedge below.



23.3.2 Belt Friction

• Belts are a common tool for transmission of forces, motions and velocities.

• If we have a flat belt, it primarily depends on friction to hold it in place.

• The basic rules of static friction still apply for local friction between the belt and the drum, but over the length of the belt the effective normal force changes.

• If we consider one element of the belt we can see an element of friction and a differential of ten-sion.

P

A

B

C

10°

D

µAB 0.05=

µBC 0.02=

µCD 0.5=

MA 5Kg=

MB 12Kg=

MC 20Kg=

P

A

B

10°C

µAB 0.2=

µBC 0.5=

MA 10Kg=

MB 12Kg=

∆θ2

-------

∆θ2

------- ∆θ2

-------

∆θ2

-------

T ∆T+T∆FS∆FN

∆FS µS∆FN=

Fx∑ T∆θ2

-------cos – T

∆θ2

-------cos ∆T

∆θ2

-------cos ∆FS–+ + 0= =

∆T∆θ2

-------cos∴ ∆ FS=

+

Fy∑ T∆θ2

-------sin– T∆θ2

-------sin– ∆T∆θ2

-------sin– ∆FN+ 0= =+

∆FN∴ 2T∆θ2

-------sin ∆T∆θ2

-------sin+=

∆FS µS∆FN=

Next combine the equations to get a single expression,

First, find the equations for forces acting on the belt element,

∆T∆θ2

-------cos µS 2T∆θ2

-------sin ∆T∆θ2

-------sin+ =∴

Now find the differential form of the equation,

∆T∆θ2

-------cos µS2T∆θ2

-------sin µS∆T∆θ2

-------sin+=∆θ 0→lim

dT∴ µ S2Tdθ2

------ µSdTdθ2

------+ µSdTdθ2

------= =

0

dTT

------ µSdθ=

• An example of belt friction is given below (8.123, pg. 40, Beer and Johnston). In this problem the upper drum is moving slowly (this means the belt sticks with static friction), the lower drum allows the belt to slide (the belt slides with dynamic friction). We need to find the force W that will balance the 150lb load on the other side.

1T---

TdT1

T2

∫ µS θdθ1

θ2

∫=

Tln[ ] T1

T2∴ µ Sθ[ ] θ1

θ2=

T2ln T1ln–∴ µ Sθ2 µSθ1–=

T2

T1-----

ln∴ µ S θ2 θ1–( )=

T2

T1-----∴ e

µS θ2 θ1–( )=

• Consider the simple problem below,

θA

θB

θ

8”

Both drums have a radius of 2”.

150lb W

T1T2

µS 0.3=

µk 0.25=

θ 24---

asin 30° 0.524rad= = =

First we can find the various angles,

θA 0.524rad=

θB π 2 0.524( )+ 4.19rad= =

Next we can develop the equations for relative tensions in cables,

T1

W----- e

µkθA=

T2

T1----- e

µsθB=

150lbT2

-------------- eµkθA=

Finally, we combine the equations into a single expression,

W150lb-------------- W

T1-----

T2

T1-----

T2

150lb--------------

eµkθA( )

1–e

µsθB( ) eµkθA( )

1–= =

W150lb-------------- e

µsθB 2µkθA–( )=∴

W 410lb=∴

1 lb

A

B

FANS.2.57lb.

A belt is wrapped about drums A and B, both have a radius of 3”. How much force F will have to be applied to lift a weight of 1 lb. if the drums are not moving, and the coefficient of static friction is 0.2?

• V-belts use the same principle as flat belts, except the friction is increased by the angle of the sides.





α

T2

T1----- e

µSθα2---sin

-----------

=

BELT

PULLEY

READ

SLI 8.5

PROBLEMS

SUGGESTED

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REQUIRED

24. MASS PROPERTIES

• In addition to basic statics this topic will lead in to estimating strengths of beams, dynamic behavior, mass, etc.

• Important terms,

• A simple way to illustrate the moment arm effect of gravity about a centroid is by considering an exaggerated T shape that is supported by two cables, one at either end.

Centre of Mass - This will indicate where the effective gravitational force should be applied to take into account the moment arm effect.

gravity

Centre of Gravity - If the gravity vector is not uniform over an object the force exerted by a volume of mass will change, this takes that into account. Centre of gravity is commonly used to describe the point of application for the gravitational force.

Centroid - typically where distributed forces, such as gravity are applied (this is the same as the centre of mass, if the mass density and gravity is uni-form). This is the easiest property to calculate. This is also called the first moment of area.

1 m

1 m

1 cm

1 cm

total mass 200 kg

As shown, the cross section has two major sections, each of equal area (and mass of 100 kg), One section hangs beneath the left cable, The mass of this section will largely be held up by the lefthand cable, but the long section that stretches from left to right will be supported on both ends with about the same force. Therefore an intuitive estimate of forces would suggest that the left cable would support 150 kg, and the right cable would support 50 kg.

MIDDLE OF MASS - lets consider the effect if the middle of mass were used to apply the force of gravity to the two cables. First, the centre of gravity would be 1cm from the left of the section. This would be the same as lever shown below,

T1 T2

M∑ T1 1cm( ) T2 100cm( )– 0= =+

Fy∑ T1 T2 200Kg–+ 0= =+

centre of mass

T11cm

100cm------------------

T1+ 200Kg=∴

T1∴ 198Kg= T2∴ 2Kg=

These numbers are clearly INCORRECT, but in the next example we will see that the numbers are much closer.

INCORRECT!

• A note of interest, when the object is symmetrical the centre of mass and centroids will be the same. Another term of significance is the centre of gravity, but unless we have a part with a significant gravity gradient across it, this will be the same as the centre of mass. Unless dealing with very sensitive equipment, or astronomically significantly distances, the centre of gravity, and centre of mass can be considered equivalent.

24.1 CENTRE OF MASS

• This property is simply the point at which the mass on one side of the point is equal to the mass on the other side of the point.

• This property can be found with a summation of mass, weighted by distance.

CENTROID - The centroid uses area weighted (multiplied) by its distance from the centroid. As a result, the centroid is the natural point to use for the appli-cation of gravitational forces.

T1 T2

M∑ T1 25.5cm( ) T2 74.5cm( )– 0= =+

Fy∑ T1 T2 200Kg–+ 0= =+

centroid

T125.5cm74.5cm--------------------

T1+ 200Kg=∴

T1∴ 149Kg= T2∴ 51Kg=

These numbers do match the intuitive derivation of the cable tensions given before.

0.255m

xximi∑mi∑

------------------= yyimi∑mi∑

------------------= zzimi∑mi∑

------------------=

• A simple example can illustrate the calculation of this property. A 2D example is given, but extension to 3D is trivial.

• Consider the example of the T-shape from before,

x 6m 5Kg( ) 8m 6Kg( )+5Kg 6Kg+

--------------------------------------------------------- 7.1m= =

2 m

2 m3 m

3 m

6 kg

5 kg

(6m, 3m)

(8m, 2m)

x

yFind the effective cen-

tre of mass for the two plates shown.

y 3m 5Kg( ) 2m 6Kg( )+5Kg 6Kg+

--------------------------------------------------------- 2.5m= =

x

d1

d2 d3

F1

F2 F3

d1 x 0.012

-----------–=

F1 100kg=

d2x 0.01–

2--------------------=

F2 100kg x 0.01–100

-------------------- =

d3101 x–

2------------------- x+=

F2 100kg 101 x–100

------------------- =

MO∑ d1F1– d2F2– d3F3+ 0= =

O

+

x 0.012

-----------– 100kg( )– x 0.01–

2--------------------

100kg x 0.01–100

--------------------

– 101 x–

2------------------- x+

100kg 101 x–100

-------------------

+

x 0.012

-----------– – x 0.01–

2--------------------

x 0.01–100

-------------------- – 101 x–

2------------------- x+

101 x–100

------------------- + 0=

200x˜ 0.01–( )– x 0.01–( ) x 0.01–( )– 101 x– 2x+( ) 101 x–( )+ 0=

200x˜ 0.01+ x2

0.02x– 0.0001+ – x

2– 10201+ 0=

2x2

– 199.98x– 10201+ 0=

x 199.98–( )– 199.98–( )2 4 2–( ) 10201( )–±2 2–( )

------------------------------------------------------------------------------------------------------------------ 137 37,–= =

24.2 CENTROIDS

• This is also known as the geometric centre, or first moment of area.

• The fundamental definition for Centroids is given below, although a more efficient method for problems with simple geometries is discussed later.

• When finding centroids, look for symmetry, the centroids will be in the center of symmetrical sections.

x

x Vd

V∫

Vd

V∫

---------------= y

y Vd

V∫

Vd

V∫

---------------= z

z Vd

V∫

Vd

V∫

---------------=VOLUME

x

x Ad

A∫

Ad

A∫

---------------= y

y Ad

A∫

Ad

A∫

---------------= z

z Ad

A∫

Ad

A∫

---------------=AREA

x

x Ld

L∫

Ld

L∫

-------------= y

y Ld

L∫

Ld

L∫

-------------= z

z Ld

L∫

Ld

L∫

-------------=LINE

• A simple example of application for these relationships is given below,

h

dx

dA= (h) (dx) = y(x) dx

h = y(x)

ASIDE: Recall that to set up an integral we must pick a suitable element (slice). As we can see below a vertical rectangular slice is selected. We find an area for the slice by multiplying height by width. The bound for integration are taken in the differenetial (eg. dx) direction. We can use the same elements to find the y centroid.

0a

x

x'˜ Ad( )0

a

∫

Ad

0

a

∫------------------

x y x( )( ) xd

0

a

∫

y x( ) xd

0

a

∫---------------------------= =

x'˜

y'˜ h2--- y x( )

2----------= =

y

y'˜ Ad( )0

a

∫

Ad

0

a

∫------------------

y x( )2

---------- y x( )( ) xd

0

a

∫

y x( ) xd

0

a

∫-----------------------------------------= =

• Lets try a more practical problem using the centroid, ([Hibbeler, 1992], prob. 9-16, pg. )

xx 4 x 5–( )–( ) xd

5

7∫

2 2( ) 12--- 2 2¥( )+

------------------------------------------------ 16--- 9x x

2–

xd5

7∫

16--- 9

2---x

2 x3

3------–

5

7

5.889m= = = =

x

y

4 m

2 m

5 m 7 m

12 kg

Find the centroid, and cen-tre of mass for the x-axis of the object pic-tured. Assume that the 12 kg mass is evenly distributed by area.

First integrate for the centroid,

Next, use weighted sums of areas for the composite shape (covered later),

xxsquareAsquare xtriangleAtriangle+

Asquare Atriangle+----------------------------------------------------------------------------------------------------------=

x∴6m 2

3---

12Kg 5m 1

3--- 7m 5m–( )+

13---

12Kg +

23---

12Kg 13---

12Kg+------------------------------------------------------------------------------------------------------------------------------- 5.889m= =

*** Calculations for y are similar.

δ 5 lb

ft2

-------=

4 ft

A

B

The quarter of a circle steel plate to the left is to be supported by two cables. The tension in each cable is to be determined. The density per unit area is also provided to aid in the calculations.

1. In this problem it is obvious that the problem involves mass properties, so the basic properties will be calculated first. These include mass (M) and area (A). Area is used when finding the centroid and mass, and mass is needed to find the total force exerted on the cables.

A πr2

4--------- 12.57ft

2= =

M Aδ 12.57ft2

5 lb

ft2

-------

62.85lb= = =

2. Next, we are prepared to find the centroid, so we must set up the integral,

xx 16 x

2– xd

0

4∫

A----------------------------------------

23---

12---–

16 x2

–

32---

0

4

12.57------------------------------------------------------- 16 16–( )

32---

16 0–( )

32---

–3 12.57( )–

---------------------------------------------------------- 1.70ft= = = =

And, by symmetry,

y x 1.70ft= =

24.2.1 Finding Centroids Using Composite Shapes

• This method basically uses tables, or obvious centroids for basic shapes in a complex shape. By using a summation of centroids, weighted by areas/volumes divided by total area.

• A simple example for a two dimensional problem is given below, ([Hibbeler, 1992], prob. 9-48, pg. )

TA3. Finally, we are prepared to find the

tensions in the support cables. To do this, a free body diagram is drawn first, and since all of the forces are in line, a simple sum of moments is used.

TB

1.70 ft

4 ft

62.85 lb

+ MA∑ 62.85lb 1.70ft( ) TB 4ft( )– 0= =

TB∴ 27lb=

Fy∑ TA TB 62.85lb–+ 0= =+

TA∴ 36lb=

• Next, lets consider a problem using the composite method in 3D volumes, ([Hibbeler, 1992], prob 9-68, pg. )

15 mm

150 mm15 mm

150 mm

15 mm 100 mm

A

B

C

centroid

y

yAAyA AByB ACyC+ +

AA AB AC+ +-------------------------------------------------------------=

For the beam cross section shown, find the centroid.

First, it is obvious by symmetry that the x centroid will be in the centre of the beam, but the height will be somewhat more difficult to determine because of the lack of symmetry top to bottom. This method can actually be done in one step, but the beam must be divided into simple parts. I will divide this beam into three rectangular sections.

y∴15 150¥( ) 15

2------

150 15¥( ) 15 1502

----------+ 15 100¥( ) 15 150 15

2------+ +

+ +

15 150¥( ) 150 15¥( ) 15 100¥( )+ +-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------=

y∴ 16875 202500 258750+ +2250 2250 1500+ +

--------------------------------------------------------------------- 79.9mm= =

C

a

h

z

z

V113---πr

2h=

Given the 3D volume shown, find the cen-troid. This shape is essentially a cylin-der with a cone removed.

First, the only difficulty in finding the cen-troid is the height, the x and y centroids are at the centre by symmetry. To begin the solution we must look up the vol-umes and centroids for each element (look in the math section).

For the cone:

z134---h=

V2 πr2

h=

For the cylinder:

z2h2---=

zV1z1 V2z2+

V1 V2+-----------------------------------=

Then, calculate the centroid,

z∴

13---πr

2h

34---h

πr2

h h

2---

+

13---πr

2h πr

2h+

---------------------------------------------------------------------

h2--- h

4---–

1 13---–

------------- 38---h= = =

• An example of a rouhly distributed load can be seen below,egr20914.jpg


1. The 500 Kg plate below is to be used in an industrial machine. The basic shape is rectangular, but there is an angled arch cut in the bottom. Considering the centroid, and the applied force, determine the magnitude of the force (P) that is required to pull out the wedge if the both sides of the wedge have a coefficient of friction of 0.2.

ASIDE: We will often encounter distributed forces in real problems. For the convenience of calculation we want to simplify these to point forces. The distributed forces are typically given a force per unit length/area, and the total equivalent force applied can be found by integrating the force over length/area. We can also find the point to apply this concentrated force by integrating the force weighted by the area. (only force per unit length is shown, but for force per unit area the methods are similar)

x

F x( )

x

FR

FR F x( ) xd

L∫=

x

xF x( ) xd

L∫

F x( ) xd

L∫

-------------------------=

2. The plate below has a triangular hole. We are asked to select a base width ‘b’ for the triangle so that the reaction at roller B is 200N. The plate material weighs 1Kg per square meter.

3. We want to design two similar supports for a new sculpture. This means that the supports should be positioned to carry the same load. Find the distance of the second support from the first, and indicate the loads to be supported by each if the plate weights 1000kg per square

20°

15KN

20mm

30mm

60mm

40mm 50mm

100mm

P

M = 500Kgµ = 0.2

2m

4m

6m

4m

12m

b

AB

meter.

4. An object is perched on the far end of a lever and we want to determine if it will tip, or slip off, or remain in place if the coefficient of friction is 0.2. (Note: the 2m distance to the right is to the centroid of the block)

x

y

2cmd

y x( ) 0.1x2

3x– 100cm x 0 10cm,[ ]∈,+=

d= 5.62cm

5. The 500 Kg plate below is to be used in an industrial machine. The basic shape is rectangular, but there is an arch cut in the bottom. Considering the centroid, and the applied force, determine the magnitude of the force (P) that is required to pull out the wedge if both sides of the wedge have a coefficient of friction of 0.2.

object

2m2m

k=100KN/m

1m

1m

0.25m

0.5m

0.25m

0.75m

The object:

δ 5000 kg

m3

------=

θslip 11.3= θactual 8.83= θtip 49=

6. A thin flat plate is suspended by means of a screw eye and 2 cords, AC and BC, as depicted in the diagram below. Given that the plate has a mass of 30kg per square meter of surface area, determine the reaction force at point O and the tension forces in the cords.

7. Find the x-y centroid of the shape below which is essentially rectangular, except for the shape cut out of the bottom (the function is given).

20°

15KN

20mm

30mm

60mm

40mm 30mm

120mm

P

M = 500Kgµ = 0.2

A B

C

O

1m

0.6m

0.3m 0.3m

g=9.806 m/s2

8. What is the x-y centroid of the shape below if the plate is homogenous?

24.3 MOMENTS OF INERTIA

• When we considered the centroids/centre of mass we assumed that the force acting on each chunk was the same (uniform gravity).

x

y

1 m

1 m

yx

2

3-----= ANS.

x = 0.469 my = 0.550 m

x

y

0 2” 4” 7” 9”

2”

3.5”

6”

5.5”

ANS.

x = 4.6”y = 2.96”

• If we have a force that varies over a surface linearly we use the second moment of inertia.

• Consider a simple case of a beam (not quite accurate). The material in the beam acts as small springs. As the beam is bent the material in the centre is almost neutral, but towards the sides it is compressed or stretched.

MKS1

KS2

KS3

KS4

KS5

4m

δ1 0.01m=

A

δ5 0.01– m=

F1 KS1δ1 0.01KS1= = + MA1 F1 2m( )– 0.02KS1–= =

F2 KS2δ2 0.005KS2= = + MA2 F2 1m( )– 0.005KS2–= =

F3 KS3δ3 0.0KS3= = + MA3 F3 0m( ) 0= =

F4 KS4δ4 0.005– KS4= = + MA4 F4 1m( ) 0.005KS4–= =

F5 KS5δ5 0.01– KS5= = + MA5 F5 2m( ) 0.02– KS5= =

+ MA∑ M 0.02KS1– 0.005KS2– 0.005KS4– 0.02KS5– 0= =

Assume all spring constants are equal to Ks,

M 0.05KS=

Note: the moment effect increases by the distance squared because of the varying force. In this case the moment was balanced about the neutral axis (centroid) of the beam.

• If we consider the beam above as a continuous force, instead of distributed springs it changes the solution (by analogy you might recognize that with centroids we could lump forces, but with second moments we cannot).

• Another example of a linearly varying force is hydrostatic pressure based on water depth. Basi-

M

4m

δ4 0.01m=

A

δ0 0.01– m=

y

F(y)

F δ( ) K δ( )=

First we can find a relationship for the force exerted by the material, assuming a linearchange in force,

where,

delta = the deflection of the material from neutralK = a spring constant of the material (per unit length)F = the force exerted by the material (per unit length)

F y( ) K y4--- δ4 δ0–( ) δ0+

=

where,

y = the distance from the bottom of the beamK = a spring constant of the material (per unit length)F = the force exerted by the material (per unit length)

Next lets find the effective moment. Integration will be required here,

MA∑ M y( ) F y( )( ) yd∫– 0= =+

M yK y4--- δ4 δ0–( ) δ0+

yd

0

4

∫ K y2

4----- δ4 δ0–( ) yδ0+

yd

0

4

∫= =

M K y3

12------ δ4 δ0–( ) y

2

2-----δ0+

0

4

K 6412------ δ4 δ0–( ) 16

2------δ0+ 0.0267K= = =

cally we add the mass of the water directly above to find the pressure on an area.

• The general forms of moments resulting from linearly varying forces are,

• We generally calculate second moments of inertia using,

• as an example we can consider a triangle. The calculations will be to find the second moments of inertia about the x and y axis - not the centroid.

M K y2

Ad∫=

Beams

where,M = moment in pure bending (assuming a deflection of 1m for every 1m of y)K = spring coefficient for the material per unit areay,dA = geometrical values

M γ y2

Ad∫=

Fluids

where,M = moment caused by fluid over varying depthy = depth in waterdA = area element of submerged surfacegamma = specific weight of fluid

Ix y2

Ad∫=

Iy x2

Ad∫=

Jo r2

Ad∫=

where,Ix, Iy = the rectangular moments of inertia for planar problems,

where the value is about a line parallel to the x and y axes respectivelyJo = the polar moment of inertia for a rotational second moment

• The polar moment of inertia may be simple to calculate with rotational parts, but it can also be found using a simple relationship (note: the polar ‘J’ and the ‘o’ for the origin)

a

a/2

b

x

y

Ix y2a y b–

b-----------

yd0

b

∫ ay

3

b----- y

2–

yd0

b

∫ ay

4

4b------ y

3

3-----–

0

bab

3

12--------= = = =

Iy x2b

xa2---–

a2---

-----------

xd0

a2---

∫ x2b a x–

a2---

-----------

xda2---

a

∫+ x2b 2x a–

a---------------

xd0

a2---

∫ x2b 2a 2x–

a------------------

xda2---

a

∫+= =

2bx4

4a----------- abx

3

3a-----------–

0

a2---

2abx3

3a--------------- 2bx

4

4a-----------– a

2---

a

+ etc= =

• After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem.

Ix y2

Ad∫=

assume A=dA is the area at one point at y=Kx from the x-axis

Kx

Ix

A----=

Ix Kx2A=

likewise, y is found the same way, and for the polar moment we simple use a

Ky

Iy

A----=

Finally, we relate the radius to the x,y position,

Ko

Jo

A-----=

Ko2

Kx2

Ky2

+=

Jo

A-----

2 Ix

A----

2 Iy

A----

2

+=

radius,

Jo Ix Iy+=

• Consider the application of the parallel axis theorem to the triangle seen before, To find the moment of inertia about the y centroid, when all we have is the y moment of inertia about the x axis.

I y2

Ad∫ y y'+( )2

Ad∫= =

where,

y distance from first axis to element=

y distance from first axis to centroid=

y' distance from centroid to element=

I y2

2yy' y'2

+ +( ) Ad∫ y2

Ad0

A

∫ 2y y' Ad0

A

∫ y'2

Ad0

A

∫+ += =

Ad0

A

∫ A=

y' Ad0

A

∫ 0 (becasue the distance is from the centroid)=

y'2

Ad0

A

∫ Iy

=

I y2

A( ) 2y 0( ) Iy

( )+ + y2A I

y+= =

• NOTE: when using the parallel axis theorem, the centroid should always be used as a reference.

• Like centroids, we can calculate moments of inertia for simple areas using weighted sums (a slightly different technique to finding centroids). The basic steps are,

1. For each simple shape find the moment of inertia about some global axis. This may require the use of centroids and areas to move the axis.

2. Add moments of inertia (or subtract for cut out areas.

• As an example, let’s consider a problem (9.26 pg. 64 Beer and Johnston)

a

b

x

y

Iyab

3

12--------=

yb3---=

Aab2

------=

Iy y2A I

y+=

ab3

12-------- b

3---

2 ab2

------ I

y+=

Iy

b3---

2 ab2

------ ab

3

12--------–

ab3

36--------= =

x

y

R20mm

40mm

Find the 2nd moment of inertiaabout the x and y axes.

First we can break down the shape into primitive shapes, relative to the x-axis.

1

2

3

4


Ixbh

3

3-------- 40( ) 20( )3

3------------------------ 107 10

3mm

4×= = =


x

x'

x

20mm Ix'πr

4

8-------- 63 10

3mm

4×= =

Aπr

2

2-------- 628mm

2= =

x x'–4r3π------ 8.5mm= =

20mm

Ix' Ix

x x'–( )2A+= I

x∴ 17.6 10

3mm

4×=

Ix Ix

x'( )2A+ 528 10

3mm

4×= =

Finally we find the combined moments of inertia using symmetry for areas 3 & 4.

Ix 2 107 103× 528 10

3×+( ) 1.27 106× mm

4= =


1. Find the Second Moment of Inertia about the centroid for the beam section shown below.

24.4 PRODUCT OF INERTIA

• When we want to find the maximum maximum moment of inertia of an area we can start by finding the product of inertia.

1

2

1

3

4

3

Next to find the second moment of inertia about the y-axis,

Iy 2πr

4

8-------- bh

3

3--------+

2π 20( )4

8---------------- 40( ) 20( )3

3------------------------+

0.34 106mm

4×= = =

x

y

6 in

1 in

1 in 7 in

• This value can be either positive or negative.

• The value will be zero if either x or y is an axis of symmetry. This will allow us to find the prod-uct of inertia if we knoe the area and location of the centroid.

• If we want to find the maximum and minimum moment of inertia we can use the derived rela-tionship below. We can also find the axis of these maxima and minima,

• We can also shift the moment of inertia values to other arbitrary axes using the following rela-tionships,

• If we plot out the values as we rotate the axis, we generate a plot called Mohr’s force circle.

Ixy xy AdA∫=

Or in a simpler form,

Ixy I˜x'y' xyA+=

Imin max,Ix Iy+

2--------------

Ix Iy–

2--------------

2Ixy2

+±=

2θtan2Ixy

Ix Iy–--------------=

Ix'

Ix Iy+

2--------------

Ix Iy–

2-------------- 2θcos Ixy 2θsin–+=

Iy'Ix Iy+

2--------------

Ix Iy–

2--------------– 2θcos Ixy 2θsin+=

Ix'y'Ix Iy–

2-------------- 2θsin Ixy 2θcos+=

25. INTERNAL FORCES IN MEMBERS

• Up to this point beams have been treated as completely rigid, in truth they are not.

• Even a simple beam with only two forces applied can have a variety of forces and moments internally. These include,

Normal force - This is the tension/compresion force along the axis of the member.Shear force - This is the force acting across the axis of the member.Bending moment - This moment attempts to bend the beam.

• To start, we need to know the forces and moments acting on the beam, as well as the geometry of the member. The forces on the beam are often found using techniques such as the method of members.

• Internal forces and moments will be found by cutting a beam at the point of interest, and using the internal forces as a reaction to external forces. Note that these can be found at any location in the beam as they vary between pins and other supports.

• Consider problem 7.12a) in [Beer and Johnston],

A B C

D

a a

a

P

45°

First, lets try to find the internal forces and moments between points B and D on the beam. First we need to draw a complete FBD of the member, and find the reaction forces.

FDx

c

b

P

FDy

FAx

FAy

F E

FDxFD 45°cos

FD

2-------= =

FDyFD 45°sin

FD

2-------= =

FDxFDy

FD

2-------= =∴

FAxFDx

–= (by insp.)

MA∑ a P( ) a FDx( )– 2a FDy

( )– 0= =+

P∴ 3FDx3FDy

= = FDx∴ FDy

P3---= = FAx

P3---=∴

Fy∑ P– FAyFDy

+ + 0= =+FAy

∴ Pp3---–

2P3

-------= =

Next, lets cut the beam at an arbitrary point, and draw the FBD. We will add in the internal forces at this point as if they are now external loads.

FDx

b

FDy

a

FEx

FEy

ME Fy∑ FEyFDy

+ 0= =

FEy∴ P

3---–=

(At all points)

+

Fx∑ F– ExFDx

+ 0= =

FEx∴ P

3---=

(At all points)

+

ME∑ ME a FDx( )– b FDy

( )– 0= =+ MEaP bP+

3--------------------= b 0 a,[ ]∈

As we can see, the internal force FEy (called shear) is constant, as is the tension/compres-sion FEx. But, we see that the bending moment ME varies up to a maximum value of -0.666aP at the point where load P is applied.

• We can be a little more formal when describing the applied forces. Consider a small part of a beam as shown below. The external shear and bending moments are indicated. Both the shear force V and the bending moment M are drawn in equilibrium. And by convention, both are shown as positive. This is best remembered by looking at the reactions on the left hand side. If we are to draw the internal forces at some point in the beam they would look like those below.

Now consider the internal forces between points A and B.

P/3

a

P/3

a

FFx

FFy

MF Fy∑ FFyP–

P3---+ 0= =

FFy∴ 2P

3-------=

(At all points)

+

Fx∑ F– Fx

P3---+ 0= =

FFx∴ P

3---=

(At all points)

+

MF∑ MF P c a–( ) aP3---– cP

3---–+ 0= =+ MF P

4a3

------ 2c3------–

= c a 2a,[ ]∈

ASIDE: Consider that these forces and moments calculated will eventually be used to determine how much the member bends, stretches, and how it fails

P

c

RECALL: When we look at these beams we can have only 3 or fewer unknowns for the problem to be statically determinate.

• It can be useful to draw diagrams for both shear and bending moments. These can be done easily by calculating the internal forces for an arbitrary point on the beam.

• As an example lets consider problem 7.38 in [Beer and Johnston]

M

V

M

V

M

V

M

V

F F

FF

50 lb 100 lb 50 lb

10”

8”

6”10”

10”

6”

4”

10

712.2

A B

C

D

E F G

HFHx

T

FHy

MH∑ 50lb 26in( )– 1012.2----------T

4in( )– 712.2----------T

20in( ) 100lb 10in( )– 50lb 6in( )+ +=

T∴ 244lb=

+

Fy∑ 50lb– 712.2----------T 100lb– FHy

50lb–+ + 0= =+ FHy∴ 60lb=

Fx∑ 1012.2----------T FHx

+ 0= =+ FHx∴ 200– lb=

We can draw the beam in free body form,

A E

F

G B

50lb 100lb 50lb

60lb

712.2----------T 140lb=

1012.2----------T 4in( ) 800lb in( )=

FHx8in– 1600lb in( )=

We are now ready to start drawing the shear force, and bending moment diagrams. To do this we will draw a FBD for the beam left of the joint, and continue across to the right (keeping in mind the conventions for positive values). It is commonly not done, but full equations will be developed to illustrate this solution.

A

50 lb

L

V

M

x

x [0 6in ),∈

ML∑ 50lb x( )– M– 0= = M∴ 50xlb–=+

Fy∑ 50lb– V– 0= =+ V∴ 50lb–=

We then look at the next segment of the beam,

50lb

140lb

800lb(in)

L

V

M

x

x 6in 16in,( )∈

ML∑ 50lb x( )– 140lb x 6in–( ) 800lb in( ) M–+ + 0= =

M 40lb in( )– 90lb x( )+=∴

Fy∑ 50lb– 140lb V–+ 0= =

+

+ V∴ 90lb=

• NOTE: don’t section at a point load or couple on a beam. At these points the function is unde-fined. Instead section just before or after.

• You may recognize in the last example that the slope of the bending moment diagram was the corresponding value on the shear force diagram.

This continues across the beam, but notice two details seen so far,

i) the shear force is the slope of the bending moment diagram.ii) applied moments simply cause a jump in the bending moment graph.

Keeping these points in mind, let us proceed to the shear force graph first.

V(lb)

90

50

-10

-50

6” 16” 26” 32” x

NOTE: As a check the values at both ends should go to zero.

x

M(lb in)

1400

1300

500

-300

SHEAR FORCE DIAGRAM

BENDING MOMENT DIAGRAM

• Consider a simpler technique for developing shear force and bending moment diagrams. First draw the shear force diagram from left to right. The direction of the forces should be added directly to the diagram. Then using the values on the shear force diagram, draw the bending moment diagram.

• Solve the following example using the relationship between shear and bending moments,

dMdx-------- V=

12ft

1ft 1ft 3ft

3ft100lb 200lb/ft

900lb

200lb.ft.

5100lb.ft.

A B C D EF

• When considering distributed forces, you can lump the force to find reactions, but you must not lump the forces when drawing the shear force diagrams.

26. STRESS

• Stress is the force per unit area. This is because the geometry of the parts are not fixed, and we must accout for the size of the cross section that is resisting a force or moment.

• So far we have dealt with tensions and compressions in members somewhat simply. If we con-sider a simple member as shown below, we use the force divided by the area to give us the force per unit area.

• The material at the section of interest acts like numerous supports, each exerting internal forces. As a result the stress is statically indeterminate, but a reasonable estimate is made by assuming uniform distribution. In the internal forces are not constant over the cross section of the part.

P

P

Aσave

PA---=

where,

σave average normal stress (psi, ksi, Pa)=

A cross sectional area=

P applied load=

NOTE: This assumes that the material deforms uniformly across the section and each section of material pulls the same amount.

• For two dimensional objects the internal forces of interest are shear force, normal force and bending moment.

• Using stress we can calculate deformations (strain) and failure loads. This is done by refering to tables of values for particular materials..

26.1 TYPES OF STRESS

• Consider an example of using a pair of scissors (“shears”) to cut a piece of metal,

Shear Stress

NormalStress

bendingmoment

bendingmoment

• Consider the simple example given below,

P

P

A

τAVEPA---=

where,

τAVE the average shearing stress (psi, ksi, PA)=

P The applied load=

A The area of the shear=

NOTE: like axial stress, shear stress is also not uniform in distribution.

• These calculations are commonly used when examining rivets and bolts. Consider the figure below with the bolt in single shear.

P

P

Given the 1” diameter bar to the left, what is the inter-nal stress if the load P is 100KN? If the bar is made of gray cast iron, what is the difference between the ultimate tensile strength and the induced stress?

• Now consider a bolt that is exposed to double shear. Below we can see a section view of a pin through two members.

• We may also consider how the force on a pin effects the surrounding material. In the figure below we are finding the bearing stress. In this case the bearing stress will be a normal stress.

UNC-8-1”

P1

P1

P2

P2

dia. = 1” A∴ πd2

4---------=

σP2

A------= τ

P1

A------=

PP

PP/2

P/2

τave

P2---

A--------- P

2A-------= =

τave

τave

• The baering stress as shown only needs to be considered when the member is in tension.

26.2 STRESS ANALYSIS

• Let’s consider an example problem [1.15, pg. 16, Beer and Johnston]. For this problem we will examine all links for maximum stresses.

PA1 + A2

σ PA1 A2+------------------=

30°

45°

1.2 kip 1.2 kip

A

B

C

0.5”0.5”

0.5”

1.6”

1.6”

* all pins are 0.6” diameter

First, find the tension/compression in each member.

TAB

TBC

60°

45°

1.2+1.2kip

B

FBD B:

Fx∑ TAB 60° TBC 45°sin–sin– 0= =

TAB∴ 0.816TBC–=

+

Fy∑ TAB 60° TBC 45° 2.4kip–cos–cos 0= =+

TBC∴ 2.15kip–=

TAB∴ 1.76kip=

Now consider the pin A,

τA

TAB

2---------

A--------------- 1.76kip

2 0.283in2( )

----------------------------- 3.11ksi= = =

τA τATAB

TAB

2---------

TAB

2---------

Next consider the member AB at the pins A & B,

AABAAABB

1.6in 0.6in–( )0.5in 0.5in2

= = =

σABAσABB

TAB

AABA

----------- 1.76kip

0.5in2

------------------ 3.52ksi= = = =

Now consider the stress in the centre of AB,

AAB 1.6in( )0.5in 0.8in2

= =

σAB

TAB

AAB--------- 1.76kip

0.8in2

------------------ 2.2ksi= = =

Consider the pin at B now, we can do this from left to right,

A 0.283in2

=

τB1τB2 τB2

τB1

1.2kip

τB1A τB1

1.2kipA

--------------- 1.2kip

0.283in2

--------------------- 4.24ksi= = =

1.2kip

τB2A

TCB

2---------

TCB

2---------

1.2kip

τB2A

45°

Fx∑TCB

2--------- 45°sin– 0.76kip= =

Fy∑ 1.2kip–TCB

2--------- 45°cos– 0.44kip= =

τB2

0.762

0.442

+0.283

------------------------------------ 0.88ksi= =

Next consider the shear on pin C,

TCB

TCB/2 TCB/2

τ

TCB

2---------

A---------------

2.152

----------–

0.283------------------- 3.80ksi= = =

Now, lets consider member BC. At pin C the stress will be,

ACBc1.6in 0.6in–( ) 0.5in( ) 0.5in

2= =

σBCc

TCB

ACBc

----------- 2.15kip–

0.5in2

---------------------- 3.30ksi–= = =

Likewise for the support at B,

ACBB2 1.6in 0.6in–( ) 0.5in( ) 1.0in

2= =

σBCB

TCB

ACBBc

------------- 2.15kip–

1.0in2

---------------------- 2.15ksi–= = =

Finally, the stress in the member BC will be,

ACB 1.6in 0.5in( ) 0.8in2

= =

σBC

TCB

ACB--------- 2.15kip–

0.8in2

---------------------- 2.69ksi–= = =

ASIDE: If doing a real design we could then use the maximum stress values to select materials, or to select components to redesign.

Note: Because the member is in com-pression, the reac-tions at the supports will not be a prob-lem.

26.3 STRESS ON OBLIQUE PLANES

• Consider the arbitrary definition of a force, and how any force can be replaced with components. If we take a force causing an axial stress on a beam, we could replace it with two components that give the same resultant. The only complication being that as we shift the plane of applica-tion, the effective section area changes.

σA σA

PP

A

-- IS EQUAL TO --

Pθ τ'A'

σ'A'

A'A

θcos------------=

Fy∑ τ'A' θ σ'A' θsin–cos 0= =+

τ' θcos( ) Aθcos

------------ σ' θsin( ) A

θcos------------

– 0=∴

τ'∴ σ ' θsinθcos

------------ σ' θtan= =

Fx∑ P– τ'A' θ σ'A' θcos+sin+ 0= =+

P θcosA

----------------∴ τ ' θ σ' θcos+sin=

P θcosA

----------------∴ τ ' θτ' θcos

2( )θsin

-----------------------+sin=

P θcosA

----------------∴ τ ' θsin( )2 θcos( )2+

θsin--------------------------------------------

τ'θsin

-----------= =

PA--- θ θsincos∴ τ ' θ θσsincos= =

σ'∴ τ' θcosθsin

---------------- θ θσsincos( ) θcosθsin

------------ σ θcos( )2

= = =

26.4 GENERALIZED STRESS

• We can consider the more general case of the 3D element,

• If we assume the element is square, then we can replace the stresses with forces,

σy

τyx

τyz

x

y

z

σx

τxy

τxz

σz

τzx

τzy

Here we can see 3 of the 6 faces of the square element. The normal stress, and the two shear forces define the stress on each face. All of the normal and shear forces must balance for the element.

∆Aσy

∆Aτyx

∆Aτyz

∆Aσx

∆Aτxy

∆Aτxz

∆Aσz

∆Aτzx

∆Aτzy

• Finally, for simplicity, we can look at the element from a single side,

• For statics, the sum of the forces and moments must total zero. As a result we would see a shear stress (τxy) induce another shear that is perpendicular (τyx).

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

∆Aσy

∆Aτyx

∆Aσx

∆Aτxy

x

y

• We also consider the case where we rotate the virtual element,

P

P

+

∆Aτyx

∆Aτyx

∆Aτxy

∆Aτxy

NOTE: in the stress element shown abovethe directions of the stresses will be used as the positive convention for the other stress problems. As a memory tool, consider that the upper and right-most stresses are in the direction of the positive x and y axis.

τxy τyx= τyz τzy= τzx τxz=

• The value of these relationships show how stress is spread throughout an object, and will become valuable as stress is examined in greater detail.

****** Caucy’s theorem

26.5 FACTOR OF SAFETY

• When we do basic design work we typically use a process of the basic format,

P P

45°

σ1

σ1σ2

σ2

τ1

τ1 τ2

τ2

Here we can show that,

τ1 τ2P

2A-------= =

σ1 σ2P

2A-------= =

* This will reappear later

• In particular the design process often involves a factor of safety that allows for,- variations in material properties- variations in expectations load- the non-uniform distribution of stress- cyclic loading- etc

• The factor of safety is applied as,

START: select aneeded design

determinefunctional elements

pick membersgeometry, etc.

select valuesand properties

analyze forsuitability

decide on

DONE: Approvedesign

deficiencies

major changes e.g.replace cable with beam

small changese.g. diameter

no deficiencies

e.g. a support cable

e.g. maximum tension

e.g 1/2” steel cable

e.g. find stress and compareto ultimate strength

e.g. use factorof safety

• The values vary for various systems. The values given below are reasonable for static systems.

• For example let us consider a two member mechanism 1.44 from [Beer and Johnston]

F·S· σULTIMATE

σESTIMATED-----------------------------

PULTIMATE

PESTIMATED------------------------------= =

where,FS the factor of safety=

σULT ultimate shear stress before failure=

PULT ultimate load before failure=

σEST the working,mormal, stress using assumed conditions=

PEST the working, normal load using assumed conditions=

FS < 1 for a system that should failFS = 1 for a critical system (likely to fail)FS = 2 for a reasonable designFS = 5 for a safe designFS >10 is overdesigned

*NOTE: many values are given in design tables


P = 8kips

A

B C

D

30°

12”

6”

w

given,FS = 3σULTIMATE 60ksi=

find the required width of BC.

ABC thickness( ) w( ) 14---in

w= =

σBC

σBCULTIMATE

FS--------------------------

TBC

ABC---------= =

TBC∴ 60ksi3

------------- w

4----

=

MD∑ TBC 12in 30°cos( )– P 18in 30°sin( )+ 0= =+

w4----

20ksi( )12in 30°cos∴ 8kip 18in 30°sin( )=

w∴ 4( ) 8kip( ) 18in 30°sin( )20ksi( ) 12in 30°cos( )

---------------------------------------------------------- 1.39in= =

27. STRAIN

• If we apply a force to a material, we create a stress, as shown before. The the effect is that it stretches or compresses somewhat. This deformation is know as strain. In very brief terms, stress causes strain.

27.1 STRAIN CAUSED BY AXIAL LOADS

• Under simple axial loads metals and other materials act like simple springs. This behavior is known as elastic

• If the material becomes permanently deformed this is known as plastic deformation.

• Consider a generic load-deflection curve,

P necking

rupture

linearelasticregion

plasticregion

PULTIMATE

δ

∆δ

∆P

P

P

A

L

(Hookes law)

∆δ

K∆P∆δ-------=

Test Specimen

• This curve is produced by forcing the deformation, then measuring the holding force using a special tensile tester. The specimen pictured above is fastened into a hydraulic or screw based machine. The specimen is then stretched and the applied load measured. The curves can be directly drawn from using load and deflection.

• Secondly, the graph above is often changed as shown to make it independent of geometry.

• Note that Young’s modulus is a spring constant for a unit volume of material. To get it into a tra-ditional spring constant we must multiply it by area, and divide by lengh.

27.2 STRESS STRAIN CURVES

• For low carbon steel,

σULTIMATE

∆ ∈

∆σ

∈

σ where,

σ PA--- stress= =

∈ δL--- strain= =

E∆σ

∆ ∈------------- Youngs modulus( )= =

σ E∈=

δ LELσE

------- LPEA-------= = =

or

or

• While necking, the cross section at one point decreases, thus increasing the stress. In turn this continues rapidly until fracture.

• The strain hardening of some materials occurs as they are stretched, the Young’s modulus value increases.

• Each material will have it’s own stress-strain curve and these are determined experimentally, and found in abundance in handbooks.

• If a material is brittle, it does not deform much and simply breaks. Or simply the ultimate and rupture strengths are the same.

• Ductile materials deform quite a bit before the ultimate stress, necking typically occurs before rupture.

σU 60=

σ ksi( )

σy 37=

∈

elastic yield strain hardening necking

rupture

0.0012 0.02 0.2 0.25

σ

∈

• Creep is an effect that can lead to permanently elongated specimens,

27.3 ANALYSIS OF MEMBERS


σ

∈

σ

∈

1

23

• When a load is applied and removed in cycles the material may fatigue. This means the ultimate strength is effectively reduced. Design curves that can take this into account are found in hand-books, and look like those shown below,

P = 10,000 lb

P


6”

If the rod is 6” before the load is applied, what is the new length? What load P would result in plastic deforma-tion? What load would result in rupture?

• As another example, let’s consider problem 2.26 in [Beer and Johnston]

# cycles

σ ksi( )

60

50

40

30

20

10

103

104

105

106

107

108

109

1020 Hot Rolled Steel

2024 Aluminum

A

B

C

D

P

10” 15”

8”

8”

E 29 106psi×=

Link AB & CD are, 1/4” by 1”

Given,

A 0.25in2

=

Find the load required for a tip deflection of 0.01”

δ 0.01in=

MB∑ TCD 10in( )– P 25in( )+ 0= =+

δ1

8inTCD

29 106psi×( ) 0.25in

2( )---------------------------------------------------------= TCD∴ 29 10

6lb×

32in--------------------------δ1=

P∴δ1 29 10

6lb×( ) 10in( )

32in( ) 25in( )-----------------------------------------------------= δ1∴ P

362500------------------=

(1)

MC∑ TAB 10in( )– P 15in( )+ 0= =+

similarly,TAB

29 106lb×

32in--------------------------δ2=

29 106lbδ2×

32in-------------------------------

10in∴ P 15in( )=

δ2∴ P604167------------------= (2)


• An isotropic material has the same properties in all directions. (materials such as fiber glass do not).


10” 15”

δ2 δ1

δ 0.01in=

δ δ2– 10in 15in+( )δ1 δ2+

10in-----------------

+=

0.01∴ 2.5δ1 1.5δ2+=

0.01∴ 2.5 P362500------------------

1.5 P604167------------------

+=

P∴ 1066lb=

ν lateral∈

axial∈---------------------------=

where,

ν Poissons Ratio=


• In physical terms - as we stretch a bar, it becomes a bit thinner.


P = 10,000 lb

P


6”



• When we load a material in a single direction the effect of Poisson’s ratio is naturally included in Young’s Modulus. But, when there are multiple loads in multiple directions, we must uses Poisson’s ratio to determine how they interect.

27.6 SHEAR STRAIN

• While Young’s modulus is intended for the stress to strain relationship for Axial or normal

σy

σy

x

y

z

σx

σz

σx

σz

x∈σx

E-----

νσy

E---------–

νσz

E---------–=

y∈σy

E-----

νσx

E---------–

νσz

E---------–=

z∈σz

E-----

νσx

E---------–

νσy

E---------–=

strain, we will use the Shear modulus G for shear strain.

• First consider the general square cubic element below,

• Now, Assume a shear stress in the positive x direction that induces the deformation as shown below.

σy

τyx

τyz

x

y

z

σx

τxy

τxz

σz

τzx

τzy

• NOTE: these calculations are based upon the assumption of small angles of deflection.

• How a material strains in shear is related to normal strain by Poisson’s Ratio,

• Try the simple problem below,

τyxx

y

zτxy

π2--- γxy+

π2--- γxy+

π2--- γxy–

π2--- γxy–

γxy

2------

γxy

2------

τxy Gγxy= τyz Gγyz= τzx Gγzx=

Where,

τxy τyz τzx, , shear stress in z, x and y planes respectively=

G shear module (or modulus of rigidity)=

γxy γyz γzx, , shear strain in z, x and y planes respectively=

GE

2 1 ν+( )--------------------=

• Consider the example shown below from Beer and Johnson, 1992, pg. 89 #2.86,

100lb.

100lb.

The block to the left is under a shear load. Find the shear strain if it is made of 2014 aluminum. What is the poisson’s ratio for the material?

6”

3”

P

A B

a

a

b

c

Two blocks of rubber (B),connect a plate A to outer brackets. The rubber modulus of rigidity is G=1.75 ksi. Given c=4”, P=10 kip, find a and b if the stress in the rubber is not to exceed 200 psi, and the deflec-tion should be at least 3/16”.

τ 200 lbin-----

≤

A cb 4b in( )= =

τ

P2---

A---------=

200∴ lb

in2

-------

10000 lb( )2

-------------------------

4b in( )------------------------------≥ b∴ 6.25 in( )=

First lets deal with the maximum stress,

Next, lets deal with the deflection resulting from strain,

δ 316------ in( ) aγ a τ

G----

a

200 lb

in2

-------

1750 lb

in2

-------

-------------------------

0.114a= = = = =

a∴ 1.64 in( )=

27.7 STRESS CONCENTRATIONS

• Saint Venant’s Principle states that regardless of how a force is applied, when we move far enough away, the distribution becomes even. For example, if we are applying forces as point loads, they will have very high stress concentrations near the point of application. But, as we move away from the point of application, the force distribution evens out. Consider the pin in the hole where the pin applies a load of P.

• Stress concentrations are hard to predict, and this must often be done using experiments, Finite Element Analysis (FEA), or other techniques.

• During design we must pay attention to the stress concentrations. At some points the stress will be higher that the average stress.

• One technique for estimating maximum stress concentrations is to use tables derived experimen-tally.

P

Stress distribution is relatively even

KσMAX

σAVG-------------=

where,

K stress concentration factor=

σMAX the maximum stress estimate=

σAVG stress calculated with normal methods=

PP

d/2

d/2

r

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0r/d

3.0

2.5

2.0

1.5

K

Maximum stress

• Consider the sample problem below,

PP

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0r/d

2.5

2.0

1.5

1.0

KD

r

d

D/d = 2.0D/d = 1.5D/d = 1.25

D/d = 1.1

Maximum stress

27.8 TORSION

• If we consider a cylindrical shaft with torques applied it will tend to rotate an angle proportional to the torque.

10 KN 10 KN5”3”

R0.5”

Find the maximum stress that will be found in this bar with a hole in it.

• We can find the torque exerted using simple stress equations for a plane cutting through the cyl-inder,

• Next, consider the entire length of the cylinder and deformation of an element in the shaft.

TT

T=0

TT

T>0

dFT

ρ

T ρ Fd∫ ρτ Ad∫= =

TT

ργφ

L

γ ρφL

------= (for small angles) (1)

γ ρc---γMAX= (2)

where,c the outer radius=

ρ a variable radius [0, c]=

γ the strain at radius C=

φ a small twist angle at the end=

L the length of the shaft=

γMAX the strain at outer radius C=

Using the shear modulus recall the shear stress-strain relationship,

τ Gγ= γ τG----=or (3)

2( ) 3( )→∴

τG---- ρ

c---

τMAX

G------------= τ∴ ρ

c---τMAX=

T ρτ Ad∫τMAX

C------------ ρ2

Ad∫τMAX

c------------J= = =

Now recall,

Polar moment of inertia

τMAX∴ cTJ

------ γMAXG= = γMAX∴ cTJG------- cφ

L------= = φ∴ LT

JG-------=

• Consider the simple example of the beam below,

27.9 TORSION STRESS CONCENTRATIONS

• We can estimate the maximum stress for two shafts connected with a fillet using the following graph,

1000 lb.ft. 1000 lb.ft. What is the maximum shear stress in the bar? If the bar is made of cold-rolled yellow brass, what is the angle of deflection along the bar?

10”

1” dia.

• What is the maximum stress in the shaft shown below,

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1r/d

2.5

2.0

1.5

1.0

K

D

r

d

D/d = 2.0D/d = 1.33D/d = 1.2

D/d = 1.1

TT

27.10 PURE BENDING

• Consider a section of beam that has two moments applied, thus inducing bending,

0.25”

1”

The shafts shown are joined with a fillet of radius 1/16”. What is the maximum shear stress if a torque of 100 lb.in. is applied?

• Consider a cross section of the beam being bent about the z-axis,

• Now consider the overall geometry of the beam,

M=0

θ

M > 0

y

c

c

neutral axis (centroid)

σdA

Mz yσy Ad∫=

θ

ρ ρ y–( )

y

L ρθ=

L' ρ y–( )θ=

for the arbitrary height y

δ L' L– ρ y–( )θ ρθ– yθ–= = =

x∈ δL--- yθ–

L--------- yθ–

ρθ--------- y

ρ---–= = = =

Now considering the extreme dimension is ±c we can find a maximum strain,

MAX∈ cρ---=

x∈( ) yc-- ∈ MAX– σx–=

And, assuming that stress varies linearly from the maximum at the outer limit to

zero at the neutral axis,

σx

E-----∴ y

c--σMAX

E-------------–=

σx∴yσMAX–

c-------------------=

Now consider the moment of the section,

M y–( ) σx( ) Ad∫ y–( )yσMAX–

c-------------------

Ad∫σMAX

c------------- y

2( ) Ad∫σMAX

c-------------I= = = =

σMAX∴ cMI

--------=

MAX∈ cρ---

σMAX

E------------- cM

IE--------= = = ρ∴ EI

M------=

• For the beam elow find the maximum stress and radius of curvature,

27.11 TRANSVERSE LOADING

• When loads are applied to beams they tend to deflect

A

A

Section A-A100 KNm 100 KNm

4m

0.3m

0.4m

The beam above is under pure bending. If the beam has the triangular cross section shown, find the maximum stress, and the radius of curvature.

• For a beam under a transverse load the stress is a function of the distance from the centroid,

• We can consider the shear on some axis other than the neutral axis of the beam,

PP

P = 0 P > 0

y

P

x

L

y

CA B

B’’

σxMy

I--------–

PxyI

---------–= =

• Try the problem shown below from Beer and Johnson 1992, Pg. 291 #5-1,

H∴ PQI

-------- x qx= =

σxdA PxyI

---------dA–=

For an element some distance from the neutral axis the force on the element is,

Now we can integrate to find the shear H for the shaded part of the beam cross section,

Fx∑ 0 H PxyI

--------- Ad∫– H PxI

------ y Ada

b

∫– HPxQ

I-----------–= = = =

a

b

where,H = the shear force for the section

section

Q = the first momment for the section (the centroid multiplied by the area)q = horizontal shear per unit length for the section

2.5”

2.5”

2.5”

1.5”

1.5”

1.5”

3.5”

The end of the beam is loaded with a vertical shear of 250lb. This beam is made of a set of three 2x4 studs nailed together. Find the shear force in each nail.

28. MECHANISM DYNAMICS

• Basically, things in motion required forces to accelerate to a velocity, and will apply forces to other objects as they slow down.

28.1 INTRODUCTION

• Some basic topics to review,

• The basic topics of statics that should be well understood are,- units- Newton’s laws- forces- applied/reaction forces- free body diagrams/rigid body assumption- force components in 2D and 3D- equilibrium of forces- moments- moment components in 2D and 3D- equilibrium of moments- equilibrium of forces and moments- two and three force members- friction dry static, velocity- internal forces- method of members- mass properties - center of mass, centroids- moments and product of inertia, polar moment of inertia- D’Alembert’s principle

28.2 PLANAR

• The basic inertial functions are,

• D’Alembert’s principle states that when we have a dynamic system the sum of all applied forces/moments including inertia should equal zero.

• Consider the simple link below that is known to have the forces applied, at the given velocities. What are the reactions at the pin joint.

F∑ 0= M∑ 0=Static equilibrium

xx Ad∫

Ad∫-------------=Centroids z

z Ad∫Ad∫

-------------=yy Ad∫

Ad∫-------------=

Iy x2

Ad∫=Second Moments Ix y2

Ad∫=of Inertia

J Ix Iy+=

K IA---=Radius of Gyration

F∑ Ma= M∑ Jα=D’Almbert’s Principle (planar)

F M A( )=

T J α( )=

where,

M mass=F inertial force=A linear acceleration=T inertial torque(moment)=J polar moment of inertia=α angular acceleration=

Fg 500N

3m

16m

ω 50 radsec-----------=

Given the 100 Kg steel beam, what is the instantaneous force in the pin?

First, establish the kinematic velocities of the point using the given angular velocity.

P 8eiθ

m=

V 8iωeiθm

s----=

A 8iαeiθ

8ω2e

iθ–

8ωz2

–

8α z

0

m

20000m

s2

----–

8αzm

0

= = =

A P

Next, we draw a free body diagram, and then write the force and moment balance equa-tions.

F1F2 Fg

F1

0

500N

0

=Fg

0

981N–

0

=F2

F2x

F2y

0

=

F∑ F1 F2 Fg+ + Ma= =

0

500N

0

F2x

F2y

0

0

981N–

0

+ + 10KgA=

J16m 3m( )3

12-------------------------- 3m 16m( )3

12--------------------------+ 1060m

4= =

F2x

500N F2y981N–+

0

10Kg

20000m

s2

----–

8α zm

0

= F2x∴ 200000N–=

481N– F+ 2y∴ 80αzKgm=

M∑8m–

0

0

F2×8m

0

0

F1×+ Jα= =

8m–

0

0

F2x

F2y

0

×8m

0

0

0

500N

0

×+ 1060m4

αx

αy

α z

=

0

0

8mF2y– 8m500N+

1060m4αx

1060m4αy

1060m4α z

=

αx∴ 0=

αy∴ 0=

8mF2y– 4KNm+∴ 1060m

4α z=

• we can extend these principles to more complicated mechanisms.

• Consider the dynamic effects on the four bar linkage below,

We can then collect the two equations, and solve for the remaining unknown force.

1 80–

8 1060

F2y

α z

481

4000= F2y

1 80–

8 1060

1–481

4000= =

• When doing analysis of more complicated mechanisms we can use a technique called Superpo-sition. This involves the basic steps,

4”

5”

A

B C

D

Each of the rods weighs 1 lb/in., and AB is being turned with an angular velocity of 100 rad/sec, and angular acceleration of 5 rad/sec. a) Do a kinematic analysis. b) Do a static analysis. c) Do a dynamic analysis. d) What are the total forces on the bodies?

1. do a kinematic analysis to find positions, velocities, accelerations.2. do a static analysis of the mechanism to find the static forces at the pins3. using the results of the kinematic analysis find the dynamic forces at the pins4. add the results of the static and dynamic analyses for the total forces.

• Superposition assumes that the structure is linear. In other words the dynamic forces should not change the static forces, and vice versa. This will occur if the mechanism is flexible, has fric-tion, etc. (We will not use this technique in this course because computational tools available (eg Mathcad) reduce its value.)

28.2.1 Measuring Mass Properties

• This section will discuss how to find Moments of Inertia for complex solids using experimental techniques.

• The value of this material is somewhat limited, as most modern CAD systems will use the geometry to calculate centroids, moments of inertia, volume, etc.

• This method can be valuable when reverse engineering a part (i.e. no detailed designs available), or when testing a completed part.

• Basically these techniques use the part as a pendulum mass, and use the period to calculate the moment of inertia.

• The basic procedure is,1. Locate the center of mass - this can be done by looking for x, y, & z planes of balance.2. Measure the mass of the object.3. Secure the object so that it rotates about some point other than the center of mass.4. Introduce a very small displacement less than 1 degree, and allow it to oscillate.5. Record the period of oscillation.6. Calculate the moment of inertia.

• The differential equation and manipulated final form used to describe a pendulum is given below,

• other techniques employ similar approaches,torsional rod - the object is hung from a rod of known torsional stiffness. The object is

then twisted slightly and released. The period of oscillation is used to find the moment of inertia.

Fg

r

M

θ

+ MO∑ Fg θsin( ) r( ) IOddt-----

2θ+ 0= =

O

Fgr

IO--------

θ ddt-----

2θ+ 0=

First sum the moments about point O where the object is balanced,

Next, simplify the equation assuming that the total deflection is small,

The general form of the solution for this differential equation is,

θ A tFgr

IO--------

sin B tFgr

IO--------

cos+=

where,

Fg the gravitional force on the object=O the balance point=M the center of mass=r the distance between O and M=θ the angular displacement=

Io the moment of inertia about O=

The period motion and the moment about O becomes,

τFgr

IO-------- 2π= IO∴ τ

2π------

2Fgr=

The parallel axis theorem can be used to find the value about the center of mass,

IO IM mr2

+=

IM∴ τ2π------

2Fgr

Fg

g------r

2–=

τ period of oscillation=

trifilar pendulum - a platform with three wires has the object placed on the platform. The platform is then twisted in plane, and released. It then oscillates in rotation. The period is used to calculate the combined moments of inertia of the object and the platform.


1. A geneva mechanism is shown below with both components rotating.

a) Write the closed loop position equation for the configuration shown.b) Write the closed loop velocity equation.c) Write the acceleration closed loop equation.d) Draw free body diagrams (FBDs) for both links.e) Write D’Alembert’s dynamic equations for link 3, assume the link 3 has a reaction torque.

IM Ktτ

2π------

2=

where,

Kt torsional stiffness of rod=τ period of oscillation=

2

3

L2

l3

Given,

M1 M2, Component masses=

J2 J3, Polar moments of inertia=

ω· 2 constant angular velocity of link 2=

A

B

C

g) Expand the vector equations for link 3 to show vector components, or into individual equa-tions, so that they could be solved. (Do not solve for numerical values)

28.4 REFERENCES

Erdman, A.G. and Sandor, G.N., Mechanism Design Analysis and Synthesis, Vol. 1, 3rd Edition, Prentice Hall, 1997.

Shigley, J.E., Uicker, J.J., Theory of Machines and Mechanisms, 2nd Edition, McGraw-Hill, 1995.

29. VIBRATION

• Every moving system has some amount of vibration. This is a natural result of unbalanced com-ponents, rubbing, reversals, etc.

• Some vibrations are very predictable, and can be estimated mathematically, and their effects compensated for.

• Mathematically a vibration is a displacement/force/velocity/acceleration of small amplitude that will ‘shake’ and object.

29.1 VIBRATION MODELLING

• The most significant vibration in engineered systems is periodic. In these systems there is often an approximate spring-mass-damper system that gives us a second order response to distur-bances.

• In vibration modeling we typically assume that all components are linear. In a linear system the forcing (input) frequencies are directly related to response (output) frequencies.

• In non-linear vibration systems we end up with the frequency of the forcing function being transformed to other frequencies. This tends to make the vibrations seem less clear, and appear more chaotic.

• There are a few types of descriptive terms for these systems,Damping Factor - The damping factor will indicate if vibrations will tend to die off. If the

damping factor is too low the vibrations may build continually until failure.Forced Vibration - When a periodic excitation is applied to these systems they will tend to

show a steady state responseFree Vibration - When displaced/disturbed and released there is an oscillation at a natural

frequency for any system. This is one measure of a system, and is typically induced by displacing a system and letting it go.

Natural Frequency - Each system will have one or more frequencies that it will prefer to vibrate at. When we excite a system at a natural frequency the system will reso-nate, and the response will become the greatest.

Response - This is a measure of how a system behaves when it is disturbed. For example, this could be measured by looking at the position of a point on a mechanism.

Steady State Response - After a system settles down it will assume a regular periodic response, this is steady state. The steady state excludes the transient.

Transient Response - When a forcing function on a system changes, there will be a short lived response that tends to be somewhat irregular. The transient will eventually die off, and the system will settle out to a steady state.

• These systems can be modeled a number of ways, but we typically start with a differential equa-tion.

29.1.1 Differential Equations

• In modeling any linear system we are best to start by developing a differential equation for the system components.

• Consider the simple spring-mass-damper system shown below. A free body diagram can be drawn for the mass ‘M’, and a sum of forces can be written, and expanded with the values for the mechanical components.

• We may also consider a torsional vibration. We will assume that the vertical shaft has a stiffness of Ks and a damping coefficient of Kd. There is an applied torque ‘T’ and a moment of inertia ‘I’.

x

F

Kd Ks

Mx

F

M

Fd Fs

+ Fy∑ F– Fd Fs M ddt-----

2x+ + + 0= =

F∴ Kdddt-----

x Ksx M ddt-----

2x+ +=

29.1.2 Modeling Mechanical Systems with Laplace Transforms

• Before doing any sort of analysis of a vibrating system, a system model must be developed. The obvious traditional approach is with differential equations.

KsKd

I

T

I

T

θ

Fs

Fd

O

MO∑ T Fs– Fd– I ddt-----

2θ– 0= =+

T∴ Ksθ Kdddt-----

θ I ddt-----

2θ+ +=

29.1.3 Second Order Systems

• Basically these systems tend to vibrate simply. This vibration will often decay naturally. The contrast is the first order system that tends to move towards new equilibrium points without any sort of resonance or vibration.

KdKs

M x

F Md2x

dt2

-------- Kddxdt------ Ksx+ +=

F t( )x t( )---------- M d

2

dt2

------- Kdddt----- Ks+ +=

L F t( )x t( )----------

F s( )x s( )----------- Ms

2Kds Ks+ += =F

ASIDE: An important concept that is ubiquitous yet largely unrecognized is the use of functional design. We look at parts of systems as self contained modules that use inputs to produce outputs. Some systems (such a mechanisms) are reversible, others are not (consider a worm gear). An input is typically something we can change, an output is the resulting change in a system. For the example above ‘F’ over ‘x’ implies that we are changing the input ‘x’, and there is some change in ‘F’. We know this could easily be reversed mathematically and practically.

ANOTHER ASIDE: Keep in mind that the mathematical expression ‘F/x’ is a ratio between input (displacement action) and output (reaction force). When shown with differentials it is obvious that the ratio is not simple, and is a function of time. Also keep in mind that if we were given a force applied to the system it would become the input (action force) and the output would be the displacement (resulting motion). To do this all we need to do is flip the numerators and denominators in the transfer function.

•

• Some basic relations,

• These generally have an effects on the Bode plot that are very evident.

KsKd

M

F(t)

x

** Assume gravity negligible

+y

Fy∑ F t( )– Kddxdt------ Ksx+ + Md

2x

dt2

--------–= =

F s( )–∴ Kdsx Ksx Ms2x+ + + 0=

F s( )x

-----------∴ Kds Ks Ms2

+ +=

xF s( )-----------∴ 1

Ms2

Kds Ks+ +--------------------------------------=

xF s( )-----------∴

1M-----

s2 Kd

M------s

Ks

M-----+ +

---------------------------------=

ωn natural frequency=ξ damping factor (values > 1 tend to absorb vibrations)=

xF--- A

s2

2ξωn( )s ωn( )2+ +

----------------------------------------------------= Transfer function for second order systems

ωn

Ks

M-----= ξ

Ks

2Mωn---------------=

Therefore the two relationships below can be derived

(examining exact bode plots will confirm this)

• Under the influence of damping, the natural frequency will shift slightly,

• To continue the example with numerical values

If ζ is less than 1/root(2) = 0.707 the roots become complex, and the Bodeplots get a peak. This can be seen mathematically because the roots of the

Gain

dB = A=0.5

=0.1

-40dB/decade

ξ 0.707=

ωn

Resonant Peak

freq

transfer function become complex.

ωd 1 ξ2– ωn=

when vibrating freely, this is the frequency the system will assume.

M 1kg= Ks 2Nm----= Kd 0.5Ns

m------=

Assuming an input of,

Assuming component values of,

F t( ) 5 6t( )Ncos=

F s( )∴ 5s

s2

62

+----------------=

This means,

x s( ) F s( ) x s( )F s( )-----------

5s

s2

62

+----------------

11---

s2

0.5s 2+ +------------------------------

= =

x s( )∴ 5s

s2

36+( ) s2

0.5s 2+ +( )----------------------------------------------------------=

x∴ s( ) As 6j+------------- B

s 6j–------------- C

s 0.5– 1.39j+---------------------------------- D

s 0.5– 1.39j–----------------------------------+ + +=

A s 6j+( ) 5s( )s 6j–( ) s

236+( ) s

20.5s 2+ +( )

----------------------------------------------------------------------------s 6j–→lim

30j–12j–( ) 36 3j– 2+( )

-----------------------------------------------= =

A∴ 30j–432j– 36– 24j–

----------------------------------------- 30j36 456j+----------------------- 36 456j–

36 456j–-----------------------× 13680 1080j+

209 232,----------------------------------- 0.0654 0.00516j+= = = =

--Continue on to find B, C, D same way

x∴ s( ) 0.0654 0.00516j+s 6j+

-------------------------------------------- 0.0654 0.00516j–s 6j–

-------------------------------------------- …+ +=

x t( ) 2 0.06542

0.005162

+ e0t–

0.00516t0.005160.0654-------------------–

atan+ cos …+=

Do inverse Laplace transform

29.1.4 Phase Plane Analysis

• When doing analysis of a system that has both a steady state and transient response it can be handy to do a phase plane analysis to help separate out the components.

• To construct a phase plane graph we plot the value of a response variable against it’s first or sec-ond derivative.

• The shape of the graph exposes the phase between the displacement, and one of the derivatives. Here we see the system start to spiral out to an outer radius. The change in the radius of the spi-ral is the transient, the final radius is the steady state. If the forcing function changed, the path would then shift to a new steady state position.

29.2 CONTROL

• Vibrations are the natural result of many engineered systems.

• These vibrations can become significant when they shake carefully designed structures, or induce sounds in the air.

• As an engineer attempts to suppress or negate vibrations and sound, one of the most powerful weapons is a good analytical understanding of the phenomena.

29.2.1 Vibration Control

• If this displacement is induced by a machine in the next room, and it travels through the floor,

f t( )

ddt-----f t( )

ωn--------------

we want to isolate the noise source for the high frequency (2000Hz) that will be noticeable as a whine.

• Your boss asks you to design a mounting for the machine that gets rid of the high frequency whine.

Vibration travels through the floor (and other paths) andannoys workers.

• In this case the bode plot would reduce the noise by over 40dB for the high pitch sound. For the frequencies below 1000 Hz there would not be much reduction, but since this is 1/100th the frequency of the original sound, it should not be a problem.

x

F

Kd Ks

M

we choose the common spring-damper pair,and draw a Bode plot that should providethe desired performance.

GAIN(dB)

ω

40dB/decade

2π 100( ) 2π 1000( ) 2π 10000( ) 2π 100000( )

The gain must be reduced in this region

-40dB

In this case the corner frequency has been set to 1000Hz, but the damping factor must beselected, and 1 seems to give this shape.

ωn 2π 1000( )=

GAIN∴ A

s2

2ωnζs ωn2

+ +----------------------------------------=

∴ A

s2

2π 1000( ) 1( )s 2π1000( )2+ +

------------------------------------------------------------------------------=

∴ A

s2

12566s 39.56×10+ +

---------------------------------------------------------=

• We check the manuals and find the machine weighs 10000kg, and has 8 mounting points.

• The result would look something like,

Recall the spring damper, (above)

F∑ F– M d2

dt2

------- KDddt-----x Ksx+ + + 0= =

Fx---∴ Ms

2KDs KS+ +=

xF---∴ 1

Ms2

KDs KS+ +---------------------------------------

1M-----

s2

sKD

M-------

KS

M------

+ +

---------------------------------------------= =

ωn2∴

KS

M------=

39.56×10∴

KS

M------=

KS

M------∴ 39.5

6×10=

KS∴ 100008

--------------- 39.5

6×10( ) 4.9410×10 N

m----

= =

Mass per spring

And,

2ζωn

KD

M-------=

12566∴KD

100008

--------------- --------------------=

KD∴ 15.76×10 Ns

m------

=

We can then try to find materials orcomponents with these values.**Please note the size of the valueshere, they are not really practical.

• The force transmitted to the floor is,

• To find gain,

xF---

110000---------------

s2

s 12566( ) 39.56×10( )+ +

-------------------------------------------------------------------=

x s( )F s( )----------- 0.0001

s2

s 12566( ) 39.56×10( )+ +

-------------------------------------------------------------------=

FF KSx KDddt-----x+=

FF

F------∴ KS KDs+( ) 4.94

10×10 15.76×10 s+( )= =

FF

F------∴

FF

x------

xF---

4.9410×10 15.7

6×10 s+( )0.0001

s2

s 12566( ) 39.56×10( )+ +

----------------------------------------------------------------------------- s 3146+( )1570

s 6283+( )2-------------------------------------= = =

FF jω( )F jω( )----------------- jω 3146+( ) 1570( )

jω 12566+( )2---------------------------------------------- jω 3146+( ) 1570( )

ω2– 12566( )2jω 12566

2+ +

-----------------------------------------------------------------------= =

∴ 1570 jω 3146+

j 25132ω( ) 125662 ω2

–( )+-------------------------------------------------------------------

j 25132ω( )– 125662 ω2

–( )+

j 25132ω( )– 125662 ω2

–( )+------------------------------------------------------------------------

=

∴ 157025132ω2

jω 125662 ω2

–( ) 3146j25132ω– 3146 125662 ω2

–( )+ +

251322ω2

125662 ω2

–( )2

+-------------------------------------------------------------------------------------------------------------------------------------------------------------------

=

∴ 157021986ω2

3146 125662( )+( ) j 12566

2ω ω3– 3146 25132ω( )–( )+

251322ω2

125662 ω2

–( )2

+--------------------------------------------------------------------------------------------------------------------------------------------------------------

=

Do on board--------- for phase and gain

• We can develop a table of gains and phase angles for the isolator

• Consider damping at various frequencies, but consider that with damping the isolation was reduced at the high frequency, but the resonance was also reduced.

29.3 VIBRATION CONTROL

• types

ω (rad/sec)

01010010002000500010000120001500020000300004000050000600007000062800100000

Gain

111

Gain (dB)

000

θ (deg.)

000

Before andafter theresonantfrequency

Isolation - attempts to stop the flow of vibrations from a source with simple components

source

otherisolator

other

source

- these isolators are made from pads, or spring damper combinations- e.g. rubber pads under the feet of computers are one example

Inertial - Uses an added mass to reduce vibration transfer

source

other

inertial mass

other

source

- e.g.

and springssource

other

inertial massand springs

29.3.1 Isolation

• We have already done an example of using springs for damping, but, what are the practical options,

Springs - good for low frequencies because they are not massless as assumed. In fact, they have a damping ration of about 0.005. They are also well suited to harsh (long life) environments. They can have problems with “rocking”.

Elastomeric Mounts - Rubber compounds best used in compression, also good for shear.

Active - Systems that compensate for vibrations. linear examples include hydraulic/

source

other

adjustable

other

source

- e.g. expandable bellows for trains suspension

isolator

pneumatic/electric systems. Non-linear systems allow changing system parameterssuch as spring ‘K’ value changed by air pressure.

- e.g. motors that slide mass back and forth at top of tall building

Damping - Structures have a natural damping component. This is often the result of

source

other other

source with

- e.g.

heat generation, along with other factors.

new materials

Common ranges are 30-durometer for soft, low K rubber, to 80-durometer high K rubber (damping ratio about 0.05). These materials are good for high frequencies.

Isolation pads - Cork, felt, fiberglass. Frequencies start at low values (18Hz and up for fiberglass), common damping coefficients for cork and felt are .06.

• Elastomers are not linear, so the spring constant will vary as they are loaded. Graphical solutions work well when finding spring constants. Some example curves are given below.

• A solution can be done entirely with graphical methods. Manufacturers will provide graphs for specific materials and thicknesses.

40

35

30

25

20

15

10

5

0.02 0.04 0.06 0.08 0.10 deflection (cm)

Load (N)

• Note: When designing you should always attempt to get the natural frequency at least three times lower than the frequency to be damped.

• The same type of design techniques can be done with cork. (Note: they would have similar graphs to those for elastomers)

• A set of specifications for an elastomer isolator are summarized below,

100

80

60

40

20

13 16 19 22 25natural frequency (Hz)

Load (N)

fd

fn---- 3≥

29.3.2 InertialInertial Blocks - Increase the mass of the object to decrease vibration amplitude, and

decrease natural frequency.

Absorption - A secondary mass is added to draw off, and hopefully cancel out, vibrations

Load Per IsolatorMass of isolatorMaximum Dynamic DeflectionRough dimensionsNatural Freq.uency Dynamic Axial Spring RateDynamic Radial Spring Rate

3 lbs., 1.4 Kg..2 oz, 6g0.036”, 0.91mm1” by 0.5”31 Hz284 lbs/in.237 lbs/in. 1”

1.0

10.0

Tra

nsm

issi

bili

ty (

T)

Freq. (Hz)

5 20 50 100

Loa

d (l

bs.)

0

10

20

Deflection (in.)

0 0.1

xF---

1M-----

s2

sKD

M-------

KS

M------

+ +

---------------------------------------------=

2ζωn ωn2

As M increases, overall system gain decreases

ωn

KS

M------= M ∞ ωn 0→,→

29.3.3 Active

Active Systems - These systems are becoming very popular in new cars, etc. The example below uses a bellows with an adjustable pressure ‘P’. This pressure over area ‘A’ gives a spring constant ‘K’ and height ‘h’. If the pressure in the bellows is adjusted by the addition of gas, the spring constant will rise, tending to damp out different vibration frequencies (remember the ideal gas law ‘PV=nRT’).

x

M1

Ks1

Gain

ω

ωn

KS1

M1---------=

M1

Ks1

Gain

ω

ωn

KS1

M1---------

KS2

M2---------= =

M2

Ks2

• The ‘K’ values vary significantly for Elastomers and Isolation pads as the load varies, therefore graphical values are often required to find the spring constants.

• An example of a practical active vibration control system is piezo electric actuators mounted on the skin of an airplane wing. When unwanted vibrations occurred in the wings the actuators could have voltages applied to counteract the vibrations (both axial and torsional). [Mechani-cal Engineering, 1995]

• Try problems V14, V15, V16, V17, V18, V19, V20, V21, V22, V23

References

Mechanical Engineering, “Controlling Wing Flutter With Miniature Actuators”, Mechanical Engineering Magazine, ASME, 1995.

29.4 VIBRATION MEASUREMENT

x

Kd

F

Inlet valve

Vibration Source - hammers can be used to generate impulse/step function responses. Load cells/vibrators can be used to excite frequency responses (Bode plots and phase shift plots)

Sensors - Velocity Pickups/Accelerometers - lightweight devices that are mounted on structures. They produce small voltages (approx. 10mV). Velocity meters are not as accurate as accelerometers. Accelerometers are very common, and are used for vibrations above 1KHz. Many other sensors are possible.

Preamplifiers - Can power sensors, filter and amplify output.Signal Processor - Many types used, from software packages, to older pen based plotters,

or tape recorders

• Consider the example of an amplitude and phase plot measured for a real device,

pre-amp

signal processor/recorder

Source of vibrations,or site for vibrationmeasurement

Sensor

29.5 VIBRATION SIGNALS

• Some of the wave properties of interest,

freq (Hz)

acce

lera

tion 0 dB

30 dB

-30dB

Phase

0°

-90°

-180°

1000 2000 4000

• It is worth recalling the discussion of signal spectrums

29.6 VIBRATION TRANSDUCERS

29.6.1 Velocity Pickups

• Output voltage is proportional to velocity (V/(cm/s))

assuming a sinusoidal wave,

x t( ) ωt φ+( )sin=

x t( ) x0 xi iωt φi+( )sini 1=

∞

∑+=

or a more complicated periodic wave,

for the velocity,

x' t( ) xiiω iωt φi+( )cosi 1=

∞

∑ xiiω iωt φi 90 2π360---------

+ + sin

i 1=

∞

∑= =

for the acceleration,

x'' t( ) xii2ω2

iωt φi+( )cosi 1=

∞

∑ xii2ω2

iωt φi 180 2π360---------

+ + sin

i 1=

∞

∑= =

also recall the RMS values of a wave are,

XRMS1T--- x

2t( ) td

0

T

∫=

for sinusoidal waves this gives,

XRMS

xpeak

2------------ (sinusoidal waves)=

we can also check for periodic signals with a specific function, In this case thetime-average autocorrelation function,

ψ τ( ) 1T--- f t( )f t τ+( )

0

∞∫T ∞→

lim dt=

• These devices have low natural frequencies, and are used for signals with higher frequencies.

• well suited to measuring severe vibrations, but it may be affected by noise from AC sources.

• because signals are velocity, some form of integration must be done, making these devices bulky, and somewhat inaccurate

• There are two common methods for mounting velocity pickups,- Magnetic mounts allow fast and easy mounting, but the magnetic mount acts as a slight

spring mass isolator, limiting the frequency range.- Stud mounted transducers have a thin layer of silicone grease to improve contact

29.6.2 Accelerometers

• Compared to velocity pickups- smaller- more sensitive- wider frequency range

• electronic integrators can provide velocity and position

• The accelerometer is mounted with electrically isolated studs and washers, so that the sensor may be grounded at the amplifier to reduce electrical noise.

velocity pickup

stud

magnetsilicone

surface

hookup wire

velocity pickup

surface

hookup wire

grease

• Cables are fixed to the surface of the object close to the accelerometer, and are fixed to the sur-face as often as possible to prevent noise from the cable striking the surface.

• Background vibrations in factories are measured by attaching control electrodes to ‘non-vibrat-ing’ surfaces. (The control vibrations should be less than 1/3 of the signal for the error to be less than 12%)

• Piezoelectric accelerometers typically have parameters such as,- -100to250°C operating range- 1mV/g to 30V/g- operate well below one forth of the natural frequency

• Accelerometer designs vary, so the manufacturers specifications should be followed during application.

• There is often a trade-off between wide frequency range and device sensitivity (high sensitivity requires greater mass)

• Two type of accelerometers are compression and shear types.

accelerometer

isolatedisolated

surface

hookup wire

waferstud

Sealant to prevent moisture

*************** Include copy of figure

• Mass of the accelerometers should be less than a tenth of the measurement mass.

• Accelerometers can be linear up to 50,000 to 100,000 m/s**2 or up to 1,000,000 m/s**2 for high shock designs.

• Typically used for 10-10,000 Hz, but can be used up to 10KHz

• Temperature variations can reduce the accuracy of the sensors.

• typical parameters are,

• These devices can be calibrated with shakers, for example a 1g shaker will hit a peak velocity of 9.81 m/s**2

29.6.3 Preamplifiers

• The input can be either current or voltage

• sensor signals often have very low values.

• the output of preamplifiers is typically voltage

• these devices can also provide isolation both to and from the sensor

• current amplifiers generally are more costly, but they are more immune to noise.

29.6.4 Modal Analysis

• Basically, excite a vibration, and measure how it is transmitted through a structure

sensitivity

4.5 pC/(m/s**2).004

resonant f (Hz)

22 KHz180KHz

29.7 DEALING WITH VIBRATIONS

• Vibrations are basically the result of cyclic applications of forces. After the vibrations have been identified as frequencies, the phenomenon can be associated with physical design features.

29.7.1 Sources

• unbalanced rotating masses - these can be overcome by addition of counterweights

• rubbing will result in partial or full contact during some repetitive motions. Rubbing often wors-ens and leads to failure.

accelerometer

plate

load cell

Fourier Analyzer

Modal data (used forBode and Phase Shift Plots)

********* Include Figures from pg. 349

• misaligned couplings can lead to displacement or bending forces that induce vibrations.

• loose fittings will knock, rock, rub, etc.

• resonance caused by lack of damping.

• oil whirl and whip - oil films in hole shaft bearings can flow in a sporadic manner, causing the shaft to vibrate, sometimes catastrophically.

• try problems V24, V25, V26

• For advanced practice try problems V27, V28

29.8 RESOURCES

29.8.1 Library

• There are a number of resources available to the student. In many cases older textbooks will contain valuable information.

29.8.2 Computer

29.9 PRACTICE QUESTIONS

V14. A machine contains a 60Hz source of vibration that disturbs other machines in the same room. Find the spring coefficient, and natural frequency of an elastomer (with a damping coef-

ficient of .05) that will isolate the vibration source.

V15. There is a large machine that weighs 1000 Kg, and has three legs. We will mount some elas-tomer under each leg. The graph below shows the characteristics of the isolator. From the graph determine a spring constant (hint: a slope), and determine the natural frequency, and damping ration of the mount.

V16. A piece of electronic equipment is to be isolated from a mounting panel which is vibrating at 8Hz. If 90% isolation is specified what static deflection would you expect? (ans. 0.042 m)

V17. A piece of mechanical equipment contains a 60Hz electric motor driving a reciprocating mechanism which generates motion excitation at 12Hz. The equipment has a total mass of 450 Kg and is mounted on isolators. To establish some criteria regarding the actual isolation an accelerometer is mounted along the vertical axis of the machine. First, the static deflection is measured and found to be 11mm. When the machine is switched off after operation, the output from the accelerometer is captured as a trace, on a storage oscilloscope. The response ratio between two adjacent positive maxima on the trace (i.e., one cycle separation) is 1.65,

a) find the damped natural frequency of the equipment. (ans. 29.77 rad/s)b) determine the percentage isolation (interpolating fig 9.8is adequate if you note your

entry figures). (ans. 72%)c) if the equipment was operated on a 50Hz supply (motor speed reduced by 17%), explain

M = 1000kg

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

briefly what changes you would expect in the above results. (ans. <60%)

V18. It is required to provide 80% isolation for a machine using isolation pads conforming to den-sity C specifications in the figure below (note the damping ratio is 0.05). Mounted equipment will generate dynamic forces at 2600 rpm.

a) What would be the minimum mass of the machine if the total isolator pad area is limited to 750 cm2?

b) for the same pad area and a machine mass of 275 kg, what isolation efficiency would be afforded if density A material were used?

V19. Using the data shown in the figure above a total of 4 isolators were selected to isolate a small piece of equipment having a mass of 10kg. The dominant forcing frequency is 60Hz and damping can be neglected.

a) determine the isolation afforded by Isolator B.b) What would be the result if Isolator C were substituted?

V20. Using the figure above, select an isolator to assure 96% isolation efficiency at a forcing fre-quency of 112Hz (assume no damping). What would be the static deflection of the isolator selected for a load/isolator of 15N (ans. 0.075cm)

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

V21. For a system weight/per isolator of 35N, use the figure above determine what isolation effi-ciency would be obtained if isolator D were used. System forcing frequency is 80Hz. Damping ratio = 0. (ans. 95%)

V22. Cork isolation pads (damping ratio 0.05 use the figure above) are used at the 4 corners of the base of a small machine that weighs 2 KN and generates a forcing frequency of 105Hz. If the pads are 10cm wide by 20cm long, what isolation efficiency would be afforded if they are made of a slab of density C? (ans. 95%)

V23. Four springs, each having a spring constant of 2400 N/m are placed at the 4 corners of a cen-trally loaded baseplate. What is the system weight if the static deflection is 10mm? (ans. 96N)

V24. An accelerometer is attached to a piece of equipment and connected via a preamplifier to an oscilloscope. The equipment is mounted on vibration isolators. Assuming a single d.o.f. model which we can see is underdamped, what is the system damping ratio if the measured ratio of the amplitude response at two consecutive positive peaks on the decaying harmonic waveform is 1.8? (ans. )

V25. In the design of an accelerometer the requirement is for the maximum measurement error to be limited to 6% at 1/3 of the resonant frequency (i.e. magnification factor 1.06). What would be the maximum damping ratio to meet these requirements? (ans. 0.48)

V26. Using an oscilloscope the ratio of two subsequent maxima was found to be 1.6. Determine the damping ratio. (ans. 0.075) -- EXPLAIN

V27. A motor (mass M1) and controller (mass M2) are mounted on a heavy base (mass M3). Iso-lators are used to mount all masses as shown below. Dynamic forces generated by the motor are forcing the system so it will be necessary to determine the forces on M2 and the support structure below M3.

a) show the lumped parameter model which represents the system.b) using the Force/Current Mobility Analogue show the equivalent electrical circuit.

Mass (Kg)

displacement(m)

1000 2000

.001

.002

.003

.004

V28. (A long problem) Find the force transmitted by the unbalanced load in the washing machine, to the floor. The rotating mass is 1kg at a distance of 20cm from center, turing at a speed of 30rpm. The assembly consists of the upper drum and motor assembly with mass M2, is rested on a spring, that in turn rests on a large mass. This mass is suspended on a solid floor using a spring/damper combination.

a) Develop the transfer function for the force applied by the eccentric mass to the ground.b) Determine the input forcing function (from the eccentric mass)c) Develop the time based reaction using Laplace transforms.d) Use Fourier transforms to find the effect of the system in steady state.e) Draw Bode plots for the system.

V29. A machine stands on 6 legs in a corner of a room. in total the machine weighs 10,000 kg, and a vibrational force of 50N is applied at 120Hz by a rotating mass, and a force of 2 N is

Motor (M1)Controller (M2)

Base (M3)

R1, Ce1R2, Ce2

R3, Ce3

Support Structure

motor drives an unbalancedmass

M2=50kg

M1=100kg

x

y

Ks1=10KN/m

Ks2=50KN/m Kd=20KNs/m

applied at 60Hz by an AC motor. It has been decided that an isolator will be added to reduce the vibration passed to the floor. 6 isolators will be attached to the legs of the machine. The isolators will be spring damper pairs connected in parallel. (Note: assume the floor movement is negligible). The spring constant is 100KN/m, and the damper is 200KNs/m.

a) Draw a Free Body Diagram of the system.b) Develop a transfer function for the force input to the machine mass, to the force output

applied to the floor.c) Find the isolation for the two vibrations using the results in b).d) Find a Laplace input function for the vibration, and determine what the Laplace output

function will be.e) Determine the time based response of the function in d).f) Draw a Bode Plot for the transfer function in b).g) Use the Bode plot in f) to find the steady state forces applied to the floor.h) Use the Bode plot in f) to find the isolation of the vibrations.i) Design an elastomeric isolator (instead of the spring-damper) to get 90% isolation for

the 60Hz force.

V30. Given the car wheel modelled below, relate a change in the height of the wheel to a change in the height of the car. The final result should be a Laplace transfer function of ‘ycar/yroad’.

V31. Find the time response ‘x(t)’ of a system with a transfer function G(s) that is excited by the force ‘F(t)’.

yroad

ycar

CAR BODY

(Mper wheel)

Ks Kd

V32. A large machine weighs 1000kg and vibrates at 20Hz, design an inertial damper.

V33. A 10kg machine is set on isolation pads and vibrates at 60Hz, what should the natural fre-quency of an elastomer isolation pad be? If there are 3 pads, what should their spring constant be?

V34. A machine stands on 6 legs in a corner of a room. in total the machine weighs 10,000 kg, and a vibrational force of 50N is applied at 120Hz by a rotating mass, and a force of 2 N is applied at 60Hz by an AC motor. It has been decided that an isolator will be added to reduce the vibration passed to the floor. 6 isolators will be attached to the legs of the machine. The isolators will be spring damper pairs connected in parallel. (Note: assume the floor movement is negligible). The spring constant is 100KN/m, and the damper is 200KNs/m.

a) Draw a Free Body Diagram of the system.b) Develop a transfer function for the force input to the machine mass, to the force output

applied to the floor.c) Find the isolation for the two vibrations using the results in b).d) Find a Laplace input function for the vibration, and determine what the Laplace output

function will be.e) Determine the time based response of the function in d).f) Draw a Bode Plot for the transfer function in b).g) Use the Bode plot in f) to find the steady state forces applied to the floor.h) Use the Bode plot in f) to find the isolation of the vibrations.i) Design an elastomeric isolator (instead of the spring-damper) to get 90% isolation for

the 60Hz force.

29.10 SOUND/VIBRATIONS TERMS

absorption lossambient noiseattenuationaudible rangeband pressure levelbeats

G s( ) 102s 4+( )s

---------------------- x s( )F s( )-----------= =

F t( ) 5 t 0sec≥,=

F t( ) 0 t 0sec<,=

broad band noisecontinuous noisecontinuous spectrumcycledecibel (dB)diffuse fieldeffective sound pressurefree fieldfrequencyimpulse noiseinsertion lossintermittent noisemicrobarmicrophoneoctavepeak-to-peakpeak levelpink noisepitchplane sound waverandom noisereverberationsoundsound absorptionsound analyzersound levelsound level meterspectrumspherical wavestanding wavewavelength

29.11 REFERENCES

Irwin, J.D., and Graf, E.R., Industrial Noise and Vibration Control, Prentice Hall Publishers, 1979.

30. INTERNAL COMBUSTION ENGINES

• Internal combustion engines contain a wide variety of kinematics and dynamics problems.

• Some of the criteria for differentiating engines includes,Fuel Air Mixing

- Otto cycle - air/fuel mixed before compression, equal fuel-air ratio. A highly vol-atile fuel is required.

- Diesel - air/fuel mixed after compression, excess air. Compression is the main source of ignition.

Cylinder Exhaust- two stroke - within one cycle of the piston a compression and exhaust occur.

Basically the mixture is ignited, the gas expands, and near the end of compres-sion the cylinder is opened to vent the gas, at the same time new fuel-air is injected. The valves are then closed, the cylinder advances, and ignition occurs again. These engines are simpler, but less efficient.

- four stroke - In this engine the piston cycles twice. One cycle is to intake fuel-air mixture and combust. The second cycle is to vent the cylinder. The four strokes are expansion, exhaust, intake, compression.

Cylinder Arrangement- single cylinder - only one cylinder- in-line - all of the cylinders are side by side in a line. - V block- there are two rows of cylinders that form a V-shape- opposed-piston - there are two rows of cylinders that are 180° away from each

other- radial - cylinders are arranged at regular intervals about a single cam shaft

Connecting Rods- Single crank per pair of cylinders

fork and blade designarticulated rod

- separate crank throws for each piston

30.1 POWER

• We can estimate the power generated in the combustion cycle using the gas pressure and change in volume.

• Basically in an engine the volume of gas is constant, except during ignition, where the volume of gas is almost instantly increased. This results in a corresponding increase in pressure, that drives the displacement of the piston.

• For a typical engine the cylinder pressure curves will look something like those below from Shi-

gley and Uicker for a 401 cubic inch V6 truck engine.

• We can also estimate actual horsepower and torque of the motor by braking the engine, and cal-culating powers over a range of speeds. [Shigley and Uicker]

• In an engine typical component massed are, [Shigley and Uicker]

pres

sure

(ps

i)

600

300

0

volume(cubic in.)

0 40 80

pres

sure

(ps

i)

600

300

0

crankshaft angle (deg.)-105 0 105

net

gross

horsepower

torque

brak

e ho

rse

pow

er

engine speed (rpm)

50

100

150

200

1000 2000 3000 4000

• We can typically relate engine power to displacement volume, with some variance considered for engine type,

• The ideal pressure volume diagram for an engine is shown below. It does not have the rounding seen on the practical graph before (the rounding is the result of slow valve openings, and the finite time for combustion).

Motion

Reciprocating

Rotating

Component

pistonpiston pinpiston ringretainersrod

rodbearings

Mass

1.560 kg0.3175 kg0.1270 kg0.00034kg0.3600 kg

0.92600 kg0.10128 kg

Engine Type

AutomotiveMarine Diesel

horsepower/cubic in.

0.55-1.00 (0.70 avg.)0.10-0.20

• We can find the work done by the engine by looking at the change in pressure and volume on the expansion stroke.

30.2 KINEMATICS AND DYNAMICS

• Basically the engine has rotating parts that generally behave as multiple crank slider mecha-nisms. The force is applied on the slider, and the crank is rotated as a result.

• This analysis begins with a model of the mechanism. We can use this to do a static and kinematic analysis of the mechanism. We can find the driving force on the piston as a pressure of the gas

p2

p3

p4

p1

v1 v2

exhaust

expansion

compression

suction

pressure

volume

em

pb

pi-----

pb

fcpi'( )--------------= = r

v1

v2-----=

where,

em mechanical efficiency=

pb brake mean effective pressure=

pi indicate mean effective pressure=

pi′ theoretical indicated mean effective pressure=

fc card (correction) factor (0.9-0.95 typ.)=

r compression ratio=

es it expands, and the area of the piston.

• Next, to do the dynamic analysis of the engine we need dynamic parameters for each of the com-ponents. In practice we would also need to consider the bearings, pins, etc.

• The results of this analysis will be a time variant function for each cylinder. These can then be combined for all of the cylinders in the engine to get an overall dynamic function for the engine. The total mass of the engine will tend to act as an inertial dampener for the reciprocat-ing forces in the engine.

CylinderPiston

RodCrankshaft

30.3 REFERENCES

Shigley, J.E., Uicker, J.J., “Theory of Machines and Mechanisms, Second Edition, McGraw-Hill, 1995.

1. SOUND CONTROL

1.1 BASIC PROPERTIES OF SOUND

• Pressure waves

• an approximation for sound propagation is a sphere as a wavefront.

• an ideal sound wave travels as a uniform wave in a radial direction (thus having uniform distri-bution over the sphere). If we watch the pressure at point ‘P’ as a single sound wave (soliton) travels by, it will look like,

• The wave equation for sound in air is given below,

r

P

P

t

Ppeak

Patm

Prms

• If a sound is a pure tone, it will look like,

cx∂

∂ 2

ut∂

∂ 2

u=

For transmission in a continuous rod,

In solids,

c Eρ---=

In air,

cγp0

ρ--------

cpp0

cvρ----------= =

where,

c speed of sound=

E Youngs modulus=

ρ density=

c speed of sound=

x displacement along solid=

p0 ambient density=

cp specific=

cv specific=

• For air the speed of sound can be approximated with,

• Velocities in other materials (at 21.1°C) are, [Source Irwin and Graf]

Prms

Ppeak

t

P

T

f 1T--- Hz( )= 1 Hz( ) 1 cycle

sec-------------

2π radianssec

-------------------- = =

λ cT=

PressureThe peak pressure of the wave above normalThe Root Mean Squared (RMS) effective pressure of the wavePeriod of time for one cycle of the waveThe effective frequency of the wave

P

Speed of sound in the propagating mediumThe length of one wave cycle in a medium

Ppeak

Prms

Tfc

λ

PRMS1

2-------Ppeak= (for sinusoidal waves)

PRMS1T--- P

2t( ) td

0

T

∫=

Typical pressures values given are not RMS

c 20.05 K ms----

=

degrees kelvin = °C + 273.2°C

• Sound powers cover such a wide range that it makes sense for the values to be expressed in a logarithmic scale, a few values are given for reference. Note: this value measures the sound source itself, not what the listener hears. [source Irwin and Graf]

• The sound power level is useful when considering the source, but as we move farther away from the source the sound pressure our ears can detect will drops off. Another logarithmic scale can be used to express power levels we hear.

material

airwaterconcreteglassiron

c (m/s)

344137230483658

λ (m) for 1000Hz tone

0.3441.3723.0483.658

leadsteelhardwoodsoftwood

51821219518242673353

etc...

Lw 10 WWref----------

dB( )log=

Lw = sound power levelW = power of the soundWref = 10-12 W (a power level suited to human perception)

sound

threshold of hearing

Lw

0background in sound studiowhisper at 2mconversation at 1moffice environmentrestaurant/storepneumatic drillpunch presstruck hornpropeller airplanepropeller cargo airplanesaturn V rocket

204050506090110110120140195

quiet

noisy

very loud

intolerable

1.1.1 Adding/Subtracting/Averaging dB Values

• recall that log (or dB) values can’t be added directly, they must be converted back to normal val-ues first.

LP 20 PPref---------

dB( )log 10 P2

Pref2

---------

dB( )log= =

Lp = pressure levelP = RMS sound pressure (Pa or N/m2)Pref = 20*10-6 Pa

because (pressure)2 is proportional to sound power. Also, add and subtract as squared values because of relationship to power.

Lpt10 10

Lpi

10------

i 1=

n

∑

dBlog 10Pi

Pref---------

2

i 1=

n

∑

log= =

LWt10 10

LWi

10--------

i 1=

n

∑

log 10Wi

Wref----------

2

i 1=

n

∑

log= =

LP1n--- 1

n--- 10

LPi

10-------

i 1=

n

∑

log=

LPs10

P1

Pref---------

2 P2

Pref---------

2–

log 10 10

LP1

10--------

10

LP2

10--------

–

log= =

P 1n--- Pi

2

i 1=

n

∑=

• Obviously averaging values must observe the same restrictions as addition,

• Other log properties are,

• Try problems S1, S2, S3, S4

e.g. Add, then subtract the sound power levels Lw1 = 5dB and Lw2 = 10dB

Lw1 Lw2+ 10 10

Lw1

10--------

10

Lw2

10--------

+

log 11.2dB= =

Lw2 Lw1– 10 10

Lw2

10--------

10

Lw1

10--------

–

log 8.35dB= =

e.g. Add, then subtract the sound power levels Lp1 = 20dB and Lp2 = 15dB

Lw1 Lw2– 10 10

Lp1

10-------

10

Lp2

10-------

–

log 18.3dB= =

e.g., average the values Lp1 = 2dB, Lp2 = 6dB, Lp3 = 3dB

Lp 10 10

210------

10

610------

10

310------

+ +3

---------------------------------------------

log 4.0dB= =

10xlog

x=

xy( )log y x( )log=

xy( )log x( ) y( )log+log=

xy--

log x( ) y( )log–log=

1.2 SOUND MEASUREMENTS

• Previous discussions centered on simple sound measurements, but actual machines are more complicated.

• Sound levels are unequal in different directions.

• The unequal distribution of sound energy can be expressed with a directivity factor,

• A directivity Index can be helpful when the directivity factors have a wide range

r

Lp=55dB

Lp=72dB

Lp=90dB

Lp=80dB

Lp=92dB

Lp=51dB

power variesat an equalradius allaround the source

Qi

Iθ

I----

r r1=

Pθ

P------

2

r r1=

10

LPiLP–

10-------------------

= = =

Iθ = intensity measured at one point at r1I = average intensity over spherical radius r1Q = a unitless measure of sound in a particular directionPθ = same as Iθ, except for pressureP = average pressure over spherical radius r1

LP 10

10

LPi

10-------

i 1=

n

∑n

---------------------

log=

We deal with it this way because energy is additive, pressure is a resultant property.

• Try problems S5, S6

1.2.1 Measurement Techniques

• It is impossible to measure sound at all points around a source, so some points must be selected.

• One technique is to divide the surface of the sphere into a number of flat polygons.

• It is best to have an anechoic environment

anechoic - an environment that (effectively) eliminates reflected noise.

• In some cases only half of the measurement sphere can be used because of object size, or

DI 10 Q dB( )log LPiLp–= =

e.g. for an 8 sided full sphere

x

0.000.000.820.820.000.00-0.82-0.82

y

0.820.820.000.00-0.82-0.820.000.00

z

0.58-0.580.58-0.580.58-0.580.58-0.58

Multiply values by the radius‘r’ of the sphere of measurementto get the actual points in space.

Commonly foam pyramids onwalls all around a test objectto absorb sound that wouldnormally be reflected.

requirements.

• The basic procedure is,1. Select number of points to measure2. Get list of points by selecting measurement radius, and with point table (only the top

half is a hemisphere)3. Measure sound pressure at each point (Lp)4. Correct for points on plane if half sphere5. Find Lp (correct for -3dB if hemisphere)6. Find Q or DI for each point is required

• try problems S7, S8, S9, S10, S11, S12

1.3 THE HUMAN EFFECTS OF SOUND

• The human ear does not hear all sounds as equal.

• One simple measure is the ‘phon’, it is based on experimental data. (Note that Lp = phon at 1KHz). [Source Irwin and Graf]

- When this is done, it is best to have the origin of the sound sourceat the (0,0,0) of the measurements.

- Correct for half sphere measurements by subtracting 3dB from themeasurement points taken at ground level, and from average. Note,if a point is on the floor, but not on a polygon that crosses it, the 3dBis not subtracted.

• Another measure that is adjusted for loudness is the ‘sone’. This compensates for what we would consider a perceptual loudness level.

100 200 500 1000 2000 500050 10000 20000

120

110

100

90

80

70

60

50

40

30

20

10

0 Phons

120

110

100

90

80

70

60

50

40

30

20

10

Soun

d Pr

essu

re L

evel

freq (Hz.)

e.g., Given Lp = 70dB at 400Hz, how much louder would it seem at 100Hz.

at 400Hz, LLl = 75phonsat 100Hz, LLl = 63 phons

therefore, it would seem (63-75)=-12 phons louder, or 12 phons quieter. This isconsistent with human inability to hear lower frequencies.

• As we saw before, sound is made up of a variety of frequencies, so we must determine how all of the components are combined into a single value.

• The loudness index contours are given below. [Irwin and Graf]

S 2

LL 40–10

-----------------

= LL 33.3 Slog 40 phons( )+=

Note that it is relative to 40 phons

e.g., for the previous example,

at 400Hz

at 100Hz

S 2

75 40–10

------------------

11.3 sones( )= =

S 2

63 40–10

------------------

4.9 sones( )= =

Step 1: Measure sound pressure at a number of frequencies, generally 50-10000Hz.The frequency values are typically separated by 1, 1/2, 1/3 of an octave.

1decade = *101 octave = *21/2 octave = *1.411/3 octave = *1.26

Step 2: Enter the center frequencies and measured loudness levels in a table and addloudness index.

e.g. 1 octave

center freq. (Hz)

100200400800160032006400

Lp (dB)

69707177747572

Loudness Index

46813121615

*2

Im=16

*2*2*2*2*2

100 200 500 1000 2000 500050 10000 20000

120

110

100

90

80

70

60

50

40

30

20

10

0

Lou

dnes

s In

dex

Ban

d Pr

essu

re L

evel

(dB

)

Freq. (Hz)

150

80

20

10

5

3

1.5

0.8

0.25

• Try problems S13, S14, S15

• We can use a similar method for the perceived noise level using the ‘noys’ graph below, and use this in place of the graph in the previous method. [source Irwin and Graf]

Step 3: Calculate the total loudness in sones

St Im 1 K–( ) K Ii

i 1=

n

∑+=

St = total loudness (sones)Im = maximum loudness index valueIi = individual loudness index valuesK = constant K = 0.3 for 1 octave

K = 0.2 for 1/2 octaveK = 0.15 for 1/3 octave

St 18.0 1 0.3–( ) 0.3 7.7 8.0 9.0 18.0 13.0 15.0 10.0+ + + + + +( )+=

St 36.81 sones( )=

1.3.1 Background Noise

• The previous measurement methods are suited to measuring for specific noise sources, but there is also ambient background noise.

• Noise Criteria Curves, use the given curves for Noise Criteria (NC) or Preferred Noise Criteria (PNC) [Source Irwin and Graf]

100 200 500 1000 2000 500050 10000 20000

120

110

100

90

80

70

60

50

40

30

20

10

0

Freq. (Hz)

Ban

d So

und

Pres

sure

Lev

el (

dB)

250

125

65

35

16

8

4

2

1

Noys

100 200 500 1000 2000 500050 10000 20000

120

110

100

90

80

70

60

50

40

30

20

10

0

Octave-band Center Freq. (Hz)

Oct

ave-

band

Pre

ssur

e L

evel

(dB

)

NC-65

NC-50

NC-35

NC-20

NC-15

• To use the curves, plot on values, and the highest NC curve that is touched is the NC level

100 200 500 1000 2000 500050 10000 20000

120

110

100

90

80

70

60

50

40

30

20

10

0

Octave-band Center Freq. (Hz)

Oct

ave-

band

Pre

ssur

e L

evel

(dB

)

PNC-65

PNC-50

PNC-35

PNC-20

PNC-15

• There are some basic suggested levels for background noise indoors,

• try problems S16, S17, S40

1.3.2 Weighting Sound Values to Compensate for the Human Ear

• Previous methods use graphical techniques to adjust sound response.

• We can also use an adjusted weighting [source Irwin and Graf]

NC-45

NC-40

NC-35

NC-30

freq(Hz)

Lp(dB)

Highest point determines rating,here it is NC-38

conditions

sleepinglivingofficeaudio studiostore/restaurant

NC level

25-3535-4530-3515-20

PNC level

25-4030-4030-4010-20

laboratorycomputer room

35-5040-4545-60

35-4540-5045-55

dB x+ dBA→adjusted value (also dBB, dBC)

a correction coefficient related to frequencya measured sound pressure level

Frequency(Hz)

1012.516202531.540506380100125160200250315400500630800100012501600200025003150400050006300800010000125001600020000

A weighting(dB)

-70.4-63.4-56.7-50.5-44.7-39.4-34.6-30.2-26.2-22.5-19.1-16.1-13.4-10.9-8.6-6.6-4.8-3.2-1.9-0.800.61.01.21.31.21.00.5-0.1-1.1-2.5-4.3-6.6-9.3

B weighting(dB)

-38.2-33.2-28.5-24.2-20.4-17.1-14.2-11.6-9.3-7.4-5.6-4.2-3.0-2.0-1.3-0.8-0.5-0.3-0.10000-0.1-0.2-0.4-0.7-1.2-1.9-2.9-4.3-6.1-8.4-11.1

C weighting(dB)

-14.3-11.2-8.5-6.2-4.4-3.0-2.0-1.3-0.8-0.5-0.3-0.2-0.1000000000-0.1-0.2-0.3-0.5-0.8-1.3-2.0-3.0-4.4-6.2-8.5-11.2

• Try problem S18

1.3.3 Speech Interference

• A simple measure to estimate interference with human speech is the Speech Interference Level (SIL) and more recently the Preferred Speech Interference Level (PSIL)

• The values can be correlated to speech levels and distances that they may be heard at, [Source Irwin and Graf]

e.g., given measured values find the total pressure

Band centre freq. (Hz)

1252505001000

Lp (dB)

71616075

correction factor

-16.1-8.6-3.20

dBA

54.952.456.875.0

Lt 10 10

Lt

10------

i 1=

n

∑

dBAlog=add values,

Lt∴ 10 10

54.910

----------

10

52.910

----------

10

56.810

----------

10

75.010

----------

+ + +

dBAlog 75.1dBA= =

PSILLP 500Hz( ) LP 1KHz( ) LP 2KHz( )+ +-----------------------------------------------------------------------------dB=

100

90

80

70

60

50

400 2 4 6 8 10 12

PSIL (dB)

distance between talker and listener (ft)

shout

loud voice

raised voice

normal voice

e.g. given measured values below, find the distance levels for different voice levels

f (Hz)

50010002000

Lp

807291

PSIL80 72 91+ +

3------------------------------ 81dB= =

voice level

talkingloud talkingvery loud talkingshouting

maximum distance

03 inches1 foot2 feet

• Try problem S19

1.4 NOISE CONTROL REGULATIONS

• WHY? to give formal standards so that engineers can guarantee excessive limits are not exceeded. Without formal numbers, we could “bend-the-rules”

• The Occupational Safety and Health Act of 1970 (USA)- for prolonged worker exposure- can be measured with device attached to worker for a period of time (these measure total

sound exposure).- safe exposure levels [source Irwin and Graf]

- When calculating, consider that noise levels change all day long,

duration/day(hrs)

864321.510.50.25

Sound Level(dBA)

90929597100102105110115

DC1

T1------

C2

T2------

C3

T3------ …

Cn

Tn------+ + + +=

Ci,Ti - actual time level ‘C’ and the legal maximum time ‘T’ at each power levelD < 1 for satisfaction of safety considerations


1.4.1 Other Hearing Effects of Noise Exposure

• A table of international acceptable noise levels is given below (from [Hay, 1975])

e.g., Bob is in a factory where he works in stamping (110dBA) for 3 hours beforelunch time, and for 4 hours after lunch in assembly (95dBA)

3 hours at 110dBA, therefore C1 = 34 hours at 95dBA, therefore C2 = 4

from table T1 = 0.5T2 = 4.0

D3

0.5------- 4

4---+ 6 1+ 7= = =

6 times acceptable limits in stamping alone, 7 times in total

therefore, Bob needs hearing protection, and some measuresshould be taken to reduce noise.

• Typical requirements for exposure to noisy environments

1.4.2 Measuring Noise Revisited

• From the Noise Control Act of 1972 (USA) a basic procedure was developed for estimating noise exposure for the general public.

country

GermanyFranceBelgiumLuxembourgNetherlandsUKIrish RepublicItalyDenmarkSwedenNorwayUSA (federal)Canada (federal)AlbertaQuebecBritish ColumbiaOntarioSaskatchewanNew BrunswickAustraliaVictoria

Steady

90

NoiseLeveldBA

Time

8

Exposurehours

Overriding

-

LimitdBA

Impulse

-

PeakSPLdB

Impulse

-

numberper day

9090--9090909085-90908590909090909090

4040--8-84040-888888--88

-110--135*-115115115-115115115-115115--115105

*UK over-riding limit is 135dB (SPL), on fast response

-140--150-140---140140140-------

-100--------100---------

Sound Level (dBA)

<8585-8990-9596-105106-110>110

Protection Recommended

none requiredClass CClass BClass AClass A plug and Class A or B muffClass A plug and class A or B muff - limited exposure

• Given a number of noise readings at the perimeter of a facility over a day, find the noise levels.

A.1 An example list of measurements sampled over 1 day

time

8am-10am10am-1pm1pm-5pm5pm-7pm7pm-12am12am-3am3am-5am5am-8am

dBA

7565707560657550

duration (hrs)

23425323

A.2 Make up a table as shown

dBA

5060657075

total hours

35646

proportion at level cumulative Pj 1-Pj

3/24 = 0.1255/24 = 0.2080.2500.1670.250

0.1250.3330.5830.7501.000

0.8750.6670.4170.2500.000

A.3 Draw a graph of individual and cumulative probabilities

probability of exceeding a certainsound level

dBA

probability

A.4 Calculate the total RMS sound energy over some period of time.

Leq 10 Pi( ) 10

LPi

10-------

i 1=

n

∑ dBAlog=

Leq 10 0.125

5010------

×10 0.208

6010------

×10 0.250

6510------

×10 0.167

7010------

×10 0.250

7510------

×10+ + + +

log=

Leq 70.2dBA=

A.5 Calculate the Standard Deviation

SD PjLj( )2PjLj

j 1=

n

∑ 2

–j 1=

n

∑12---

dBA=

SD =

A.6 Calculate Noise Pollution Level

Lnp = Leq + K(SD) K = 2.56 by convention (95% sample)

Lnp = ??? dBnp

B.1 Make up a table as shown (Assume time penalty of 10dBA from 10pm to 7am)

dBA

5060657075

total hours

13344


1/15 = 0.0673/15 = 0.2000.2000.2670.267

0.0670.2670.4670.7331.000

0.9330.7330.5330.2670.000

B.2 Calculate the total Sound energy for the non-penalty daytime hours

Ld 10 Pj( ) 10

Lj

10------

j 1=

n

∑ dBAlog=

Ld 10 0.067

5010------

×10 0.200

6010------

×10 0.200

6510------

×10 0.267

7010------

×10 0.267

7510------

×10+ + + +

log=

Ld dBNP=

C.1 Make up a table as shown (Assume time penalty of 10dBA from 10pm to 7am)

dBA

50606575

total hours

2232


2/9 = 0.2222/9 = 0.2220.3330.222

0.2220.4440.7771.000

0.7880.6660.3330.000

C.2 Calculate the total Sound energy for the penalty night time hours

Ln 10 Pj( ) 10

Lj

10------

j 1=

n

∑ dBAlog=

Ln 10 0.222

5010------

×10 0.222

6010------

×10 0.333

6510------

×10 0.222

7510------

×10+ + +

log 10+=

Ln dBNP=

D.1 Find the combined day/night values

Ldn 10 924------10

Ln

10------

1524------10

Ld

10------

+ dBNPlog=


1.4.3 Noise Levels

• The American EPA developed some suggested noise levels. A list (of one at this point) is,

1.5 SOUND ANALYSIS INSTRUMENTS/TECHNIQUES

• The basic setup

1.5.1 Noisy machines • will be covered later, but for now we want to determine how much sound pressure they are cre-

ating at different frequencies.

Source

Trucks

Maximum

75dBA

- a noisy machine- an engineer- sound equipment- a room- complicating factors

• See spectrogram figure on page 95, figure 4.12 Irwin and Graf

1.5.2 A Room (or not) • The measurements are done most often in imperfect environments (perfect environments dis-

cussed later)

A specific noise source

general noise

Lp

f(Hz)

1. Walls allow sound from therunning machine to reflect, thereforecomplicating measurements.

Solutions for this problem:a) measure sound with machine off, then on. Subtract

basic sound level.

e.g., Lpoff = 90dB, Lpon = 95dB

b) Make walls more sound absorbing

LPmachine10 10

LPon

10----------

10

LPoff

10-----------

–

log 93.3dB= =

2. Standing wave patterns create a formof resonance with exaggerated sound levels

Solutions for the problem:

a) Take a few measurements along the standing wave and averageb) Change length so that the distance will not support wave

3. Other sound sourcesinterfere with measurementsfor other sources

1.5.3 Equipment (Microphones)

• Convert sound pressure to electrical signals

Solutions for problem:

a) same approach as in #1a)

b) Build a barrier between the unwanted noise source and measurement site

sound absorbing barrierunwanted noise source

source of interest

L l

H

h

front

top

right

H hλ4---+=

L l 2 λ4---

+=

λ cf--=

• A microphone has a finite area to measure the sound with. As the diameter ‘d’ approaches the wavelength of sound, the sound becomes distorted.

• When wind blows across a microphone, it generates a low frequency turbulence noise. This is often corrected by using dBA measurements, and foam balls.

• Corrections must also be made for changes in temperature and altitude

type

1. condenser2. ceramic3. dynamic

sense element

capacitivepiezo-electricmagnetic coil

freq. range

2Hz-20KHz20Hz-10KHz25Hz-15KHz

temp. & humidity stability

fairgoodgood

d << c/f = λ for good measurement

CTP 10 460 F+528

-------------------

12---

30B------

dBlog=

Barometric pressure (in Hg)temperature (F)

correct level to measurement

1.5.4 Noise Measurement Form

1.5.5 Equipment - Spectrum Analyzers

Noise Measurement Survey

Place:

Date/Time:

Barometric Pressure:

Temperature:

Client:

Survey Personnel:

Instrumentation:

Weight: Response:

Background Noise Level:

Remarks:

Sketch of Measurement Site: Height:

• Signal spectrum is a fundamental concept, these machines allow us to measure this

• Some possible spectrums are,

• Filters can be designed using Bode Plots. These filters can reduce or increase the amplitudes of input frequencies in certain ranges.

1. This sort of level indicates sound at all frequencies (white noise).

Lp

ω

** This spectrum could be from a fan

2. A peak in the spectrum indicates a distinct source of sound

Lp

ω

** This spike could be a 60Hz hum froman AC powered device.

3. A complex spectrum with many peaks indicates a complicated sound

Lp

ω

** This spectrum could be from a human voice

source. We may also find that the positions and magnitudes of peaks changes.

• How filtering works

ωf1 f0 f2

straight-line and actual

Eout

LP----------

The spectrum analyzer has a transfer function like that in the Bode plot.The function is a result of the electronic design.

f1 = the lower cutoff frequency (-3dB below)f2 = the upper cutoff frequency (-3dB below)f0 = the centre frequencyb.w. = bandwidth = f2-f1

-40 or -60 dB per decade3dB atcorners

(express in octaves or decades)

• Octave band analyzers (and others) are mainly distinguished by the difference between the upper and lower cutoff frequencies.

Lp input at microphone

80dB

ω

Gain of spectrum -80dB

ω

Eout

LP----------

analyzer (andmicrophonetogether)

Eout voltage output

80dB

ω

Side-bands effectivelycut off

• The narrower the bandwidth of the analyzer, the more expensive the instrument

• octave, 1/2 octave, and 1/3 octave analyzers are more common.


• The analyzers so far only look at specific frequencies but we can use them to look at a variety of frequencies.

• A typical mode of operation for older analyzers requires some interpretation. The graph is drawn by switching each channel on for a short period of time, the voltage from a certain frequency band is plotted (this allows variations at that frequency to be seen). After some period of time the centre frequency is shifted to another value. This means the bands are wider for an octave analyzer,or the bands of the analyzer are narrower for a 1/3 octave bandwidth. The wider bands hide noise when using wider bandwidths. For example the peak near 200Hz is more obvious with the 1/3 octave analyzer. The noise appears to be a multiple of 60 (3 * 60 = 180), suggest-ing that it is the motor speed, driving a geared up impeller.

f1

f0

2-------= f2 2f0= f2 2f1=

f1

f0

2m

----------= f2 2m

f0= f2 2m

f1=

Octave band

‘m’-octave analyzer

A special case

The frequency analyzerwill have four channels(or bands), or it might

In this example:

have one channel that isswitched between thefour frequency bands.In either case the effectivecoverage is shown in thebottom graph.

channel 1

channel 2

channel 3

channel 4

1.6 EQUIPMENT GENERATED NOISE

• Noise is generated by some common methods

• Estimation factor for noise generation [source Irwin and Graf]

• There are other sources that are noisy, but not suited to simple factors,- fans- electric motors- pumps- air compressors- construction equipment- household appliances

• A table of relationships for common equipment is given, [Irwin and Graf]

category

combustionimpactelectromechanicalgas streammetal on metalfluid in metal cavitiesmetal on fluidunbalanced rotation

example

furnacepunch pressmotorjet (on barbecue)gearsvalves (bathroom faucets)pumpswashing machine

FnP

Pm-------=

power of sound (W)

power of machine (W)

noise source

air compressor (1-100hp)gearsloudspeakerdiesel motorelectric motor (1200rpm)pump (>1600rpm)pump (<1600rpm)gas turbine

low

3*10-7

1.5*10-8

3*10-2

2*10-7

1*10-8

3.5*10-6

1.1*10-6

2*10-6

medium

5.3*10-7

5*10-7

5*10-2

5*10-7

1*10-7

1.4*10-5

4.4*10-6

5*10-6

high

1*10-6

1.5*10-6

1*10-1

2.5*10-6

3*10-7

5*10-5

1.6*10-5

5*10-5

Fn

Fans - Centrifugaland axial

Lw 10 Frlog 20 Pslog Kf+ +=

Fr = volume flow rate (ft3/min, m3/s)Ps = static pressure (in H2O or cmH2O)Kf = constants for fan types, [Irwin and Graf pg 117]

Equipment Equation/Source

Fans - Induced draft Lw 10 hplog 10 Pslog Kfd+ +=

hp = rated horsepower (750-7500 hp)Ps = static pressure in H2O (50-80) or cmH2O(125-200)Kfd = 90dB for Ps in H2O, or 86dB for Ps in cmH2O

fan type

a) Axial, tube, or vane and cen-trifugal, radial

b) Centrifugal, airfoil blade, or forward or backward curved blade

c) Centrifugal, tubulard) propeller

Imperial(dB)

47

34

4252

Metric(dB)

72

59

6777

• Some general levels of sound power for industrial equipment are given below for frequency ranges from 500-4000Hz.

Equipment Equation/Source

Electric Motors Lw 20 hplog 15 RPMlog Km+ +=

hp = rated horsepower (1-300 hp)RPM = rated speedKm = typically 13dB

Pump Noise Lw 10 hplog Kp+=

hp = rated horsepowerKp = 95dB for centrifugal

100dB for screw105dB for reciprocating

for speeds <1600 rpmsubtract 5dB

Air CompressorfBRC

NrNs

KBRC-------------

RPM60

------------- =

fBRC = Blade rate component (Hz)Nr = number of blades in rotorNs = number of blades in statorKBRC = the bigger of Nr or NsLw 10 hplog Kc+=Kc = 86 dB (for 1-100 hp)

If the frequencyis in the hearingrange the soundwill increase by afew dB.

Sound generated by equipment found in typical buildings [Irwin and Graf, pg. 129] as measured between 500 and 4000 Hz.

Equipment

a) air compressorb) air-cooled condenserc) boilerd) cooling towere) pneumatic transport systemf) pumpg) rooftop air conditionerh) steam valvei) transformerj) unit heater

Low

85906595705580708055

Medium

10010080110908090858570

High

1151051001201101051001059090

Sound Power Level (dB) Lw

Typical Sound Power Levels for Construction Equipment [Irwin and Graf, pg.130], as measured between 500 to 4000 Hz.

Equipment

a) backhoeb) stationary compressorc) concrete mixerd) concrete pumpe) movable cranef) stationary craneg) front loaderh) stationary generatori) jack hammerj) paverk) pile driver - peak valuel) pneumatic wrenchm) stationary pumpn) rock drillo) scraper of graderp) tractorq) truck

Low

105110110110110115105105115115130115100115115110115

Medium

120115115115115120115110125120135120105125120120120

High

130120125120120125120120135125140125110135130130130

Sound Power Level (dB) Lw

• Sound levels are given below that are typically generated by home appliances (500 to 4000Hz)

1.7 ROOM ACCOUSTICS

• The basic idea is to determine how sound energy is distributed, absorbed and reflected in a closed space.

Typical sound power levels for household appliances [Irwin and Graf, pg. 132] as measured between 500 and 4000Hz

Equipment

a) air conditionerb) clothes dryerc) clothes washerd) dish washere) electric can openerf) electric shaverg) fanh) food blenderi) food mixerj) food waste disposalk) hair dryerl) home shop toolsm) refrigeratorn) vacuum cleanero) toilet

Low

555555606055457055756585407055

Medium

707070757570658580907097508075

High

807585858580801009010575110659585

Sound Pressure Level (dB)

• Sound is absorbed by walls, floors, etc.

there is an energy density throughoutthe volume of the room

sound source

δ P2

ρoc2

-----------=

where,for a simple volumewith no absorption δ energy density J

m3

------ =

P RMS sound pressure Pa( )=

ρoc characteristic impedance (raleigh number mks rayls)=

c transmission speed of sound ms----

=

***** See Irwin and Grafpg. 140 for derivations.

ρ0 static gas density kg3

------ =

• the ‘α’ absorption coefficients are determine experimentally, and can be found in the table below, [sources ??????]

δ t( ) 4WαcS---------- 1 e

αcS4V

---------- t–

–=

where, δ t( ) energy density as a function of time J

m3

------ =

W power of the sound source (W)=

V Volume of room=

t time (sec)=

S surface area of ro o mm2( )=

c speed of sound transmission ms----

=

α absorption coefficient of room (covered later)=

as t approaches infinity the steady state noise level becomes,

δSS4WαcS----------=

*** See Irwin and Graffor Derivations

Material Description

acoustical plaster, average

125

0.07

250

0.17

500

0.50

1000

0.60

2000

0.68

4000

0.66Acoustic steel deck, 6” ribsAcoustone space tile 32” OC per unitAir, per 1000ft3, rel. humidity 50%Audience in upholstered seatsAudience, empty upholstered seatsAudience, empty leather seatsAudience in wooden pewsAudience, musicians with seat and instrumentsBrick, exposed unpainted unglazedCarpet, heavy on concreteCarpet, heavy on underpadConcrete block, unpaintedConcrete block, paintedConcrete, unpaintedFabrics, medium velour 14oz./sq.yd. draped to half areaFloors, concrete or terrazzoFloors, vinyl, linoleum, rubber, cork tiles on concreteFloors, woodFloors, Raised wood

0.580.22

0.390.190.150.37

0.030.020.080.360.100.010.070.010.020.150.40

0.640.81

0.570.370.250.44

0.030.060.240.440.050.010.310.010.030.110.30

0.71

0.800.560.360.67

0.030.140.570.310.060.020.490.0150.030.100.20

0.63

0.940.670.400.70

0.040.370.690.290.070.020.750.020.050.070.17

0.47

0.920.610.370.80

0.050.600.710.390.090.020.700.020.100.060.15

0.40

0.870.590.350.72

0.070.650.730.250.080.030.600.020.050.070.10

f(Hz)

cork, 1” wall panels 0.25 0.55 0.70 0.75 0.75 0.75

• The absorption coefficient ‘α’ for a room can be calculated using the ‘α’ for each surface,

• ‘α’ can also be corrected for temperature and humidity using factors in the table below,

Material Description

Geoacoustic tile, 32” OC

125

0.13

250

0.74

500 1000 2000 4000

Glass, heavy plate 32” OCGlass, ordinary windowGypsum board 1/2” on 2”by4” stud 16”OCPlaster/Gypsum/Lime on brickPlaster/Gypsum/Lime on concrete blockPlaster/Gypsum/Lime on lathPlaster/Gypsum/Lime on lath with airspacePlywood, 1/4” with 3” airspace and 1” insulation

0.180.350.290.0130.120.140.300.60

0.060.250.100.0150.090.100.150.30

0.040.180.050.020.070.060.100.10

0.030.120.040.030.050.040.050.09

0.020.040.090.050.040.030.050.09

0.020.040.090.050.040.030.050.09

Brick, unglazed and painted 0.01 0.01 0.02 0.02 0.02 0.03Panels, 1.5” fiberglassPanels, perforated metal 4” thickPanels, perforated metal and fiberglass 2”Panels, perforated metal and mineral wool 4”Panels, 3/8” plywoodPolyurethane foam, 1”Tile, mineral fiber ceilingTile, marble or glazedWood, solid 2”

0.860.700.210.890.280.160.180.010.01

0.910.990.87

0.220.250.450.010.05

0.800.99

0.170.450.810.010.05

0.890.99

0.090.840.970.010.04

0.620.94

0.100.970.930.020.04

0.470.83

0.110.870.820.020.04

f(Hz)

αSiα i∑Si∑

-----------------=

Si α i, surface area and absorption coefficient for all room surfaces=

• previous assumptions were based on rooms with uniform power distribution, but this is not the case.

• Sound pressure is more useful than energy density, and happens to be quite similar in formula-tion.

RelativeHumidity

30%3030505050707070

temp°F

687786687786687786

temp.°C

202530202530202530

k(ft-1)

0.00090.00090.00090.00070.00070.00070.00070.00070.0006

k(m-1)

0.00300.00290.00280.00240.00240.00230.00210.00210.0021

k(ft-1)

0.0029

k(m-1)

0.0095

k(ft-1)

0.0103

k(m-1)

0.03400.00240.00220.00190.00180.00180.00160.00160.0016

0.00780.00700.00610.00590.00580.00530.00530.0052

0.0065

0.0046

0.0215

0.0150

2000Hz 4000Hz 8000Hz

television

loud

quiet

δ WQ

4πr2c

-------------- 4WcR--------+=

where,

W power of the source (W)=

r distance from source=

c sound transmission speed=

R αS1 α–------------

=

Q directivity factor=

• try problems S27, S28, S29, S30, S31, S32, S33

1.7.1 Sound Source Power Measurements Revisited Again

• Step 1: Measure sound on surrounding sphere,- r > 2 times largest dimension- over frequency ranges, e.g. octaves 31.5, 63, 125, 250, 500Hz

• Step 2: Calculate R (via α) or experimentally with sound decay equation from before• Step 3: Calculate sound power level for each measurement position, and each frequency with,

• Step 4: Calculate Lw for each frequency band.

• Estimation of R from sound decay is based on equations,

LP LW 10Q

4πr2

----------- 4R---+

(add 10dB if done using imperial units)log+=

LP sound pressure level relative to 20microPa=

where,

Lw sound power level at source referenced to 1012–

W=

LWi LPi 10Qi

4πri2

----------- 4R---+

(subtract 10dB when using imperial units)log–=

LW 10 1n--- 10

LWi

10--------

i 1=

n

∑

log=

1.7.2 How Sound Propagates

• When sound reflects off a flat surface it keeps the same spherical surface, but the center of prop-agation is mirrored across the reflecting surface. Like light the angle of incidence is equal to the angle of reflection.

• When the sound is incident on a narrow slit or edge, it regenerates, as if the slit is a new source.

R S

TS0.161V----------------- 1–

--------------------------------m

2= R S

TS0.049V----------------- 1–

--------------------------------ft

2=

S surface area=

V volume=

T time for sound to decay to -120dB of original value ( 106–

120dB )–= =

Metric Imperial

R αS1 α–------------

=

Hard surface

Sourceimage

• various other combinations of openings in wall, and half slits produce different sound propaga-tion patterns.

1.8 ENCLOSURES, BARRIERS AND WALLS

• Previous notes concentrated on sound from local sources

• But, another class of problems involve walls, and barriers between sound sources, and listeners.

• The effects of walls are measured as a Transmission Loss (TL)

• Barrier Transmission Loss is a function of frequency

Slit is much narrower than wavelength

TL 10WαW2--------

log 10 1τ---

log= =

Wα The incident power on a wall that is absorbed (W)=

W2 The power that is radiated out the other side of the wall (W)=

τ transmission coefficient=

Wall stiffness region - hard to predict exactly, the stiffer the wall, the greater the TL. Res-onances in this region are a function of structure.

Wall Mass Region - The mass of the wall controls sound transmission [source Irwin and Graf]

Coincidence Region - (not common) Sound waves hit wall at angle - effectively changing frequency. Because of the lowering of the high frequency to a lower value, the wall can now be excited.

• A table of TL values for walls are, [source ????????]

TL

ω

resonance

wall stiffnessdominates

wall massdominates

coincidenceregion

fc

20dB/decade=6dB/octave

TL 20 flog 20 Dlog C–+=

f frequency (Hz)=

D surface density lb

ft2in

----------- or kg

m2cm

-------------- =

C = 33 for imperial, or 47 for metric

A table of D values material

brickdense concretewoodleadglass

D(imperial)

111236515

gypsumcinder block

58

D(metric)

21236125291015

• Another list of transmission loss figures is given below for typical building materials, [source ???????]

Material 125Hz 250Hz 500Hz 1000Hz 2000Hz 4000Hz

Brick, 4”Block, 7 5/8” hollow cinderBlock, 6” concrete lightweight paintedCurtains, lead/vinyl 1.5 lb/ft2

Door 2 5/8” hardwoodFiber Tile, filled mineral 5/8”Glass plate, 1/4”Glass laminated, 1/2”Panels, perforated metal with mineral 4”Plywood, 1/4”, 0.7 lb/ft2

Plywood, 3/4”, 2 lb/ft2

Steel, 18 gauge, 2 lb/ft2

Steel, 16 gauge, 2.5 lb/ft2

Sheet metal laminate, 2 lb/ft2

3033382226302523281724152115

3633362333322931341522193025

3733402540393338402027313428

3739453143433640482428323732

3745503548532647562825354039

4351564251603552622727484742

f(Hz)

• A table of typical NR values for walls are, [source ??????????]

Wall/Door/Floor/Ceiling Type

2 by 4” studs

2 by 4” studs offset2 by 4” studs with isolation clipsGlass - single glazed

Glass - double glazed

Homogenous Material

Hollow Core

Door

Floor/Ceilings

Material

1/4” plywood1/2” fiber board1/2” plasterboard3/4” plaster and metal lathegypsum, lathe and plastergypsum, lathe and plaster1/8”1/4”1/2”1/4” glass, 1/4” space1/4” glass, 1/2” space1/4” glass, 1” space1/4” glass, 6” spaceInsulated wall panel2” plasterunfinished cinder block (unfinished)4’ brick8” brick12” brick16” brick4” cinder block (plaster finish)concrete block4” clay tile8“ clay tile6” porous rockwool4” cinder block, 2” space4” cinder block, 4” space8” cinder block, 4” space3/16” wood panelHollow core flush1 3/4” normally hungWeather Stripped 2 1/2” solidplywood rough finish floorplywood rough finish floor/plaster ceiling4” concrete slab4” concrete slab and suspended ceiling4” concrete slab and resilient suspend. ceil.

Avg. T.L.

25303540484826303333374555303720475053554046384015525558151520273037435258

• The Sound Transmission Class (STC) is the Transmission Loss (TL) at 500Hz

1.8.1 Noise Reduction Through Walls

• We can use the transmission loss of the wall to estimate sounds heard in adjoining rooms.

Wall type

2”by4” stud wall with 1/2” drywall 6lb/ft2

2”by4” staggered studs with 1/2” drywall 7lb/ft2

Metal studs with 1/2” drywall, 5lb/ft2

2”by4” stud wall with 1/2” plywood 1 side 2lb/ft2

125 250 500 1000 2000 4000

22362510

30373015

35403817

40474719

41524820

40454426

frequency (Hz)


NR LP1 LP2–= Lp values taken right beside wall

Lw1

Lp3

source

For room #1:

LP1 Lw1 10 4R1------

log+=

*note: only reverberation is consideredas a simplification

R1 = room constant, room 1

For sound through wall:

Sw = Area of the wallR2 = room constant of room 2

LP2 LP1 TL– 1014---

Sw

R2------+

log+=

*note: transmission loss figures can beobtained from the tables that follow

For sound away from wall in Room 2:

LP3 LP1 10Sw

R2------

log TL–+=

NR TL 1014---

Sw

R2------+

dBlog–=

1.8.2 Walls with composite layers

• Sometimes wall are built up of layers to improve sound absorption

• Most walls have areas covered by different materials. When this is the case we can combine the coefficients

• Holes in the wall will increase the sound that is transmitted, but these are often needed for wires, airways, etc. Offsetting holes can help maintain sound isolation properties.

Outside of wall quiet side of wall

air gaps help

τ1 S1, τ2 S2, τ3 S3,

τ1 S1,

τ2 S2,

τ3 S3,

τ i Si, transmission coefficient and surface area of each material=

ττ iSi∑Si∑

---------------=

• Another problem with sound isolation is that other paths may be ignored. For example a barrier may be built between two rooms, but raised floor is ignored, and sound continues to travel through this route. Other well known sound paths are ventilation ducts.

1.8.3 Enclosures

• Hoods are a common method of dampening noises

Cable is not lead directlythrough the wall.

• The NR and TL are for reduction of the new reverberant field Lp1, but we are concerned with the reduction of Lp0. This is the Insertion Loss (IL).

• The noise reduction for hoods is very good at higher frequencies Isolation pads can be used to reduce the noise at lower frequencies. When used together the total noise levels will be decreased.

1.8.4 Acoustic Barriers

• Sometimes barriers (or free standing walls) can be placed in the direct path of the sound.

source

Lp0

Lp2Lp1

source

hood

add a hood todecrease sound

Calculate as normal,

NR TL LP1 LP2–= =

* Adding a hood increases the reverberant sound, and thus Lp1 > Lp0,therefore Lp2 is raised slightly.

* the design must avoid standing waves.

IL LP0 LP2–= IL 10 ατ---

log=OR

• Try problem S38

1.8.5 Panel Absorbers

• In effect we can increase absorption in a room by adding suspended additional surface area.

• suspended panels can flex, best absorption characteristics at low frequencies. typical values for alpha are .2 to .5 for 125Hz.

• These can be tuned for specific frequencies

• Locating these absorbers next to a wall with absorbing material will increase high frequency absorption.

• Typical materials are glass, plywood, metal.

1.9 MATERIALS

• Noise Reduction Coefficient (NRC) for a material takes care of frequency dependance of absorption

source

- 2 main effects- reverberant- diffraction about edges

• For reduction of reverberant noise we can use a Noise Reduction (NR) coefficient

1.10 MUFFLERS

• An addition to a sound pathway can reduce the sound levels that pass. These are commonly used with internal combustion engines to reduce sound, along with other devices. An example of muffler systems is given below,

NRCα250 α500 α1000 α2000+ + +

4------------------------------------------------------------------=

α250 α500 α1000 α2000, , , sound absorption coefficient for a material=

T C VA---

=

V volume of room=

A units in Sabines=

C = 0.049 (imperial), or 0.161 (metric)

A α iSi∑=

NR 10A2

A1------

log=

*note: keep in mind that NR will vary with thefrequency components of sound.

NR

f

• After the fan there is typically a length of 5 to 10 diameters to allow the flow to settle. This makes the fan more efficient, and reduces noise from fan turbulence. This length between ele-ments will also help improve performance.

1.10.1 Lined Ducts and Baffles

• Absorbs a portion of the sound

• The more material, the better these devices absorb sound. Typical materials include mineral wool, and may use a stainless steel or mylar cover to protect the materials from high velocities and corrosive gases.

• These are good at high frequencies.

air inletfan pipe

lined pipe

90° bends

180° bends

plenum

expansion chamber

cavity resonator

Liner Thickness

1”2”

125 Hz

0.060.20

250 Hz

0.240.51

500 Hz

0.470.88

1 KHz

0.710.99

2 KHz

0.850.99

Absorption Coefficient

4 KHz

0.970.99

• Typical duct liner sound absorption characteristics are given below,

1.10.2 Bends

• As sound must turn corners, it attenuates.

Lined duct (end view) parallel baffles (cross section)

blocked line of sight baffles(cross section)

Duct Type

5-10” Round, 1” lining18-24” Round, 1” lining18-24” Round, 2” lining18-24” Round, No lining5-10” Rectangular, 1” lining18-24” Rectangular, 1” lining18-24” Rectangular, 2” lining18-24” Rectangular, No lining

125

0.300.220.290.060.300.250.330.12

250

1.200.781.050.031.200.500.870.06

500

2.001.302.000.022.001.201.800.05

1000

3.001.802.800.023.001.802.400.04

2000

3.001.802.700.023.001.802.300.04

4000

2.401.242.300.022.401.401.500.04

Attentuation at Frequency (Hz) - in dB/ft.

• For a 90° bend in a pipe we can expect attenuation as given below. These numbers assume that the bend section is round or rectangular with the diameter or widest dimension being that given. A lining is assumed with a thickness of 10% of D.

• For a 180° bend we can find the attenuation by looking at the absorption coefficient, diameter of duct, and length of bend section.

90°

10dB

1dB

attenuation

100Hz 4KHz

f

good soundadsorption here

D (in.)

5-1011-2021-40

63 Hz

001

125 Hz

012

250 Hz

123

500 Hz

234

1 KHz

345

2 KHz

466

4 KHz

688

Attenuation of Bend (dB)

1.10.3 Plenum

• An offset in a pipe of the right dimensions can reduce the sound well. The plenum itself is a large volume compared to the duct.

• The reduction is a function of both the absorption and reflection path.

• Typical attenuations are 10 to 20 dB.

Atte

nuat

ion

(dB

)

Absorption Coefficient

N=10N=5

N=2

N=L/D 0 0.5 1.00

10

20

30LD

• Try problem S39

1.10.4 Expansion Chamber

• When a particular frequency of sound is to be muffled, a tuned cavity can be used.

• These use a change in impedance at the inlet and outlets.

l w

hd

θ l

attenuation 10 Sθcos

2πd2

------------ 1 α–αSw------------+

dBlog–=

α for the plenum lining=S plenum inlet or outlet area=

Sw plenum wall area=

1.10.5 Cavity Resonator (Helmholtz)

• Another type of tuned resonator is a Helmholtz resonator.

• This is basically a cavity added onto the side of a pipe.

• The air inside the cavity acts as a spring. When the quantity of air is a specific volume, it will resonant with a specific frequency.

• These are typically best at low frequencies.

• The device can be used at higher frequencies, but it is typically lined with fiberglass or other porous materials.

A1 A1

L

A2

Otherconfigurations

mA2

A1------=

TL 10 1 m1m----–

214--- kL( )sin( )2

+log=

k wave number2πλ

------ 2πfc

--------= = =

1.11 SOUND AND VIBRATION CONTROL STUDIES

• Attached articles have been provided so that the student may see the concepts and approaches discussed so far used in actual control problems.

1.12 AKNOWLEDGEMENTS

• Thanks to John Lea for pointing out needed material the fist time I taught this course.• Thanks to Paul Short for giving me feedback on these notes, and pointing out a number of errors.

1.13 PRACTICE QUESTIONS

S1. What is the frequency of a wave with a wavelength in air of 0.026m at 10°C (in air)? (ans. 12973 Hz)

S2. What is the sound power radiated by a machine with specified sound power level (Lw) of 130 dB? (ans. 10 W)

S3. Three machines of equal sound power are to be placed next to two existing machines each

cylindrical

d

V

2r

Lsections

frc

2π------ A

lV-----=

fr the frequency to be canceled by the muffler=

A the cross sectional area of the neck=

V volume of the cavity=

l L 0.8 A+=

having sound power levels of 82dB. It has been determined that the combined sound power of all 5 machines must not exceed 63mW. What maximum sound power level must each of the new machines be limited to? (ans. 103.2dB)

S4. A cylindrical tube is to be used for sound propagation tests with a loudspeaker located at one end pointing down the tube. The range of test frequencies to be examined is from 300Hz to 3000Hz under temperature extremes of 5 degrees Celsius to 25 degrees Celsius. Determine the minimum tube length to ensure one full wavelength at the lowest frequency and the maximum diameter to limit the highest frequency transverse (across the tube) wave to one half wave-length. The above to hold for all temperature extremes. (ans. L=1153mm, D=55.7mm)

S5. Given the loud speaker test results given below, (ans. Q=1.35)

S6. The following data was recorded for a non-isotropic source. What would be the directivity factor (Q) in direction θ = 60° (use appropriate approximation for calculation of the average value). (ans. 0.7)

Lp(dB): 77, 71, 69, 70, 69, 73θ(deg): 0, 60, 120, 180, 240, 300

S7. You are given two loudspeakers to be mounted in the passenger section of a commercial air-craft. For testing, a speaker is mounted in a test setup, and powered with a 40 watt stereo. Mea-surements are taken at the points listed below. From the data, a) calculate dB pressure level for each measurement, b) calculate directivity for each point, c) calculate the total power, and d) select the speaker best suited to the application (or neither) and explain your choice.

what would the directivity factor at θ = 30 degrees be?90

30

0

150

210

270

330

θ03090150210270330

Lpi

120116110105105110116

* All measured at the same radius

S8. What will be a) the total sound power level of 5 machines rated as follows 93dB, 91dB, 89dB, 92dB, 88dB. b) If the least noisy machine is turned off, by how much does the dB power level drop? (ans. a) approx. 98dB, b) -0.5dB)

S9. Three machines are individually measured against a background level of 78 dB. Sound pres-sure levels are 83dB, 81dB, 80dB. What would be the total sound pressure level with all three machines on at a new location where the background level was 80dB? (ans. 85.3dB)

S10. Calculate the average sound pressure levels for the following recorded values: 89dB, 93dB, 97dB, 85dB, 92dB. (ans. 92.9dB)

S11. Two identical sources have a combined amplitude (Lp) of 104dB. What would be the value of the root mean square pressure for just one of the sources? (ans. 2.24Pa)

S12. The sound pressure level measured at a distance of 3m from a machine including back-ground noise was recorded as 96dB. The ambient level was then measured with the machine off and found to be 93.4dB at the same location. If the ambient level was to be raised by 2dB what would the combined machine and background level at the same measurement location? (ans. 97.2dB)

S13. Given a set of three sound pressure measurements (.05Pa, .03Pa, .04Pa) taken at 1/2 Octave intervals, starting at 200 Hz, find the total loudness in sones.

S14. A machine has a loudness rating of 35 sones, what will be the loudness level (phons) - give approximate (graphical) and exact (calculated) answers. (ans. 91.4 phons)

speaker mounting point

passenger

150 feet

10 feet

cabiny

0.9358

-.58-.93.36

-.36.58

0-.58

z

1.33.33.33.33.33.33.75.75.75

x

0.13.75.75.13

-.87-.87-.33.67

-.33

speaker 1

1.3.15.08.18.20.12.07.11.10.05

speaker 2

.75

.35

.47

.25

.36

.24

.31

.89

.42

.72

Measurement Point (m) Sound Pressure (Pa)

S15. Based on a limited one octave (1/1) analysis calculate the loudness in sones for the following noise and loudness level. (ans. 43.9 sones)

S16. You have been approached by a business that has just signed a 5 year lease for office space. But, they want to cancel the lease because they claim the background noise is too great for an office. You enter the building and take some readings at every octave from 100Hz. The values you get are 35dB, 42dB, 48dB, 42dB, 35dB, 31dB. Is the sound level suitable for office space?

S17. For the same spectrum as in #S15, calculate the NC rating (use the graph). (ans. NC > 65)

S18. Using the graph to calculate the total level in dBA for the same spectrum. Use fig 2.8 (A-weighted curve) to find dBA levels then add. (ans. approx. 85 dBA)

S19. You have decided to quit engineering, and go into the restaurant business. And, your new concept is to design an environment where the patrons don’t have to shout. You are using 5 foot round tables throughout the room, with a 1000W sound system that gives even music throughout your establishment. a) Calculate the general range of pressure levels from 500Hz to 2000Hz that would interfere with shouting. b) Calculate the general range of pressure levels from 500Hz to 1000Hz that would interfere with normal speech.

S20. The sound pressure readings below were taken at 1/3 octave intervals beginning at 500 Hz; 54, 72, 89, 87, 92, 94, 95, 104, 88, and 74 dB. These readings were taken in the same room that houses 3 workers. We must determine if the room sound levels are safe.

a) Calculate the loudness index.b) Calculate the PNC.c) Calculate the total pressure Lt.d) Calculate the PSIL. (ans. 78.7 dBA)e) What is the farthest distance apart and two workers may be to hear each other shouting

using PSIL and background dBA? (ans. 2ft for both)f) What is the maximum safe exposure time allowed in this environment each day? (ans.

30 min.)

Band f0 (Hz)

2505001000200040008000

Lp(dB)

586675808075

Use the graph and St Im 1 K–( ) K Ii

i 1=

n

∑+=

S21. The Shoutalot Sound Measurement Company was hired to determine if an assembly shop was safe for the workers at the Rustyheap Car Company. To do this they took sound measure-ments and found that the workers were right at the safety limit (i.e. D = 1). But before the report was completed, Mr. I.M. Kluzt, P.E., lost the sheets of paper that the measurement data was on. But he remembered that 3 hours of the 7 hour shift are at 95 dBA. You must estimate what the sound level was for the other 4 hours of the shift (the additional 4 hours were all at the same level). (ans. 85 dBa max.)

S22. At the perimeter of a local fast food restaurant you find the noise levels are 60dBA from 2am to 8am, from 8am to 8pm the noise level goes up to 75dBA, and from 8pm to 2am the noise level goes down to 70dBA. Assuming this has a 10 dBA night-time noise penalty from 10pm to 7am, calculate the noise pollution level.

S23. a) A worker has been exposed to noise levels over a period of 5 hours as listed below. For what remaining portion of his 8 hour shift could he be exposed to a level of 90 dBA. (ans. 1hr 20min)

b) Based on a daytime Leq of 60dBA and a nighttime level of 45dBA, what would be the Ldn index for the entire day? (ans. 58.7dBA)

S24. The sound spectrum below is known to exist for an industrial machine. Keeping in mind the proximity of some of the peaks, what bandwidth would our sound instrumentation require to tell them apart? (ans. 1 octave?)

Duration(hours)

221

Exposure Level(dBA)

929095

S25. For a 1/12 octave band pass frequency analyzer what would be the percentage (B.W./f0) and the upper and lower cutoff frequencies at f0=12000Hz. (ans. 5.78%)

S26. A 1/10 octave bandpass filter has a centre frequency of 5000Hz. Calculate the half power bandwidth cutoff frequencies (f1, f2) and the local attenuation of a pure tone signal at 6000Hz by this filter.(ans.

S27. The Bouncy Rubber company uses a 100 horsepower air compressor and an exhaust fan when making vulcanized rubber balls. The axial fan creates a flow of 5 m3/sec at a pressure of 200 mmHg. The (10’ by 10’ by 10’) room they are housed in has walls of concrete blocks painted red.

a) Calculate the acoustic power generated by both machines.b) Calculate the absorption coefficient of the room.c) Find the sound energy density of the room and compare the direct and reverberant com-

ponents.d) Calculate the longest duration a worker may stay in the room.e) Suggest a new room coating to reduce the sound levels to allow 8 hour work days.

S28. A generator room contains an average 500 h.p. diesel motor (the sound is about 1000Hz). The ceiling and walls of the (3m by 3m by 3m) room are covered in plaster and the floor is con-crete. On one wall there is a 2m by 2m window. If the generator is in one corner of the room, and you are in the other corner, what sound pressure will you hear (in Pascals)? (ans. 11.3 Pa)

S29. Acoustic Ohm’s law for freely propagating states that,

a) Derive the equation which specifies

Peaks of interest

W

4πr2

-----------PRMS

ρc-------------=

b) evaluate the unknown (?) expression given, (ans. -11dB)

S30. A machine is located at position A and radiates 0.6W of sound power in the octave band cen-tered at 1000Hz with a directivity of 1.6 in the direction of an observer at location B. The machine is operated for each 8 hour shift. There is a second machine located at position C which is used infrequently. It radiates a sound power of 0.2W with a directivity of 1.0 in the same frequency range.

a) calculate the total direct sound pressure level at position B with both machines on (i.e., ignoring the reverberant field effects). (ans. 90.0 dB)

b) calculate the average absorption coefficient and the room constant in this frequency range (use the table in the notes for component values). (ans. 0.410, 263 sq.m.)

c) calculate the sound pressure level at position B with only machine A on taking room effects into consideration. (ans. 100 dB)

Lp LW 20 rlog– ?+=

PREF 206–×10 Pa=

WREF 1012–

W=

ρc 406mksRayls=

S31. The company would like to reduce the sound pressure level at the operator location and is considering laying foam backed carpet (α = 0.69) on 75% of the floor surface. What would be the new sound pressure level at position B, machine A on under these conditions. (ans. 101 dB)

S32. Predict the expected sound pressure level at a location 3m from a source radiating 100dB of sound power, assume isotropic radiation (spherical) and in a sound absorbing room. If the actual reading at this location was 83dB, what would be the Directivity Factor (Q) be? (ans. Lp=79.47dB, Q=2.256)

Note: the equation you use in the first part of this question is derived from the relationship (i) the directivity of the source modifies the numerator of this expression (ii).

S33. Under free field conditions the sound pressure level of a noise source was measured and found to be 88dB at a distance of 4m. Assuming isotropic radiation what would be the sound

5 doors1m by 2m

question#S34 only

7

2

3

6

13

15

6

4

4

8Source A

Source C

Listener B

2 windows1.5m by 2.5m

Notes:1. All dimensions in meters2. ceiling height 3m3. Floors are concrete, wall are coarse concrete block, ceiling is mineral fibre

ceiling tile, Windows are standard glass, doors are 2” solid wood.4. Absorption coefficients can be found in a table in the notes

Prms2

ρc----------- W

4πr2

-----------=Prms

2

ρc----------- WQ

4πr2

-----------=i) ii)

power level (Lw) of the source? (ans. 111dB)

S34. a) If an enclosure was made from 3/4” thick plywood to fit over the noise source in S30, what internal treatment would be required to ensure an insertion loss of 21dB. (ans. α>= .251)

b) For the room described in S30 assume acoustic treatment is applied to raise the average absorption coefficient to 0.40. What time would be required for the sound energy density to reach 60% of it’s steady state value after the machine was switched on? (ans. .02sec)

c) A noise source is located in anechoic free space and emits 1.5W of sound power. A sound level meter is located 3m from the source and the directivity factor in this direction is 4. Calculate the expected sound pressure level and derive the average sound pressure level at a radius of 3m from the machine. (ans. 101.18dB)

S35. The Boris and Natasha Detective Agency has hired you to build soundproof walls between their interrogation rooms. The dimensions and location of the two rooms are shown below. The loudest voice in either room will be 80dB, and a listener can hear voices at pressure levels above 20dB. What materials should be used for the wall?

S36. In another study the company decides to investigate the benefit of making a separate room to contain machine A and program machine B to run unattended after hours. A door between the two rooms would not be necessary, however a 1m by 2m window is considered essential (this would be heavy plate glass to reduce transmission losses α=0.03 and TL=40dB at 100Hz). The broken line shows the intended separation line which would be a 2” by 4” stud wall with 0.5” (6 lbs/ft) gypsum panels on both sides (α = 0.05 and TL = 40dB at 1000Hz). What would be the new sound pressure level at position B with the machine on and the wall in place. Assume reverberant field conditions (Lp3).

S37. The room pictured below is made from standard construction materials as noted. The machine radiates 2W of acoustic power isotropically in the octave band centered at 500Hz.

a) calculate the steady state space average sound energy density for the room. (ans.

3.213*10-4 J/m3)b) what will be the sound pressure level at position #1? (ans. Lp = 108.6dB)c) without doing the actual calculations, explain how you would go about determining the

Room 1 Room 2

1m 1m

source listener

Each room is 4m by 4m by 4mand has carpet on the floorswalls and ceilings.

sound pressure level at point #2, just outside the exterior west wall. (ans.)

S38. (See figure below) The elevation and plan views of a room are shown. A barrier is to be installed. The location and height have been determined. The required insertion loss consider-ing diffraction paths only is to be 5dB. The absorption coefficients in the octave band centered at 500Hz are .70, .30, .05 for the ceiling, walls and floor respectively. The printer generates a sound power level of 110 dB and in the direction of the worker location has a directivity factor of 1.9. Calculate the barrier length necessary to ensure the required insertion loss. (ans. L=3.57m)

3 windows 1.5m2 exterior wall

interior wall

source2W, 500 Hz

6m

10m

2 doors 2m2

#2 #1

1m 1m 7m

N

Notes:1. Room height 2.5m2. ceiling mineral fibre tile, walls concrete block unpainted, floor concrete,

windows standard glass, doors solid wood 2” thick.

S39. You think that a bend in a narrow hallway of your house reduces noise levels. This “plenum” has the dimensions below. Determine the Transmission Loss for 500 Hz.

S40. For the same noise in #S15 calculate the total perceived noise (PN) and perceived noise level (PNL) (ans. PN=54.1 noys, PNL = 97.7 PNdB)

15m

10m

L/2

L/2

1m1m

2.5m1m

T

F

worker

printer

barrier10m

5ft

2ft

4ft

2ft

• All the walls are gypsum• The floor is carpet with underpad• The ceiling is rough plaster• The ceiling height is 7ft.

PN Nm 1 K–( ) K Ni

i 1=

n

∑+=

PNL 33.3 PN( ) 40+( )log=

Use the given graph and

S41. The S.Butts Manufacturing company produces large castings. Seymour, the division direc-tor has asked that the Engineering department take sound readings about a large machine. The engineers make the following measurements with the machine turned off, then on.

a) Correct for the machine on/off.b) Find the Lp values.c) Find the average Lp value.d) Find the directivity values.e) What level of our perception would we say this is equivalent to? (e.g. whisper)

S42. A large incinerator is located in the centre of a residential neighborhood. Inside the facility the measurements are made of general background noise, these are listed below. Measure-ments are also made outside over a 24 hour period at the perimeter of the facility.

a) Find the inside values in phons and sones, and find the total loudness.b) Find the A weighted values and calculate the total pressure.c) How far apart would two employees have to stand to hear each other shouting?d) How many hours a day could a worker safely be in the work area?e) If the local municipality has an 11pm to 6am noise penalty of 15dB, what is the noise

pollution level, and the Ldn?

S43. A room, 8’ high by 10’ by 20’ has painted concrete floors, painted concrete walls, and an unpainted concrete ceiling. If a 100hp electric motor (1200rpm) is in one end of the room,

a) What is the general sound level in the room?b) Design a wall to divide the room so that on the side without the motor the sound is

20dB lower.

x(m)

0.08.20.0-8.2

y(m)

8.20.0-8.20.0

z(m)

5.85.85.85.8

Machine OFF P (Pa)

6*10-3

7*10-3

9*10-3

8*10-3

Machine ON P (Pa)

46*10-3

23*10-3

33*10-3

37*10-3

freq. (Hz)

5006308001000125016002000

Lp(dB)

98958793939772

time

7am-5pm5pm-10pm10pm-7am

dBA

857772

INSIDE:OUTSIDE:

c) Instead of a wall as in b), design a hood to sit over the motor.

S44. We take measurements about a sound source and get the following table (except the missing value). Using the following relationship, and the table of values, find the missing value.

S45. A sound is measured in octave intervals starting at 115 Hz. The pressure level values mea-sured (in order of frequency) are given below. Find the total loudness in noys. (for octaves K= 0.3)79dB, 87dB, 92dB, 105dB, 83dB

S46. A recent observation has revealed that an older type of beverage, commonly known as a ‘shooter’, has become popular in an previously unexplored market. As a result you decide to cash in on the trendy new happening and cater to this niche market. To keep the patrons drink-ing, you plan to play the music loud, but if it is too loud the waiting staff will not be able to hear the somewhat intricate orders. You have decided that in the worst case the patron will have to shout over a maximum distance of 3 feet. Tests on the music system reveal that the sound has a fairly constant level, except near 1000 Hz, where it always seems to be 20dB above all of the other frequencies. If we set the volume level of the music system using a sound pressure meter that is set to 500Hz, what is the ideal reading on the sound meter?

S47. A design is specified so that it should have a maximum preferred noise criteria value of 35. When tested, the sound values measured at half octaves from 63Hz are,

41dB, 43dB, 46dB, 51dB, 48dB, 48dB, 35dB, 33dBAt what frequencies are the sounds measured above tolerance, and which sound frequency is the worst violator. Use the graph and clearly indicate how you arrived at your solution.

S48. The S.Butts Manufacturing company produces large castings. Seymour, the division director has asked that the Engineering department take sound readings about a large machine. The engineers make the following measurements with the machine turned off, then on.

angle(degrees)

-180-120-60

060

120180

Lp(dB)

63

Directivity Index

-6.9972??71677465

1.99-2.011.00

-3.014.00

-4.99

DI LpiLp–=

a) Correct for the machine on/off.b) Find the Lp values.c) Find the average Lp value.d) Find the directivity values.e) What level of our perception would we say this is equivalent to? (e.g. whisper)

S49. A large incinerator is located in the centre of a residential neighborhood. Inside the facility the measurements are made of general background noise, these are listed below. Measure-ments are also made outside over a 24 hour period at the perimeter of the facility.

a) Find the inside values in phons and sones, and find the total loudness.b) Find the A weighted values and calculate the total pressure.c) How far apart would two employees have to stand to hear each other shouting?d) How many hours a day could a worker safely be in the work area?e) If the local municipality has an 11pm to 6am noise penalty of 15dB, what is the noise

pollution level, and the Ldn?

S50. A room, 8’ high by 10’ by 20’ has an unpainted concrete floor, painted concrete block walls, and an unpainted concrete ceiling. If a 100hp electric motor (1200rpm) is in one end of the room and whines at 990Hz,

a) What is the general sound level in the room?b) Design a wall to divide the room so that on the side without the motor the sound is

20dB lower.

x(m)

0.08.20.0-8.2

y(m)

8.20.0-8.20.0

z(m)

5.85.85.85.8

Machine OFF P (Pa)

6*10-3

7*10-3

9*10-3

8*10-3

Machine ON P (Pa)

46*10-3

23*10-3

33*10-3

37*10-3

freq. (Hz)

5006308001000125016002000

Lp(dB)

98958793939772

time

7am-5pm5pm-10pm10pm-7am

dBA

857772

INSIDE:OUTSIDE:

2. INTRODUCTION TO KINEMATICS OF MECHA-NISMS

• Consider a pair of adjustable vice grips.

Draw a picture of the vice grips

Simplify the vice grips to a stick diagram

How does the applied force vary over the range of application? (statics)

How do the jaws move relative to the motion in the handles? (kinematics)

When do these toggle/lock? (kinematics)

•Some definitions,Machine - a collection of components that will do work.Mechanism - a collection of components to transform motionKinematics - consider positions/velocities/accelerations in mechanical systemsStructure - a collection of components to make larger static structuresStatics - estimate forces in mechanisms that are in equilibriumDynamics - determine motion that results when forces are out of balanceLink - rigid body between jointsBinary Link - has two joints onlyTernary Link - has three jointsQuaternary Link - has four jointsPair or Joint - a connection between two linksDriver / Follower - the driver link will be driving the followerKinematic Chain - a sequence of links making up a mechanismOpen Loop - a snake like set of connected linksClosed Loop - a kinematic chain has one or more links that go back in the chainFrame - a grounded or fixed link in a mechanismSpatial - in 3 dimensionsRelative/Absolute - a position, velocity, etc. is measured based on a fixed (absolute) or

moving (relative) point.

• A Degree Of Freedom (DOF) is an independently controllable variable. As an example, a machine that has two degrees of freedom might need two motors to control it.

• Lower Pairs, - constrained position/orientation of both sides of the joints are identicalTurning / Revolute - basically a pin joint (R)Prismatic - a slider (P)Screw/Helix - a nut and screw pair (H)Cylindric - a shaft in a collar (C)Globular/Spherical - a ball joint (S)

• Higher pairs include, - typically other equations are needed to constrain the joints, such as gear ratios (if the joint has more than a single degree of freedom)

- flat/planar - constrained to move over a plane- belt on pulley- meshing gears- sliding wheel on a surface- etc.

• The definition of higher/lower pairs given in Shigley [1995] is, “the lower pairs, such as the pin joint, have surface contact between the pair elements, while higher pairs, such at the connec-tion between a cam and its follower, have line or point contact between the surface elements.” They go on to point out that the definition is not exact, which is somewhat disappointing.

• A better definition of a higher pair is - A higher pair is not a lower pair, where a lower pair per-mits the following relative motions between links; circular, linear, helical, cylindrical, spheri-

cal, planar.

• If a link has one joint, it is a unary link. A link with two joints is binary, with three it is ternary, with four it is quaternary, etc.

• Planar linkages use lower order pairs, and are constrained to a single plane of motion.

2.1 SOME POPULAR MECHANISMS

• Some basic mechanism types are listed below, and split into some suggested categories

2.1.1 Locking/Engaging

Snap-action mechanisms - typically bistable mechanisms, such as electrical breaker switches, or toggle mechanisms such as XXXX

Clamping Mechanisms - vices, collets, etc.Locational Devices - self alining/centering devicesRatchets and Escapements - A locking mechanism, like a ratchet wrench or winch

Bistable

Toggle

Indexing Mechanisms - e.g. the geneva mechanism

Reversing Mechanisms - A mechanism that can disengage a transmission, and reverse direction of transmission

Ratchet

Escapement

2.1.2 Motion Transmission/Transformation

Linear Actuators - produce a straight line motion. can be done with threads, or hydraulic cylinders

Fine Adjustments - screws, wedges, etc. - these can overcome imperfections during manu-facture

Couplings and Connectors - transmit rotations between rotating shafts. e.g. pulleys and belts

Sliding Connectors - transmit linear motions in different directions

2.1.3 Four Bar Linkages

Swinging or Rocking Mechanism - produce cyclic motions

Stop/Pause/Hesitation - a motion is produced that appears to come to a stop for a short period of time.

Curve Generators - mechanisms set up to follow complex paths - typically four bar link-

Crank

Rockercoupler

frame

In this mechanism the crank is turned, and the follower oscillates. But, the motions forward and then backwards are not at the same rate.

ages.

Straight Line - mechanisms are set up to generate straight line motions

2.1.4 Reciprocating

Reciprocating Mechanisms - converts a rotational motion to a linear motion

Roberts Mechanism - the geometry is such that the mechanism is made of three isosceles triangles (note the dashed lines). When actuated the bottom point ‘A’ follows an approximate straight line.

A

Offset slider crank mechanism - will generate a fast stroke of the slider in one direction. For example, if the crank is turned clockwise then the slider will move fast going right, and slower returning left. The faster stroke will provide less force, if the crank torque is constant.

advance (clockwise rotation only)

return

Qtime of advance stroketime of return stroke

------------------------------------------------------ angle of advance strokeangle of return stroke

--------------------------------------------------------= =

where,

Q = the advance to return time ratio

2.1.5 Six Bar Linkages

• These allow more complex motion, especially when ternary links are used.

• In Watt linkages there are two ternary links touching,

• In Stephenson linkages the terary links don’t touch,

Scotch Yoke - the crank rotates, and causes a lin-ear motion.

Watt I Watt II

2.2 SKELETON DIAGRAMS FOR MECHANISMS

• Many of the mechanisms drawn so far have been in a ‘stick’ or ‘skeleton’ form.

• We will use this representation to do abstract work.

• Convert the mechanism below to a skeletal form,

Stephenson III

Stephenson IIStephenson I

2.3 DOF AND THE KUTZBACH/GRUEBLER CRITERION

• Planar mechanisms can be made up of a number of joints and simple links.

• We can calculate a number that represents the mobility of a mechanism. If the value of ‘m’ is zero, then the mechanism will be rigid. If ‘m’ is less than zero, then the mechanism is rigid and overconstrained. When larger than zero, there are ‘m’ d.o.f. (degrees of freedom)

• Some examples of joints are- lower pairs (1 dof)

pin jointsliderroller with no sliding

- higher pair (2 dof)roller with sliding

- even higher pairs (3 dof)un fixed

• Consider the following examples with simple links,

where,n = number of linksj1 = number of lower pairs (1 d.o.f.)j2 = number of higher pairs (2 d.o.f.)m = mobility criterion

m 3 n 1–( ) 2j1– j2–=

n = 3j1 = 3j2 = 0m = 0

Rigid Triangle

n = 4j1 = 4j2 = 0m = 1

Four bar linkage

• Next consider a more complex system. In this case the slider acts as another link.

• In trusses we have a number of links meet at a joint. In this case we assume that one of the links provides the pin to the joint - each other link adds a new joint acting on that pin.

n = 5j1 = 5j2 = 0m = 2

Five bar linkage

n = 4j1 = 4j2 = 0m = 1

Piston/Crank type arrangement

n = 17j1 = 24j2 = 0m = 0

Rigid Truss

• If we have higher pairs, then we must include these,

2.4 KINEMATIC/GEOMETRIC INVERSION

• Although relative motions between components remain the same, we can pick different links to be the stationary frame.

• As different links are chosen to be grounded the other links will move, and this can lead to radi-cally different effects.


n = 14j1 = 20j2 = 0m = -1

Overconstrained Truss

n = 4j1 = 3j2 = 1m = 2

Crank and Roller

• For a geometric inversion we reconnect links so that they are different from their original con-figuration.

2.5 GRASHOF’S LAW

• If we want a mechanisms that will undergo continuous motion we must satisfy the Grashof crite-rion. In particular the shortest link, with length ‘s’ will be in constant relative motion if the Grashof Criterion is satisfied. (In other words we can rotate the smallest link continuously)

•. Consider the equation, and the four basic kinematic inversions below. Keep in mind that the crank will be the shortest link, with length ‘s, and in all four cases will rotate continuously.

s l p q+≤+

where,

s,l = the lengths of the shortest and longest members respectivelyp,q = the lengths of the two intermediate length members

Drag link mechanisms

Crank-Rocker mechanisms

• Consider the example problem below,

Double-Rocker mechanisms

AB

CD

The lengths of three links are, BD = 5cm, DC = 6cm, CA = 7cm. What should the length of AB if it is to have continuous motion rel-ative to the other links? What length should AB have for it to not move continuously? What is the longest length of AB if BD is the crank?

2.6 MECHANICAL ADVANTAGE

• As a mechanism moves over a range of motion its geometry changes. If we are using a mecha-nisms to transmit torque, or force then we must consider the ratio between the input and output force in various positions.

• Transmission angle is the angle between the coupling member and the output member in a mechanism. As this angle approaches ±90°, the mechanical advantage of the mechanism typi-cally increases.

• Toggle positions occur when the input crank has near infinite mechanical advantage. Note: this also applies that the follower has no mechanical advantage on the crank.

• Consider the example below, [prob. 1-3 from Shigley & Uicker],

s = 25mml = 100mm

p = 75mm

q = 90mm Find the maximum and minimum transmission angles. Find the two toggle angles of the crank AB.

100mm

75mm90+25mm

First, find the minimum toggle angle.

θ1

752

1152

1002

2 115( ) 100( ) θ1cos–+= =115mmθ1∴ 40.1°=

100mm

75mm90-25mm

Next, find the maximum toggle angle.

θ2’

752

652

1002

2 65( ) 100( ) θ2cos–+= =65mmθ2 ′∴ 48.6°°=

θ2 48.6 180+ 228.6°= =

A

B

C

D

Note: these angles are a measure of when the crank torque will create a maximum force.


1. Draw the kinematic inversions of the linkage below. Does it satisfy the Grashof Criterion? What is the mobility of the mechanism? Determine the maximum transmission angles, and the toggle angles if CD is the crank. What is the advance-to-return time ratio?

100-25mm=75mm

75mm90mm

Now, find the minimum transmission angle.γmin

752

752

902

2 75( ) 90( ) γmincos–+=

γmin∴ 53.1°=

100+25mm=125mm

75mm90mm

Finally, find the maximum transmission angle.γmax

1252

752

902

2 75( ) 90( ) γmaxcos–+=

γmax∴ 98.1°=

Note: This occurs as we get the largest angle between the links, or when the bottom two toggle.

Note: These angles show the relationship between tension/compression in the driver and the follower.

2. Draw the kinematic inversions of the linkage below. Does it satisfy the Grashof Criterion? What is the mobility of the mechanism?

3. The human arm is an open loop kinematic chain. Assuming the shoulder is the ground, what is the mobility of the human arm, including all joints down to the finger tips?

2.8 REFERENCES



10mm

7mm

7mm

5mm

A

B

C

D

10mm

7mm

7mm

5mm

3. POSITIONS/DISPLACEMENTS OF POINTS AND MECHANISMS

• The calculation of the position of mechanisms is an essential first step to modeling how they work.

• To do the calculations a number of basic mathematical tools are used.

3.1 MATHEMATICAL TOOLS

• Recall the basics,- coordinate systems

CartesianLeft/Right Hand RulesPolar/ComplexSphericalCylindrical

- vectorsmagnitudeunit vectorsdot products and projected valuescross productstriple product

- we can define locations of points using,direction cosinespolar locationcartesian valuesprojected coordinatescylindrical coords.spherical coords

- parametric values- Newton-Raphson- Complex math, including exponential form- basic graphing- rotation and translation- calculus

derivatives- polynomials- triginometry- exponentials

3.2 DEFINING POSITIONS AND DISPLACEMENTS

• A point can be defined using a variety of methods. For our purposes, cartesian coordinates will be predominantly used.

• We will largely rely upon vectors to define positions.

• We must pay attention to the fact that the vector has a direction and magnitude. The direction is relative to the orientation of the coordinates (which YOU will choose arbitrarily). In more advanced problems we will use multiple coordinate axes.

• We can add vectors easily if they are relative to the same reference coordinates,

x 1i=

y 1j=

RPO iRPOxjRPOy

kRPOz+ + P O– OP= = =

NOTE: I am drawing this figure in 2D to try and simplify the comprehen-sion. Extension to 3D is trivial, and the math terms will be set up this way.

λPO

RPO

RPO-------------

iRPOxjRPOy

kRPOz+ +

RPOx( )2

RPOy( )2

RPOz( )2

+ +-------------------------------------------------------------------------= =

∴ i Px Ox–( ) j Py Oy–( ) k Pz Oz–( )+ +=

O iOx jOy kOz+ +=

P iPx jPy kPz+ +=

Unit Vector

Note: ‘Z’ is used fordisplacement vectors inthe text with magnitude‘R’

• So far we have reviewed vectors that are all relative to the same coordinate axes.

• Now, consider what would happened if we had multiple coordinate axes.

x

y

RBA

A

B

C

RCB

RCA

RBA RCA RCB–=

RCB RCA RBA–=

RCA RBA RCB+=

• Consider the drill problem below,

O1

O2

P

x

y

x’y’

RO2O1RO2 1⁄=

RPO2RP 2⁄=RPO1

RP 1⁄=

Notice the way that relative to other coordinate axis is indicated with ‘/’, and then the reference coordinate number. For example Rp/1 is the position of ‘P’ as seen when standing at O1. Rp/2 is the position of ‘P’ when standing at O2. Note: Erdman and Sandor omit the ‘/’ but still use the order of the subscripts.

RPO1RO2O1

RPO2+=

RP 1⁄ RO2 1⁄ RP 2⁄+=

Note this notation shows the apparent posi-tion of point ‘P’ to an observer at point O2. If the axes of O2 do not line up with those of O1, we must compensate when doing the math.

RP RO2RP 2⁄+=

Notation based on absolute positions. In this case we have arbitrarily chosen one of the frames to be the global reference (typically ‘1’).

Here traditional difference vectors are used. We can expect to use the same direction axes.

x

y

z

x’y’

z’

O1

O2

RO2 1⁄ 5i 4j 2k–+=

P

If the apparent position of point P is (-1,-2,4), what is the absolute position of P? What is the absolute posi-tion difference between O2 and P?

3.3 CLOSED LOOP MECHANISMS

• If we have a kinematic chain that forms a closed loop, the sum of the vectors for the links must total zero.

3.4 SOLVING FOR POSITIONS

• We can solve kinematics problems using a variety of techniques. When the appropriate tech-nique is selected at the right time, it will simplify calculations.

• Some of the basic techniques are,- Trigonometry- Complex numbers- Numerical- Simulation

• There are also some specialized techniques for analysis of mechanisms,Chase Solutions - a closed form approach for planar mechanisms that can be used for

numerical calculations.

A

B

C

D

1

2

3

4

RBA RCB RDC RAD+ + + 0= Loop Closure Equation

3.4.1 Trigonometry

• Basically we generate a set of equations that define the shape of the mechanism. To do this we will need to call upon the methods of trigonometry. (Note: recall the sine and cosine laws, as well as the basic trig identities)

• Consider the example from the textbook. The offset crank-piston can be solved easily with care-ful observation,

• Next, consider the (crank-rocker) four bar linkage below,

Given the angle of rotation at A, find the distance Ls. Given Ls, find the joint angle at A. Assume or add variables as needed.θ

Ls

A

B

C

d

ef

We can calculate,

RBAxe θcos=

Ls RBAxRCBx

+=

RBAye θsin=

RCBy d RBAy+ d e θsin+= =

Therefore we need the x components,

f2

RCBy2

RCBx

2+ d e θsin+( )2

RCBx

2+= =

Finally combine and rearrange,

Ls e θcos f2

d e θsin+( )2–+=

A

B

C

D

Given the crank angle (at A) find the rocker angle (at D). Given the rocker angle, find the crank angle. Assume, and add, and all required dimensions.

3.4.2 Complex Numbers

• Complex numbers can be very useful when dealing with planar mechanisms.

• Recall the basic forms for representing complex numbers,

• When adding or subtracting complex vectors we can often use a calculator with complex or polar notations. This will save time and reduce errors.


R Rxi Ryj+=

x

yCartesian Vector Notation - A cartesian

approach is shown here. Note i and j are used as direction placeholders.

R Rx jRy+ Rx 1– Ry+= =

Real

Imaginary

Complex Notation - The components are represented with a complex number with a real part, and a com-plex part ‘j’.

R R ejθ

R θcos j θsin+( )= =

Real

Imaginary

Complex Polar Notation - This time the vector is represented with a direc-tion and length. Note, the complex ‘j’ is here again.

θ

A

B

C

The crank slider mechanism is in the position where BC is vertical.a) If the crank angle is 25°, find the position of C using Complex Polar Notation, b) repeat using complex numbers only. c) How can the solution be checked for accuracy?

10”

4”

3.4.3 Numerical Solutions

• At times it will be very difficult, or unreasonable to isolate a single variable in an equation.

• When this occurs we may take the resulting equation, and set it equal to zero by moving all terms to one side. Then, by trial and error, we can make guesses that will lead to a value equal to zero. One popular way for making these guesses is the Newton-Raphson technique.


y x( ) x2---

sin e0.15x

+=

1 3 5

6

Find the final resting position for the disk on the curved surface

3.4.4 Simulations

• We can use packages such as Working Model.

• A typical method for problem solution might be,1. lay out the basic components in the mechanism (not necessarily in the final configura-

tion)2. Apply a driving function to the input joint (e.g., the input angle on the crank)3. Run the simulation, and generate data/graphs.4. If motion is satisfactory, check for forces, velocities, accelerations, etc.

• Quite often it is possible to write a script that will run a simulation, adjust the mechanism, run the simulation again, and eventually by trial and error an ideal mechanism can be found.

• The simulation packages typically contain an integration technique (e.g., Runge-Kutta) that will deal with the dynamics of the problem. The basic process is,

1. Set the initial conditions of the problem, based on the user setup. A time step will also be chosen for the period of integration.

2. Determine the forces on each body in the system. In effect a free body diagram like analysis is done based on applied forces, contact forces, friction, gravity, etc.

3. For each body in the system, integrate the dynamic effects to get a new position, veloc-ity and acceleration.

4. Look at the amount of change that occurred. If too large, reduce the time step. and repeat step 3.

5. Update the model of the system. If the system is no longer changing, stop the simula-tion.

6. Update the display, and go back to step 2.

3.5 GRAPHING OF POSITIONS

• We can graph the positions of point on mechanisms using the techniques covered so far.

• Note that we have been plotting joint positions so far, but in actual practice we will be using other point on the links.

• Consider the example,

A

B

C

D

E

4”

1”

2”

5” 4”

Trace the path for point E, as the crank is rotated about joint A. Try this using calculations, mathcad, working model, and a computer program.

3.6 DISPLACEMENT, TRANSLATION AND ROTATION

• As a mechanism is moved, points on the mechanism will also move through space. The problem is that they will continue to be the same points, but we need to indicate the difference in posi-tion (displacement) for calculation purposes.

• We will indicate position using the traditional delta symbol,

• We can also consider relative positions between points on rigid bodies,

• Translation of a rigid body indicates that all points are traveling in the same direction at the same velocity. In this case the displacements of both points will be equal for pure translation.

• Rotation occurs when the displacements of two points on a body are not equal. If one of the dis-

A

B

A’

B’

∆RA RA' RA–=

A

B

A’

B’

RAB RA RB–=

RA'B' RA' RB'–=

rigid body

A

B

A’

B’

∆RAB ∆RA ∆RB– 0= =

Pure Translation

∆RA ∆RA' ∆RA–=

∆RB ∆RB' ∆RB–=

placements is zero, then it is the center of rotation.

• Apparent displacement considers that the coordinate axes are also moving. In this case the dis-placement and apparent position notations are combined.

• Consider the example below and find the motion of the point B using complex polar notation, and then matrices.

A

B

A’

B’

∆RA ∆RA 2⁄ ∆RO2+=

Apparent Displacement

O1

O2

O’2x

x

x

y

y

y

A

A’

B

B’

(3,3)

(6,2)

6”

45°

1”1/2”

x

y


1. Crank CB is 6” in length. Find the location of point D as a function of the crank angle at B using either complex numbers or complex polar notation.

2. The test pilot of the aircraft below is doing loops to evaluate the cockpit cup holders. Part way through one loop he notices (apparently) another plane one mile straight ahead from his posi-tion. If he is flying upside down at the location shown below, what is the absolute position of the other plane relative to the control tower. Note: use proper notation to solve this problem.

3.8 REFERENCES

Erdman, A.G. and Sandor, G.N., Mechanism Design Analysis and Synthesis, Vol. 1, 3rd Edition,

A

B

C

D

18”

12”

controltower

10 mi.5 mi.

1 mi.

test plane 40°

Prentice Hall, 1997.


4. MECHANISM VELOCITY

• Velocity is ‘how fast something is moving’.

4.1 THE BASIC RELATIONSHIPS

• One form of velocity is translation. In proper terms it is the first derivative of position,

• A second form of velocity is rotational. This is in terms of orientation angle,

• When we deal with 3D rotation and velocity we must consider the effects of the entire body. If we consider two points (P and Q) on a single rigid body, then we can write the two following relationships,

VP

dRP

dt----------

RP' RP–

∆t--------------------

∆t 0→lim= =

where,VP = the velocity of a pointRP = the position of a pointRP’ = the position of a point after some time delta tdelta t = some small period of time

ω dθdt------ θ' θ–

∆t-------------

∆t 0→lim= =

where,

ω rotational velocity=θ rotated position=

θ' rotated position after some time∆t=

∆t some small time period=

VP VPQ VQ+= Velocity Difference Equation

VPQ ω RPQ×=

• Consider that the apparent position equation from before can be extended to apparent velocity,

• Consider the example shown below, from Shigley and Uicker, 1995,

VP2VP1

VP2 1⁄+=

14”

3”

2”E

A

B

D

If the crank EA has been rotated 30° counterclockwise about E and has a rotational velocity of 36 rad/sec, what is the angular velocity of the angled rod BD?

• While we have considered apparent velocity in translation before we may also consider apparent angular velocity.

• Consider that in most cases seen so far the joints are of the pin type, and therefore are fixed points on each member.

• If we have a rolling contact we will have the following conditions, (Note: even though the roll-ers are round below, they could be other shapes)

4.2 CALCULATION TECHNIQUES

• We can expand the tools for analysis of position to find velocities.

• The most important factor to remember is that velocity is the first derivative of position.

4.2.1 Complex Algebra

• Recall the position vector in complex polar notation. And, after finding the first derivative we get the simple form,

ω2 ω2 1⁄ ω1+=

P

1

2

VP1 2⁄ 0=

VP1VP2

=

No Slip Condition

Rolling Contact Condition

• Consider the crank-slider mechanism below,

R R ejθ

=

V ddt-----R d

dt-----R e

jθj d

dt-----θ

R ejθ

+= =

• Next, consider a four bar linkage, in the double rocker configuration,

The crank is 20cm long, and the driver is 80cm long. If the crank is rotat-ing at 60rpm, what are the velocities of each of the links? Use a com-plex approach.

Find the velocity of the connector, if the lefthand rocker is the input. Assume that the input position and velocity is known. Use a vector approach.

3”

7”

1”

8”

4.2.2 Analytic Methods

• Basically use techniques to develop equations for position, and then differentiate

• Consider the mechanism pictured below,

Relate the position of the yoke to the left-most crank.

4.3 INSTANTANEOUS CENTERS OF ROTATION

• If we look at any body in a mechanism it will be rotating and translating.

• And, if we pick the right point to watch, it will appear to be rotating, we will call this the center of rotation.

• And, because the translation and rotation often changes as the mechanism moves through differ-ent positions we say that it is instantaneous.

• For a binary link the instant center is defined by the velocities at the joints.

• The easiest way to find an instantaneous center is to find the intersection (P) of vectors perpen-dicular to the velocity vectors at the joints (A and B).


A

B

P

VA

VB

• We can also find an instantaneous center for a chain of links.

• If we consider an entire mechanism with ‘n’ links, it will generally have ‘N’ instantaneous cen-ters.

• We can help keep track of which centers are to be found using points mapped out on a circle. In this technique we lay out a circle, and then put on a point for each of the links (including the ground). Draw on lines to connect each point where links exist. Also draw on lines for the Kennedy theorem.

1 rad/sec3”

3”6”

6”

Nn n 1–( )

2--------------------=


1

2

3

4

12

23

34

41

1 1

2

3

4

13

24

• The instantaneous center has no velocity (VP=0), but there is also a rotational velocity about this point. This can be found using the relationship,

4.3.1 Aronhold-Kennedy Theorem of Three Centers

• The theorem states that for three rigid links in a mechanisms, the instantaneous centers for each of the three mechanisms lie on the same straight line.

ωVA

RAP------------

VB

RBP------------= =

• We can consider a linkage with three members as pictured below,

4.3.2 Some Examples of Instantaneous Centers

• Consider the following mechanism - use the Aronhold-Kennedy theorem,

2

3

A

B

C

P32

ω3

VB3 2⁄

P12 P13

• Consider the following example. Find all of the instant centers.

• Find the instant centers for the mechanism below,

4.3.3 The Angular-Velocity-Ratio Theorem

• “The angular velocity ratio of any two bodies in planar motion relative to a third body is inversely proportional to the segments into which the common instant center cuts the line of centers”

• In general, after finding the instant centers, we can find rations between angular velocities using,

ωk i⁄ωj i⁄----------

RPjkPij

RPjkPik

--------------=

4.3.4 Mechanical Advantage

• Keeping in mind that energy is conserved we can develop equations for mechanical advantage.

• These two relationships can then be used with the results of earlier calculations to find the mechanical advantage.

PIN TINωiN=

PIN FINViN=

POUT TOUTωOUT=

POUT FOUTVOUT=

PIN POUT= MAFOUT

FIN-------------

VIN

VOUT-------------= =

For linear motion,

For torque,

PIN POUT=

MAFOUT

FIN-------------

RINωiN

ROUTωOUT---------------------------= =

TINωiN TOUTωOUT=

FINRINωiN FOUTROUTωOUT=

4.3.5 Freudenstein’s Theorem

• If in a 4 bar linkage we find instant centers for adjacent links (i.e. 1 & 3 and 2 & 4). These form a collinear axis if there is a perpendicular to the connecting rod then that position has the mini-mum mechanical advantage.

Find the velocity of the connector, if the lefthand rocker is the input. Assume that the input position and velocity is known. Use a vector approach.

3”

7”

1”

8”

4.3.6 Centrodes

• Instant Centers are valid for only a single point in time. If we watch the locations of these cen-ters evolve, we will see them move through space. Below the centrode for the four bar linkage shows how P13 sweeps through space as the mechanism moves.

1

2

3

4

• In the example above the centrode is called ‘fixed’. If we had used the kinematic inversion where 3 was the frame, then the instant center would trace out another called a ‘moving’ cen-trode.


1. Develop position and velocity equations for the following mechanisms in complex polar nota-tion. If other calculations are required, write these out also. Show the equations needed to calcu-late any unknown angles. Do not attempt to solve any equations, but indicate which variables are unknown by underlining them in the final equation.

a) Find all of the positions and velocities.

A Fixed centrode

P13

P13

P13

4

3

2

1

1

2

34

5

A

B

C

D

E

b) In addition to the basic equations, also write the equations for the position and velocity of point G?

b) Find all of the positions and velocities. Assume points A, C and F lie on a horizontal line.

2. Find the orientations and velocities of the joints and point G in the mechanism using trigonom-etry, or any other non complex method. Assume that link 2 is being driven at 2 rad/sec.

A

B C

DE

F

G

1 1 1

2

3

4

5

1 1 1

2

3

4

5

6A

B

C

D

E

F

4.5 REFERENCES



A

B C

DE

F

G

1 1 1

2

3

4

5

4”

3”

6”

6”

6”

5”

2”

5. MECHANISM ACCELERATION

• The amount that a point on a mechanism speeds up or slows down.

5.1 THE BASIC DEFINITION

• The acceleration of a point ‘P’ is the first derivative of velocity, or second derivative of position,

• To analyze linear acceleration we must first separate the velocity into a magnitude, and direc-tion.

• We can also consider angular acceleration,

• If we consider the relative velocity and acceleration between two points on a rigid body we get,

• For planar calculations we can write a complex equivalent,

AP

∆VP

∆t----------

∆t 0→lim d

dt-----VP

ddt-----

2RP= = =

VP VP λP=

APddt----- VP

λp VPddt-----λP

+=

α ddt-----ω=

APddt----- VP( ) d

dt----- VQ ω RPQ×+( ) AQ ω d

dt-----RPQ

× ddt-----ω RPQ×+ += = =

AP∴ AQ ω ω RPQ×( )× α RPQ×+ +=

AP AQ ω2RPQe

jθ– jαRPQe

jθ+ AQ APQ

nAPQ

t+ + AQ APQ+= = =

• As an example, let’s consider the four bar mechanism below,

A

B

CD45°

If the link lengths are AB=18”, BC=12”, CD=8”, DA=6”, AE=12”, EB=10” and the angular velocity of the crank about D is 200 rad/sec. Find the acceleration of point E in this system.

E

• there are a couple of other relationships of use,

• Apparent acceleration is found by first separating velocity into magnitude and direction.

• Note: the coriolis component above happens in systems with moving frames of reference. This can create a whip effect.


AP AQ APQ+=

APQ An

PQ AtPQ+=

Velocity Difference Equation

where,

An

PQ normal acceleration component=

AtPQ tangential acceleration component=

AP3 2⁄ An

P3 2⁄ AtP3 2⁄+

V2

P3 2⁄–ρ

------------------- ρ d

2

dt2

-------sτ

+= =

AP3AP2

AP3P2

cAP3 2⁄

nAP3 2⁄

t+ + +=

where,

AP3 2⁄ The apparent acceleration=

An

P3 2⁄ The apparent normal acceleration=

AtP3 2⁄ The apparent tangentialacceleration=

τ

ρ

where,

AP3P2

cThe coriolis component=

AP3P2

c2ω2 VP3 2⁄×=

The apparent acceleration equation

O2

O3

In polar form,

AP3AP22

eiθ

2ωieiθ

R3 2⁄ α ieiθ

R3 2⁄– ω2e

iθ+ +=

• Apparent Angular acceleration is found using,

14”

3”

2”E

A

B

D

If the crank EA has been rotated 30° counterclockwise about E and has a rotational velocity of 36 rad/sec, what is the acceleration of the rod BD?

• For rolling contact the centripetal and normal accelerations are zero, therefore we can simplify the apparent acceleration equation to,


α3 α2 α3 2⁄+=

AP3AP2

AP3P2

cAP3 2⁄

nAP3 2⁄

t+ + + AP2

0 0 AP3 2⁄

n+ + + AP2

AP3 2⁄

r+= = =

In rolling contact the rolling and coriolis effects are both zero. This reduces the apparent acceleration equation. The tangential component is relabeled the rolling component to indicate that the direction is normal to the rolling plane.

• We can also solve these problems using simple derivatives. Consider the example below,

If the crank AB has a rotational velocity of 90°/sec, what is the angular accelera-tion of the angled rod BC?

A

B

C

2

3

10”

5”

1”

The crank is 20cm long and 20° below the horizon, and the driver is 80cm long. If the crank is rotating at 60rpm, what are the accelerations of each of the links?

• Consider the accelerations of the joints in the four bar linkage below,

4”

5”

A

B C

D

AB is being turned with an angular velocity of 100 rad/sec, and angular acceleration of 5 rad/sec.

5.2 INSTANT CENTERS OF ACCELERATION

• We can find this center using normals to acceleration vectors applied at joints.

• This point will generally not the same as the instantaneous center of velocity.

• Generally there are few uses for this value.


1. Find the positions, velocities, and accelerations of all of the members in the mechanism below.

5.4 REFERENCES



A

B

CD

The link lengths are AB=14.1”, BC=20”, CD=10”, DA=10”, and the angular velocity of the crank about D is 20 rad/sec.

1

2

3

4

6. LINKAGE SYNTHESIS

• The large number of possible ways to satisfy design criteria make it hard for us to select a single useful mechanism.

• The design techniques outlined below can help narrow down, or select a specific mechanism.

• To do the design work we typically begin with some functional specifications for a path of motion. These specifications typically include one or more of the following,

- rotational motions (i.e. shaft)- differing advance/return times- small stop/dwell times- straight line segments in path- specific velocities along path- a curved path

• Some points of interest about the selection of linkages in design,- linkages can be used in place of cams to,

reduce forceseliminate loading springsimplify manufacture and tolerance problems

- the space required for these linkages may be larger than cams- in many cases ratios are specified. After we get these we will design the weakest member

to be mechanically adequate, and then design the other parts.

6.1 SELECTION OF MECHANISM TYPES

• When trying to select a mechanism type one way is, (NOTE: there is no exact way to this, and you need to exercise some creativity)

1. Examine the motion profile required.2. Determine the distinct phases of motion, and estimate the d.o.f. needs for each/all.3. Look for crosses in path, dwell spots, motion reversal, straight line segments, etc.4. Consider the various configurations you know of and how they satisfy parts or all of the

motion. The dof can be used to suggest the number of links and types of joints required (i.e. Grashof’s criteria).

5. If no satisfactory solutions are available, the look to a handbook of available mecha-nisms for other ideas.

6. Use design techniques for the specific mechanisms to determine dimensions7. Test the design on Working Model, or by building a prototype.

• Shigley and Uicker identify three types of common linkage problems to look for,Function generation - in this case we want to drive an output shaft (angular displacement)

using an input shaft (angular displacement). This function can also be performed using cams with oscillating followers.

Prescribed Path Shape - Here we follow a point on the mechanism as it follows a path through space.

Body Guidance - A combination of translation and rotation may be used to move an object.

• Erdman and Sandor also introduce the idea of choosing mechanisms into stages,1. Type synthesis2. Dimensional synthesis

6.2 DESIGN METHODS - SYNTHESIS

• This stage is somewhat inexact and should involve a few questions about the application.- Is there a starting/ending position?- What is the starting ending position?- Are there dimensional limits?- How much power must be transmitted?- Is the mechanism function generating, prescribed path or body guidance?- How fast is the mechanism?- Does the motion need to be smooth?- How many degrees of freedom does the motion require?- Is the input/output motion type known? (i.e. rotation, linear, other)- Are inertial effects desired?- Can the mechanisms be planar/3D?- How much error is permissible? Is accuracy needed at all points?- Does the mechanisms appear to have multiple modes of motion?

• There are a number of excellent books that can be used to find suitable matches. These can often be modified to suit your needs. This technique is often known as ‘cookbook’.

Chironis & Sclater, Mechanisms and Mechanical Devices Sourcebook, McGraw-Hill, 1995

Jones, F.D., Ingenious Mechanisms for designers and inventors, Industrial Press Inc., Vol-umes 1-4, 1930.

6.2.1 Mechanism Typing

• This method helps systematically generate mechanism types.

• We begin by considering the Gruebler/Kutzbach equation rewritten,

• After the # of degrees of freedom for the application are determined we can then use the equa-tion to generate alternatives.

1. Examine # dof in the application.2. Pick a # of links in the mechanism.3. List values of B, T, Q, P, etc. that satisfy the equation.4. Sketch mechanisms based on the list of B, T, Q, P, etc.5. Sketch a suitable mechanism and then do detailed design.6. If the design isn’t satisfactory choose a new mechanism.

• Consider the problem below and find possible solutions.

n m 3+( )– T 2Q 3P …+ + +≥where,

n = # of links in mechanismm = degrees of freedom needed in mechanismT = # of tertiary linksQ = # of quaternary linksP = # of pentagonal links

select a general mechanism to reach around a box and push on the back side and then retract.

6.3 DESIGN METHODS - DIMENSIONAL

• What follows is an assorted collection of independent design techniques.

• These techniques can be used after the general mechanism type has been selected.

• Learning these techniques before they are needed can help in the initial selection of mechanism types.

6.3.1 Two Position Design

• When we have an advance-return-ratio we can use the basic geometry of a linkage to specify dimensions.

• For this approach we are given the extreme angular or position displacements.

• We can then calculate the geometry of the mechanisms at the two extreme two positions.

• If we are considering a slider crank mechanism we can find the extremes as shown below. A general relationship may be derived that can be used for subsequent design.

A

B

Cd

e f

L1

L2

L2 L1– d2

e2

+=

Both extreme positions are developed when AB and BC are collinear. For this analysis we will assume that AB rotates clockwise, therefore the slider will be advancing when C is moving to the right.

Case A - leftmost position d

e

L2-L1

L2 L1+ d2

e f+( )2+=

Case B - rightmost position d

e+f

L2+L1

Now, consider the Advance-return-ratio, Q,

θ1

θ2

θ1 180° de---

atan–=

θ2d

e f+----------

atan=

Qθ1 θ2+

360 θ1 θ2+( )–-------------------------------------

180° de---

atan– de f+----------

atan+

180° de---

atan de f+----------

atan–+-----------------------------------------------------------------------= =

For design we can reduce the equations to,

L1 L2 d2

e2

+– L2– d2

e f+( )2++= =

L2∴ d2

e f+( )2+ d

2e

2++

2-------------------------------------------------------------= L1∴ d

2e f+( )2

+ d2

e2

+–2

-------------------------------------------------------------=

• The next stage in design is to select, or constrain some values, and then through trial and error estimate a design.

• Consider the following example,

• A similar case arises when we are planning to design a crank-rocker mechanism for return time.

Assuming we have decided to use a crank slider mechanism with a slide of 30cm, and a advance-return-ratio of 1.2, what would the remaining dimensions be. (Hint - you will have to select values for ‘d’ and ‘e’, and check Q for a match, and L1 and L2 for reason-able sizes). Use mathcad, or a programmable calculator.

if we consider the mechanism pictured below, the extreme angles for CD will be when AB and BC are collinear. We can use this as the basis for analyzing the mechanism. We can also assume that the design will specify the extreme angles for CD.

A

B

C

D

L1

L2

L3

L4

Case A - Leftmost angles

L1

L4L3-L2

L3 L2–

180° φ1–( )sin------------------------------------

L4

θ1 180°–( )sin------------------------------------=

φ

φ1θ1

θ1∴L4

L3 L2–-----------------

180° φ1–( )sin asin 180°+

L4 φ1( )sin

L3 L2–------------------------

asin 180°+= =

L1 L4180° 180° φ1–( )– θ1 180°–( )–( )sin

θ1 180°–( )sin---------------------------------------------------------------------------------------------

L4φ1 θ1–( )sin

θ1( )sin------------------------------

= =

Case B - Rightmost angles

L1

L4L3+L2

L3 L2+

180° φ2–( )sin------------------------------------

L4

θ2( )sin------------------=

φ2

θ2

θ2∴L4

L3 L2+------------------

180° φ2–( )sin asin=

• Consider the design problem below,

• We can also use a closed form solution based on dyads.

Next, consider the advance-return-ratio,

Qθ1 θ2–

360 θ1 θ2–( )–-------------------------------------=

With the calculations given here we would be given phi 1 and 2, and a time ratio Q. We would then select values for L1, L2 and L3 by trial and error. From these we can calculate Q and L4 to check the solution.

Design a four bar linkage that gives the following values,

Q 1.2=

φ1 160°=

φ2 70°=

• Design a 4 bar mechanism to reach the points C and D below,

Z1

Z2

β2

α2

W1W2

δ2

Wδ2 Z e

iα 2 1–( )–

eiβ2 1–

-------------------------------------=

Trial and error can be used to find a solution. This would need to be done twice for a 4 bar mechanism.

A=(8,0)B

C=(10,6)D=(5,5)

E

• Consider the design problem below,

6.4 REFERENCES


Shigley, J.E., Uicker, J.J., Theory of Machines and Mechanisms, 2nd Edition, McGraw-Hill, 1995.

Find the center of rotation for the two points marked on the part below,

7. SPATIAL KINEMATICS

• Basically mechanisms that are 3D (not planar).

7.1 BASICS

• When we deal with geometries in two dimensions we have three position variables (dof) for each rigid body (two for position, one for orientation).

• When a problem is expanded to three dimensions we then have six position variables (dof) for a rigid body (three for position, and three for orientation).

• These added degrees of freedom expand the complexity of the problem solutions. There are a few potential approaches,

- look for regularities that simplify the problem (scalar)- vector based approaches (positions)- matrix based approaches (positions and orientations)

• Consider the example of the spherical joint - all of the axes of rotation coincide.

7.1.1 Degrees of Freedom

• The scalar and vector approaches are easily extended to 3D problems. One significant difference is that the polar notations are no longer available for use.

• We can determine the number of degrees of freedom using a simple relationship that is an exten-sion of the Kutzbach criteria,

m 6 n 1–( ) 5j1– 4j2– 3j3– 2j4– j5–=

where,

m mobility of the mechanism (d.o.f.)=

n number of links=

j1 j2 …, , the number of joints with 1, 2, ... dof respectively=

• Consider the number of degrees of freedom in the linkage below,

7.2 HOMOGENEOUS MATRICES

• This method still uses geometry to determine the position of the robot, but it is put into an ordered method using matrices.

• Consider the planar robot below,

x

y

z

AB

C

D

E

10”

6”40”

3”

• The basic approach to this method is,1. On the base, each joint, and the tool of the robot, attach a reference frame (most often x-

y-z). Note that the last point is labels ‘T’ for tool. This will be a convention that I will generally follow.

2. Determine a transformation matrix to map between each frame. It is important to do this by assuming the joints are in their 0 joint positions. Put the joint positions in as variables.

(xb, yb)

(xT, yT)

1m

1m

0.2m

TCP

θ1

θ2

(x1,y1)

x

y

z

x

y

z x

y

z

x

y

z

θ1

θ2

F0

F1

F2

FT

T0 1,T1 2,

T2 T,

T0 1, = T1 2, = T2 T, =

trans ∆x ∆y ∆z, ,( )

1 0 0 ∆x

0 1 0 ∆y

0 0 1 ∆z

0 0 0 1

=

rot y θ,( )

θcos 0 θsin– 0

0 1 0 0

θsin 0 θcos 0

0 0 0 1

=

rot x θ,( )

1 0 0 0

0 θcos θsin 0

0 θsin– θcos 0

0 0 0 1

=

rot z θ,( )

θcos θsin 0 0

θsin– θcos 0 0

0 0 1 0

0 0 0 1

=

ASIDE: The structure of these matrices describe the position (P) and orientation of the x (N), y (O), z (A), axes.

NX OX AX PX

NY OY AY PY

NZ OZ AZ PZ

0 0 0 1x (N)

y (O)

z (A)

P

3. Multiply the frames to get a complete transformation matrix.

T0 1,

θ1cos θ1sin 0 0

θ1sin– θ1cos 0 0

0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

rot z θ1,( )trans 1 0 0, ,( )= =

T1 2,

θ2cos θ2sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

rot z θ2,( )trans 1 0 0, ,( )= =

T2 T,

1 0 0 0.2

0 1 0 0

0 0 1 0

0 0 0 1

trans 0.2 0 0, ,( )= =

• The position and orientation can be read directly from the homogenous transformation matrix as indicated above.

• To reverse the transform, we only need to invert the transform matrix - this is a direct result of the loop equation.

T0 T, T0 1, T1 2, T2 T,=

T0 T,

θ1cos θ1sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

θ2cos θ2sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0.2

0 1 0 0

0 0 1 0

0 0 0 1

=

T0 T,

θ1cos θ1sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

θ2cos θ2sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1.2

0 1 0 0

0 0 1 0

0 0 0 1

=

T0 T,

θ1cos θ1sin 0 0


0 0 1 0

0 0 0 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

θ2cos θ2sin 0 1.2 θ2cos

θ2sin– θ2cos 0 1.2– θ2sin

0 0 1 0

0 0 0 1

=

T0 T,

θ1cos θ1sin 0 0


0 0 1 0

0 0 0 1

θ2cos θ2sin 0 1.2 θ2cos 1+

θ2sin– θ2cos 0 1.2– θ2sin

0 0 1 0

0 0 0 1

=

T0 T,

θ1 θ2+( )cos θ1 θ2+( )sin 0 θ1cos 1.2 θ1 θ2+( )cos+

θ1 θ2+( )sin– θ1 θ2+( )cos 0 θ1sin 1.2 θ1 θ2+( )sin+

0 0 1 0

0 0 0 1

=

PositionOrientation

complete the multiplicationand simplify to get......

7.2.1 Denavit-Hartenberg Transformation (D-H)

• Designed as more specialized transforms for robots (based on homogenous transforms)

• Zi-1 axis along motion of ith joint

• Xi axis normal to Zi-1 axis, and points away from it.

• Basic transform is,1. rotate about Zi-1 by thetai (joint angle)2. translate along Zi-1 by di (link offset)3. translate along Xi by ai (link length)4. rotate about Xi by alphai (link twist)

T∴ T 0, T0 T,( ) 1–=

T0 1, T1 2, T2 T, TT 0, I=

where,

I = Identity matrix

T0 T,∴ TT 0, I=

We can manipulate the equation,

Ti 1– i, rot zi 1– θi,( )trans 0 0 di, ,( )trans ai 0 0, ,( )rot xi α i,( )=

Ti 1– i,

θicos α icos θisin α i θisinsin ai θicos

θisin α icos θicos α i θicossin– ai θisin

0 α isin α icos di

0 0 0 1

=

• We can see how the D-H representation is applied using the two link manipulator from before

zi

zi 1+

xi 1+yi 1+

xi

yi

ai 1+

di 1+

θi 1+

zi α i 1+

xi

RobotBase

7.2.2 Orientation

• The Euler angles are a very common way to represent orientation in 3-space.

• The main problem in representing orientation is that the angles of rotation must be applied one at a time, and by changing the sequence we will change the final orientation. In other words the

T0 1,

θ1cos θ1sin– 0 θ1cos

θ1sin θ1cos 0 θ1sin

0 0 1 0

0 0 0 1

=

θi θ1=

di 0=

ai 1=

α i 0=

T1 2,



0 0 1 0

0 0 0 1

=

θi θ2=

di 0=

ai 1=

α i 0=

T2 T,

1 0 0 0.2

0 1 0 0

0 0 1 0

0 0 0 1

=

θi 0=

di 0=

ai 0.2=

α i 0=

T0 T, T0 1, T1 2, T2 T,=

T0 T,



0 0 1 0

0 0 0 1



0 0 1 0

0 0 0 1

1 0 0 0.2

0 1 0 0

0 0 1 0

0 0 0 1

=

T0 T,



0 0 1 0

0 0 0 1

=

three angles will not give a unique solution unless applied in the same sequence every time.

• By fixing a set of angles by convention we can then use the three angles by themselves to define an orientation.

• The convention described here is the Euler angles.

• The sequence of orientation is,

• Therefore to reorient a point in space we can apply the following matrix, to the position vectors, or axes vectors, (there will be more on these matrices shortly)

• We can find these angles given a set of axis before and after.

In order,

rot θ( ) rot φ( ) rot ψ( ), ,

where,

θ rotation about z axis=

φ rotation about new x axis=

ψ rotation about new z axis=

Rx'

Ry'

Rz'

ψcos ψsin 0

ψsin– ψcos 0

0 0 1

1 0 0

0 φcos φsin

0 φsin– φcos

θcos θsin 0

θsin– θcos 0

0 0 1

Rx

Ry

Rz

=

Rx'

Ry'

Rz'

θ ψcoscos θ φ ψsincossin–( ) θsin ψcos θcos φ ψsincos+( ) φ ψsinsin( )θ ψsincos– θ φ ψcoscossin–( ) θsin ψsin– θcos φ ψcoscos+( ) φ ψcossin( )

θ φsinsin( ) θ φsincos–( ) φcos( )

Rx

Ry

Rz

=

7.2.3 Inverse Kinematics

• Basically we can find the joint angles for the robot based on the position of the end effector.

• This is not a simple problem, and there are few reliable methods. This is partly caused by the non-unique nature of the problem. At best there are typically multiple, if not infinite numbers of equivalent solutions. The 2 dof robot seen before has two possible solutions.

• We can do simple inverse kinematics with trigonometry.

• If we have more complicated problems, we may try to solve the problem by examining the trans-form matrix,

x0

1

0

0

= x1

0.71

0.71

0

= y0 = y1 =

7.2.4 The Jacobian

• A matrix of partial derivatives that relate the velocity of the joints, to the velocity of the tool.

T0 T,



0 0 1 0

0 0 0 1

=

xT θ1cos 1.2 θ1 θ2+( )cos+=

yT θ1sin 1.2 θ1 θ2+( )sin+=

xT θ1cos–

1.2-------------------------∴ θ 1 θ2+( )cos=

yT θ1sin–

1.2------------------------∴ θ 1 θ2+( )sin=

Combine the two to eliminate the compound angles,

1 θ1 θ2+( )cos( )2 θ1 θ2+( )sin( )2+=

Separate positions and simplify,

1∴xT θ1cos–

1.2-------------------------

2 yT θ1sin–

1.2------------------------

2+=

1.44∴ xT2

2xT θ1cos– θ1cos( )2yT

22yT θ1sin– θ1sin( )2

+ + +=

0.44 xT2

– xT2

–( )∴ 2xT θ1cos– 2yT θ1sin–=

0.44 xT2

– xT2

–

2–---------------------------------

∴ xT θ1cos yT θ1sin+( )=

ETC.....

• The inverse Jacobian is used for motion control

• Find the Jacobian and inverse Jacobian for the 2 dof robot.

ddt-----x

T

ddt-----y

T

ddt-----z

T

∂xT

∂θ1--------

∂xT

∂θ2--------

∂xT

∂θ3--------

∂yT

∂θ1--------

∂yT

∂θ2--------

∂yT

∂θ3--------

∂zT

∂θ1--------

∂zT

∂θ2--------

∂zT

∂θ3--------

ddt-----θ

1

ddt-----θ

2

ddt-----θ

3

J

ddt-----θ

1

ddt-----θ

2

ddt-----θ

3

= =

J1–

x·T

y·T

z·T

θ· 1

θ· 2

θ· 3

=

x

y

z

x

y

z x

y

z

x

y

z

θ1

θ2

F0

F1

F2

FT

T0 1,T1 2,

T2 T,

7.3 SPATIAL DYNAMICS

• The basic principles of planar dynamics are expanded up for 3D spatial problems. The added dimension adds some complexity that should be addressed.

7.3.1 Moments of Inertia About Arbitrary Axes

• Moments of Inertia are normally found for a single axis of rotation. When the object is rotating about another axis, we must recalculate the moments of inertia.

• If we take the moments of inertia for the original axes, and project these values onto new vec-tors, we can get new values,

T0 T,



0 0 1 0

0 0 0 1

=

Find the Jacobian matrix for the matrix given below. This will give a matrix that relates tool velocity to joint velocities.The joint angles are 30° and 20° for joints 1 and 2, find the joint velocities if the tool velocity is 0.05 m/s and 0.1 m/s in the x and y directions respectively (hint: inverse jacobian).

We start by defining the vector equivalencies for rotated axes,

R' R iRx jRy kRz+ + i'R'x j'R'y k'R'k+ += = =

We can project this vector to the other set of axes,

R'x i'( ) iRx jRy kRk+ +( )• Rx θi'icos Ry θi'jcos Rz θi'kcos+ += =

R'y j'( ) iRx jRy kRk+ +( )• Rx θj'icos Ry θj'jcos Rz θj'kcos+ += =

R'z k'( ) iRx jRy kRk+ +( )• Rx θk'icos Ry θk'jcos Rz θk'kcos+ += =

Next we integrate for moment of inertia for the shifted x axis,

Ix'x' i' R'x×( ) i' R'x×( )• md∫=

∴ Rx θi'icos Ry θi'jcos Rz θi'kcos+ +( )2md∫=

∴ Rx θi'icos( )2Ry θi'jcos( )2

Rz θi'kcos( )22 Rx θi'iRy θi'jcoscos( )+ + + +∫=

2 Rx θi'i Rz θi'kcos( )cos( ) 2 Ry θi'j Rz θi'kcos( )cos( )dm+

Ix'x'∴ Ixx θi'icos( )2Iyy θi'jcos( )2

Izz θi'kcos( )22Ixy θi'i θi'jcoscos( )+ + + +=

2Ixz θi'i θi'kcoscos( ) 2Iyz θi'j θi'kcoscos( )+

Next we integrate for the product of inertia for the shifted x and y axis,

Ix'y' i' R'x×( ) i' R'y×( )• md∫=

Similarly for the shifted y and z axes,

Iy'y'∴ Ixx θj'icos( )2Iyy θj'jcos( )2

Izz θj'kcos( )22Ixy θj'i θj'jcoscos( )+ + + +=

2Ixz θj'i θj'kcoscos( ) 2Iyz θj'j θj'kcoscos( )+

Iz'z'∴ Ixx θk'icos( )2Iyy θk'jcos( )2

Izz θk'kcos( )22Ixy θk'i θk'jcoscos( )+ + + +=

2Ixz θk'i θk'kcoscos( ) 2Iyz θk'j θk'kcoscos( )+

This will lead to,

Ix'y' Ixx θi'i θj'icoscos–( ) Iyy θi'j θj'jcoscos( ) Izz θi'k θj'kcoscos–( )+ + +=

Ixy θi'i θj'jcoscos θi'j θj'icoscos+( ) Iyz θi'j θj'kcoscos θi'k θj'jcoscos+( )+ +

Ixz θi'k θj'icoscos θi'i θj'kcoscos+( )

Ix'y' Ixx θk'i θi'icoscos–( ) Iyy θk'j θi'jcoscos( ) Izz θk'k θi'kcoscos–( )+ + +=

Ixy θk'i θi'jcoscos θk'j θi'icoscos+( ) Iyz θk'j θi'kcoscos θk'k θi'jcoscos+( )+ +

Ixz θk'k θi'icoscos θk'i θi'kcoscos+( )

Iy'z' Ixx θj'i θk'icoscos–( ) Iyy θj'j θk'jcoscos( ) Izz θj'k θk'kcoscos–( )+ + +=

Ixy θj'i θk'jcoscos θj'j θk'icoscos+( ) Iyz θj'j θk'kcoscos θj'k θk'jcoscos+( )+ +

Ixz θj'k θk'icoscos θj'i θk'kcoscos+( )

Rx'' Rx' dx'+= Ry'' Ry' dy'+= Rz'' Rz' dz'+=

We can define the new coordinate system in terms of translated axes,

This can be integrated for the shifted x axis,

Ix''x'' Ry''2

Rz''2

+( ) md∫ Ry' dy'+( )2Rz' dz'+( )2

+( ) md∫= =

∴ Ry'2

2Ry'dy' dy'2

Rz'2

2Rz'dz' dz'2

+ + + + +( ) md∫=

∴ Ry'2

Rz'2

+( ) md∫ 2Ry'dy'( ) md∫ 2Rz'dz'( ) md∫ dy'2

dz'2

+( ) md∫+ + +=

∴ Iy'z' 2mRMy'dy' 2mRMz'

dz' dy'2

dz'2

+( )m+ + +=

7.3.2 Euler’s Equations of Motion

• We can use Euler’s equations of motion to determine moments produced by angular velocities and accelerations.

• These can be used to examine rotating three dimensional masses. Consider the following,

This eventually leads to,

Ix''x'' Ix'x' 2mRMy'dy' 2mRMz'

dz' m dy'( )2m dz'( )2

+ + + +=

Iy''y'' Iy'y' 2mRMx'dx' 2mRMz'

dz' m dx'( )2m dz'( )2

+ + + +=

Iz''z'' Iz'z' 2mRMx'dx' 2mRMy'

dy' m dx'( )2m dy'( )2

+ + + +=

Ix''y'' Ix'y' 2mRMx'dy' 2mRMy'

dx' mdx'dy'+ + +=

Iy''z'' Iy'z' 2mRMy'dz' 2mRMz'

dy' mdy'dz'+ + +=

Ix''z'' Ix'z' 2mRMx'dz' 2mRMz'

dx' mdx'dz'+ + +=

MMijy∑ IMyy

αy IMxxIMzz

–( )ωxωz–=

MMijx∑ IMxx

αx IMyyIMzz

–( )ωyωz–=

MMijz∑ IMzz

α z IMxxIMyy

–( )ωxωy–=

7.3.3 Impulses and Momentum

• Momentum is a convenient alternative to energy in analysis of systems.

7.3.3.1 - Linear Momentum

• momentum is defined as,

The disk and shaft shown are rotated at 2000 rpm, and there is an angular acceleration of 20 rev/(sec.sec.). The steel part is held in a cantilevered bearing that can be approxi-mated with the forces shown.

2” 6”

8”

1”

8”

1”

F

F

• If no external forces are applied, momentum remains constant (is conserved). In this case L is a constant.

• An impulse is a force applied that will change momentum.

7.3.3.2 - Angular Momentum

• Angular momentum is for rotating objects. The rotation about some center tends to make these equations a bit more complicated than linear momentum.

• We can start to find this as a velocity times a distance of rotation, and this will lead to the even-tual relationships,

• These equations show the angular momentum H, along with other familiar terms.

7.4 DYNAMICS FOR KINEMATICS CHAINS

• There are a variety of common methods,- Euler-Lagrange - energy based- Newton-Euler - D’Alembert’s equations

F t( ) td

t1

t2

∫ mA t( ) td

t1

t2

∫ mV t2( ) mV t1( )– L t( )= = =

MomentumImpulse

MMx∑ IMxxαx IMxy

αy– IMxzα z– ωyHz ωzHy–+=

MMy∑ I– Myxαx IMyy

αy IMyzα z– ωzHx ωxHz–+ +=

MMz∑ I– Mxzxαx IMzy

αy– IMzzαz ωxHy ωyHx–+ +=

7.4.1 Euler-Lagrange

• This method uses a Lagrangian energy operator to calculate torques

• For a typical link,

• If we have used matrices to formulate the problem, we use the Jacobian to find velocities.


Li θi ωi,( ) Ki Pi–= L θ ω,( ) K∑ i P∑ i–=

where,L = lagrangianK = kinetic energy of link ‘i’P = potential energy of link ‘i’

ddt-----

ωi∂∂ Li θi∂

∂ Li– Qi=

Q = forces and torques

Ki

miVCiT

VCi

2----------------------- ωi

TIiωi+=

where,m = mass of link iV = velocity of center of mass of link iomega = angular velocity of link iI = mass moment of inertia of link i

Pi migTRCi=

where,g = gravity vectorR = displacement from base of robot to center of mass of link

VCi

ωi

J θ( )ω=

(xb, yb)

(xT, yT)

1m

1m

0.2m

TCP

θ1

θ2

Mlinks 5kg=

Mtool 0.5kg=

Ilinks 10=

Itool 1=

7.4.2 Newton-Euler

• We can sum forces and moments, and then solve the equations in a given sequence.

• These equations can be written in vector form,

• To do these calculations start at the base, and calculate the kinematics up to the end of the manipulator (joint positions, velocities and accelerations). Then work back from the end and find forces and moments.

7.5 REFERENCES


Fu, Gonzalez, and Lee,


Fi∑ miAi– 0=

Mi∑ Ii α i ωi2

+( )– 0=

fi 1– fi– mig miAi–+ 0=

where,

f = forces between link i and i+1A = acceleration of center of mass of link i

ni 1– ni– ri Ci, fi× ri 1– Ci, fi 1–×– Iiα i– ωi Ii ωi×( )×–+ 0=

where,

f = forces between link i and i+1A = acceleration of center of mass of link i


1. For the Stanford arm below,

a) list the D-H parameters (Hint: extra “dummy” joints may be required)b) Find the forward kinematics using homogenous matrices.c) Find the Jacobian matrix for the arm.d) If the arm is at θ1 = 45 degrees, θ2 = 45 degrees, r = 0.5m, find the speed of the TCP if

the joint velocities are θ’1 = 1 degree/sec, θ’2 = 10 degrees/sec, and r’ = 0.01 m/sec.

3. Robotics and Automated Manipulators (RAM) has consulted you about a new robotic manipu-lator. This work will include kinematic analysis, gears, and the tool. The robot is pictured below. The robot is shown on the next page in the undeformed position. The tool is a gripper (finger) type mechanism.

r

x

y

z

θ2

θ1

d1

(0,0,0)

TCP

TOP VIEW

FRONT VIEW

The robot is drawn below in the undeformed position. The three positioning joints are shown, and a frame at the base and tool are also shown.

Tool

The tool is a basic gripper mechanism, and is shown as a planar mechanism below. As the cylin-der moves to the left the fingers close.

a) The first thing you do is determine what sequence of rotations and translations are needed to find the tool position relative to the base position.

b) As normal, you decide to relate a cartesian (x-y) velocity of the gripper to joint velocities. Set up the calculation steps needed to do this based on the results in question #1.

c) To drive the revolute joints RAM has already selected two similar motors that have a maximum velocity. You decide to use the equations in question #2, with maximum specified tool veloci-

x0

y0

z0

x0

y0z0

xT

yT

zT

xT

yT

zT

r

θ2

θ1

c b

a

d

e

g

f

Finger

Pneumaticcylinder

ties to find maximum joint velocities. Assume that helical gears are to be used to drive the rev-olute joints, specify the basic dimensions (such as base circle dia.). List the steps to develop the geometry of the gears, including equations.

d) The gripper fingers may close quickly, and as a result a dynamic analysis is deemed necessary. List the steps required to do an analysis (including equations) to find the dynamic forces on the fingers.

e) The idea of using a cam as an alternate mechanism is being considered. Develop a design that is equivalent to the previous design. Sketch the mechanism and a detailed displacement graph of the cam.

f) The sliding joint ‘r’ has not been designed yet. RAM wants to drive the linear motion, without using a cylinder. Suggest a reasonable design, and sketch.

4. For an articulated robot, find the forward, and inverse kinematics using geometry, homogenous matrices, and Denavit-Hartenberg transformations.

5. Assign Denavit-Hartenberg link parameters to an articulated robot.

6. For the Stanford arm below,

a) list the D-H parameters (Hint: extra “dummy” joints may be required)b) Find the forward kinematics using homogenous matrices.c) Find the Jacobian matrix for the arm.d) If the arm is at θ1 = 45 degrees, θ2 = 45 degrees, r = 0.5m, find the speed of the TCP if

the joint velocities are θ’1 = 1 degree/sec, θ’2 = 10 degrees/sec, and r’ = 0.01 m/sec.

r

x

y

z

θ2

θ1

d1

(0,0,0)

TCP

TOP VIEW

FRONT VIEW

7. Consider the forward kinematic transformation of the two link manipulator below. Given the position of the joints, and the lengths of the links, determine the location of the tool centre point using a) basic geometry, b) homogenous transforms, and c) Denavit-Hartenberg transfor-mations.

a) For the robot described in question 1 determine the inverse kinematics for the robot. (i.e., given the position of the tool, determine the joint angles of the robot.) Keep in mind that in this case the solution will have two different cases. Determine two different sets of joint angles required to position the TCP at x=5”, y=6”.

b) For the inverse kinematics of question #2, what conditions would indicate the robot position is unreachable? Are there any other cases that are indeterminate?

8 Find the dynamic forces in the system below,

L1 = 12”

L2 = 10”

theta1 = 30 deg.

theta2 = 45 deg.

Pw(x, y)

x

y

9. Examine the robot figure below and,

a) assign frames to the appropriate joints.

x

y

z

AB

C

D

E

10”

AB rotates 20rad/s c.c.w. in the xy plane, there are ball joints at B and C, and the collar at D slides along the prismatic shaft. What are the positions, velocities and accelerations of the links?

6”40”

3”

x

y

z

x

y

z

θ1

L1

l2

l3

L4

b) list the transformations for the forward kinematics.

c) expand the transformations to matrices (do not multiply).

x

y

z

x

y

zANS.

x

y

z

x

y

z

x

y

z

F0

F1

F2

F3

FT

ans.

T1 2, trans l2 0 0, ,( )rot z 90°,( )=

T0 1, trans 0 L1 0, ,( )=

T2 3, trans l3 0 0, ,( )rot z 90°– θ1+,( )=

T3 T, trans L4 0 0, ,( )rot z 90°,( )rot x 90°,( )=

10. Given the transformation matrix below for a polar robot,

a) find the Jacobian matrix.

ans.

T1 2,

1 0 0 l2

0 1 0 0

0 0 1 0

0 0 0 1

90°cos 90°sin 0 0

90°sin– 90°cos 0 0

0 0 1 0

0 0 0 1

=

T0 1,

1 0 0 0

0 1 0 L1

0 0 1 0

0 0 0 1

=

T2 3,

1 0 0 l3

0 1 0 0

0 0 1 0

0 0 0 1

90°– θ1+( )cos 90°– θ1+( )sin 0 0

90°– θ1+( )sin– 90°– θ1+( )cos 0 0

0 0 1 0

0 0 0 1

=

T3 T,

1 0 0 L4

0 1 0 0

0 0 1 0

0 0 0 1

90°cos 90°sin 0 0

90°sin– 90°cos 0 0

0 0 1 0

0 0 0 1

1 0 0 0

0 90°cos 90°sin 0

0 90°sin– 90°cos 0

0 0 0 1

=

T0 T,

θ( )cos θ( )sin 0 r θ( )cos

θ( )sin– θ( )cos 0 r θ( )sin

0 0 1 0

0 0 0 1

=

ans.ddt-----x

ddt-----y

r∂∂x

θ∂∂x

r∂∂y

θ∂∂y

ddt-----r

ddt-----θ

θ( )cos r– θ( )sin

θ( )sin r θ( )cos

ddt-----r

ddt-----θ

= =

b) Given the joint positions, find the forward and inverse Jacobian matrices.

c) If we are at the position below, and want to move the tool at the given speed, what joint veloci-ties are required?

11. Examine the robot figure below and,a) assign frames to the appropriate joints.

θ 30°= r 3in=

ans.

J1– 0.866 0.5

0.167– 0.289=

J 30°( )cos 3– 30°( )sin

30°( )sin 3 30°( )cos

0.866 1.5–

0.5 2.598= =

ddt-----x 1in

s-----–= d

dt-----y 2in

s-----=

ans.ddt-----r

ddt-----θ

0.866 0.5

0.167– 0.289

1–

2

0.134

0.745= =

b) list the transformations for the forward kinematics.c) expand the transformations to matrices (do not multiply).

12. Given the transformation matrix below for a polar robot,

a) find the Jacobian matrix.b) Given the joint positions, find the forward and inverse Jacobian matrices.

c) If we are at the position below, and want to move the tool at the given speed, what joint veloci-ties are required?

13. Find the forward kinematics for the robots below using homogeneous and Denavit-Hartenberg matrices.

x0

y0

z0

x0

y0z0

yT

xT

zT

yT

xT

zT

r

θ2

θ1

c b

a

T0 T,



0 0 1 0

0 0 0 1

=

θ1 30°= θ1 40°=

ddt-----x 1in

s-----–= d

dt-----y 2in

s-----=

14. Use the equations below to find the inverse Jacobian. Use the inverse Jacobian to find the joint velocities required at t=0.5s.

x

y

x

y

x

y

x

y

x

y

x 4 θ1( )cos 6 θ1 θ2+( ) in.cos+=

y 4 θ1( )sin 6 θ1 θ2+( ) in.sin+=

ANS.

P 0.5( ) 3

52t

3– 3t

2+( ) 5

2+ 5.5

6= =

First, find tool and joint positions,

r 5.52

62

+= α 65.5-------

atan=

r2

42

62

2 4( ) 6( ) 180 θ· 2–( )cos–+= θ∴ ·2 180

r2

42

62

+( )–2 4( ) 6( )–

--------------------------------- acos–=

θ1 α–( )sin

6----------------------------

180 θ2–( )sin

r---------------------------------= θ∴ ·

16 180 θ2–( )sin

r------------------------------------

asin α+=

Next, the Jacobian,

J4 θ1( )sin– 6 θ1 θ2+( )sin– 6 θ1 θ2+( )sin–

4 θ1( )cos 6 θ1 θ2+( )cos+ 6 θ1 θ2+( )cos=

Substitute and solve

ddt-----

θ1

θ2

J1– 7.5

3=

8. MECHANICAL COMPONENTS

• There are a variety of mechanical components that we will include in designs. These include,- gears- cams- screws/bolts

9. CAM DESIGN

• Cams are basically shaped surfaces that are typically not round. The cam is rotated or translated, and a follower (possibly a small wheel) is displaced as it moves over the surface.

• Cams can generate complex motion profiles in a compact area.

• Engine values are a well known application of cams. They can open and close the cylinders with a large force, but will also dwell in open or closed positions.

9.1 CAM TYPES

• Typical cams are pictured below,

Plate (or Disk, or Radial) Cam

Wedge/Translating Cam

• Some types of reciprocating followers include,

End (or Face) Cam

Cylindrical (or Barrel) Cam

Roller Knife Edge flat-face

• Some oscillating followers include,

9.2 CAM MOTION

• We can often describe a cam by drawing the displacement profile on a graph.

• Consider possible displacement curves for,

roller curved shoe

Rise Dwell Return Dwell

lift

displacement

camangle

0 360°

f θ( )

θ

• The curve above can be broken into sections and described with a mathematical function,

• Of these two functions, the parabolic will allow a greater level of control, but harmonic motion will permit smooth transitions between motions.

• Some of the general design rules,1. fulfil the basic motion requirements. (cam profiles are not exact and decisions are

required.)2. The displacement, velocity and acceleration curves must be continuous, but the jerk

- wave machine

- stamping press

- washing machine

y f θ( )=

f θ( ) Aθ2Bθ C+ += Parabolic motion

f θ( ) A B Cθcos+= Harmonic motion

f θ( ) A= Constant/dwell motion

must not be infinite. This means that the positions and first and second derivatives must be equal at the segment ends.

3. Minimize the velocities and accelerations.

• The first step in developing a cam is to develop a motion profile. Consider example 5-2 from Shigley and Uicker,

We want a cam driving a reciprocating follower. The cam is rotated with an angular veloc-ity of 150 rev/min. The follower should start in a dwell and accelerate to a constant velocity of 25 in./sec. for a rise of 1.25 in. The follower should come to a rest after moving a total height of 3 in. The follower should then drop back, and dwell for 0.10 sec.

First we will draw a function that has the basic components of the motion. In total there is the initial acceleration from a dwell (parabolic segment) to a constant velocity (straight line segment). After the constant velocity there are two transitions back to a final dwell at the starting height (2 parabolic segments). This evens out to a dwell (straight line segment). The displacement magnitudes were calculated and labeled on the graph

θ

f θ( )

accel. 25 in/s decel. reverse dwellβ0 β1 β2 β3 β4

360° β0 β1 β2 β3 β4+ + + +=

β4 0.10s( ) 150 revmin---------

1min60s

------------- 360°

1rev-----------

90°= =

β11.25in

25ins-----

---------------

150 revmin---------

1min60s

------------- 360°

1rev-----------

45°= =

0.0

3.0

0.875

2.125

225°∴ β 0 β2 β3+ +=

Next, determine the obvious angles,

L0

L1

L2 f2 θ( )f3 θ( )

f4 θ( )f0 θ( )f1 θ( )

C∴ 0=f0 0( ) 0 Aθ2Bθ C+ += =

f0' 0( ) 0 2Aθ B+= =

f0 β0( ) L0 Aθ2= =

First, consider the first acceleration segment, and the maximums,

B∴ 0=

A∴L0

β02

-----=

second, consider the first decceleration segment (flipped left/right), and the maximums,

f0' β0( ) 2Aθ 2L0

β02

-----β0 2L0

β0-----= = =

f0'' β0( ) 2A 2L0

β02

-----= =

C∴ 0=f2 0( ) 0 Aθ2Bθ C+ += =

f2' 0( ) 0 2Aθ B+= =

f2 β2( ) L2 Aθ2= =

B∴ 0=

A∴L2

β22

-----=

f2' β2( ) 2Aθ 2L2

β22

-----β2 2L2

β2-----= = =

f2'' β2( ) 2A 2L2

β22

-----= =

finally, consider the return to the dwell (half the way), and the maximums,

C∴ 0=f3 0( ) 0 Aθ2Bθ C+ += =

f3' 0( ) 0 2Aθ B+= =

f3β3

2-----

L0 L1 L2+ + Aθ2= =

B∴ 0=

A∴4 L0 L1 L2+ +( )

β32

--------------------------------------=

f3'β3

2-----

2Aθ4 L0 L1 L2+ +( )

β3--------------------------------------= =

f3''β3

2-----

2A8 L0 L1 L2+ +( )

β32

--------------------------------------= =

• Now,

Now, we need to balance the velocities and accelerations. This can be done by setting all of the maximum velocities and accelerations equal,

Velocities,

2L0

β0----- 2

L2

β2-----

4 L0 L1 L2+ +( )β3

--------------------------------------= =

Accelerations,

1.75

β02

----------∴ 1.75

β22

---------- 24

β32

------= =

1.75β0

---------- 1.75β2

---------- 12β3------= =∴

2L0

β02

----- 2L2

β22

-----8 L0 L1 L2+ +( )

β32

--------------------------------------= =

At this point we can use either velocities or accelerations to find the times for each segment. I arbitrarily decide to use velocities,

β1 β0= β312

1.75----------β

0=

225° β0 β2 β3+ + β0 β012

1.75----------β

0+ + 8.857β0= = =

β∴ 0 25.4°=

β∴ 2 25.4°=

β∴ 3 174.2°=

• You may have recognized that the previous design assumed that the follower must have a point contact with the curve.

• In actual practice we will have surfaces that are in contact, the surfaces can be identified using the equations developed previously.

• Consider the flat-face follower.

Convert the cam profile in the previous problem to a circular cam profile using a knife edge follower.

• We can develop the a modified cam profile based on the flat faced follower. (Note: the proof is done as if a milling cutter is used, but this turns out to be more a matter of convenience)

Note the follower is offset, this will not change the operation, but can be used to reduce moments in the shaft for a clock-wise rotation.

90° dwell45° constantvelocity

We can use the following relationships to plot the cam profile, based on the motion profile. The first relationship will avoid undercutting - this is a case where the cam gets ‘stuck’ on a second peak.

R0 rmin f''min θ( )– f θ( )–>

R0

rmin a minimum allowed curvature for the cam=

The face width must be wider than,

face width f'max θ( ) f'min θ( )–>

• Using the derivation of the basic relationships, we can now develop a method to plot out a com-plete cam profile.

γ

R

c

rmill

φ

L f θ( ) R0+=

a

dL dθ

dθ

φ

R

LR--- φcos=

aL--- φtan= φ∴ a

L---

atan=

θθ

Fixedpositionon cam

R∴ Lφcos

------------=

• Now, develop a cam for example 3-52 from Shigley and Uicker,

1. Pick an angle of 0 for the first iteration, increment this in subsequent calculations.

2. Calculate,

3. Plot the point, pick a subsequent point and then do a new calculation for the new angle.

θi 0°…360°[ ]=

L f φ( ) R0+=

a ddφ------f φ( )=

x R θ φ+( )cos= y R θ φ+( )sin=

φ aL---

atan=

RL

φcos------------=

• Keep in mind that when designing cam-follower pairs that the radius of the follower is not zero. Therefore it may be necessary to compensate for this during the design.

• Consider the effect of a round follower on a wedge cam.

First, determine the minimum radius of the cam, if the curvature of the cam is to be greater than 0.5” at all points, and the face width.

Develop the equations for the geometry of the cam using the profile calculated earlier.

• Consider the effect of a round follower on a radial cam.

• Other arrangements are possible, and some proofs are provided in the text.

NOTE: The point of contact remains tangential, angle of the cam suggests a different point of contact. In this case the upper position of the cam has a small offset across the surface of the cam.

We can use the following relationships to plot the cam profile, based on the motion profile.

ψ mL----

atan=

x F θ ψ+( )cos rfollower φcos–=

F L2

m2

+=

φ α ψ–=

1. Pick a displacement angle

θi 0°…360°[ ]=

2. Calculate

3. Plot the point, and select the subsequent point.

m

rfollower

L f θ( ) rfollower rmin+ +=

γ

θ

x F θ ψ+( )sin rfollower φsin–=

a ddθ------f θ( )=

α La

F2

ma–-------------------

atan=

9.3 USING CAMS AS JOINTS IN MECHANISMS

• We can use cams to give complex joint motion,


1. You were recently hired as a Fuel Containment and Monitoring specialist for Generous Motors. Your first job is to design a mechanical gauge for an instrument panel. The tank holds up to 10 gallons of fuel. It has been determined that the needle on the gauge should remain steady at the full ‘F’ mark while the tank contains 8 to 10 gallons. When the tank has less than 3 gallons the gauge should read empty ‘E’. The last design was a failure ‘F’, and your boss fired the engineer responsible. It seems that he his design did not follow good cam design rules - the velocity and accelerations were not minimized - and so the gauge would wear out, and jam prematurely. Design a new cam to relate the float in the tank to the gauge on the instrument panel.

2. The motion profile curve below has 4 segments. Segments A and C are based on polynomials. Segment D is based on a harmonic/cosine function. Segment B is a constant velocity segment.

a) Write the equation for curve segment B.b) What effect does the follower shape have when converting the motion profile to a cam profile. Draw a figure to illustrate this with a round follower.c) Write the coefficients for the curve segment C.

E Fgasoline

float

linear cam

follower

gauge(old design)

tank

3”10”

needle

θ

3 6.242210

1

0.5

0.3

0

y

A

B

C

D

9.5 REFERENCES



yC ACθ3 BCθ2CCθ DC+ + +=

AC =

BC =

CC =

DC =

10. GEARS

• When forces become large we cannot count on friction for rolling contact (no slip). Gears use metal teeth that are meshed together to transmit motion between moving components.

10.1 SPUR GEARS

• Spur gears are in very wide use throughout engineering.

• These gears are flat, and either circular or straight (a rack).

• The figure below shows a typical gear with common terms marked,

egr352a0.jpg

addendum

dedendum

addendum circle

pitch circle

dedendum circle

tooth

width

circular

clearance

thickness

pitch

of space

• When gears are properly mated their pitch circles will be tangent. And the faces of the teeth will touch along the addendum and dedendum surfaces, down to the clearance circles.

• Some terms of use when discussing gears,backlash - the difference is the gap between gear teeth where they mesh. This leads to

‘play’ in the gears.pinion - a smaller gearwheel - a larger gear

• diametral pitch is defined by,

• module is defined by,

• The following relationships are also applicable,

• The ratio between angular velocities of two gears can be determined with the law of gearing,

PNd----=

where,N = number of teethd = pitch diameter, in.P = diametral pitch (teeth/in.)

mdN----=

where,m = module (mm)d = pitch diameter (mm)

pπdN------ πm

πP---= = =

where,p = circular pitch

• As seen above the law of gearing assumes that the pitch point is found at a constant radius. If this were to move the driven gear would accelerate /decelerate as the teeth mesh and separate.

• To keep the gears meshing constantly an involute profile is typically used for the shape of the teeth.

• To construct an involute profile, we need to construct a line that is tangential to both gears. The teeth on both gear will be constructed to contact only on this line.

• The involute profiles for a single tooth will trace out a line as shown below (later we will develop an equation for the point on the unwrapping string).

ωi

ωj-----

rj

ri---=

where,ωi ωj, the angular velocities of gears i and j=

ri rj, the pitch radii of two gears i and j=

base cylinders

pitchpoint

gear i

gear j

tooth on i

tooth on j

• The pressure angle is shown below,

• Standards geometries for spur gears include, (based on American Gear Manufacturers Associa-tion and ANSI standards)

To do this we start with a cord wrapped about the base cylinder of the gear. We pick a point on the cyl-inder where the tooth is to start, and mark the point on the cord. As we then unwrap the cord the point will trace out the involute profile of the tooth.

pressureangle

• Typical diametral pitches and modules include, (based on American Gear Manufacturers Asso-ciation and ANSI standards)

10.1.1 Involute Profiles

• The figure below shows the triangulated layout for the basic involute function,

Teeth type

Full depth

Stub

Pressure angle

20 (deg)22.525

20

Addendum ‘a’

1/P1/P1/P

0.8/P

Dedendum ‘b’

1.25/P, 1.35/P1.25/P, 1.35/P1.25/P, 1.35/P

1/P

CoarseP = 2, 2.25, 2.5, 3, 4, 6, 8, 10, 12, 16

FineP = 20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200

Preferredm = 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25

Less Preferredm = 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36

OR

• Next the involute curve is applied to the generation of gear teeth.

r

involute curve

x

y

r

α invϕ=ϕ

ρ

We define the involute curve such that,

ρ r α ϕ+( )=

l

ρ r ϕtan=

Therefore,

ρ r α ϕ+( ) r ϕtan= =

α∴ ϕtan ϕ–=

invϕ∴ ϕtan ϕ–=

ϕinvϕ

b/2

ρ

x

y

d/2

pitch circle

toothprofile

θr

θb

r

base circleθp

φ

• To generate points on these curves we must select values of XXXX and calculate (x, y) posi-tions. These will be correct for one face of the tooth. If these are used to generate a splined curve, or graphed, they will form the tooth profile. The upper and lower bounds are determined by the addendum and dedendum.

10.1.2 Design of Gears

• The basic steps to design a gear are outlines below,

x r θr( )cos=

θb θp invφ+π

2N------- invφ+= =

For the gear we must consider the pitch,

rd2---= ϕ φ=when

where,φ the pressure angle=

d diameter of pitch circle=

Therefore,

b diameter of given base circle=

b2--- r φ( )cos=

y r θr( )sin=

p πdN------

2 2rθp( ) 2 dθp( )= = = θp∴ π2N-------=

θrπ2--- θb invϕ–+

π2--- π

2N------- invφ invϕ–+ + π

2--- 1

1N----+

invφ invϕ–+= = =

ϕ 0deg≥

1. Calculate the desired pitch diameters for both gears, i and j. Note: gear i should be larger than gear j for the formulas given.

di

Ni

P-----= dj

Nj

P-----=

where,di dj, pitch diameters for meshing gears i and j=

P diametral pitch selected=

Ni Nj, given number of teeth on gears i and j=

2. Find the base circles of the gears by calculating the shortest perpendicular from the pressure line to the center. This will be the diameters of the base circles.

bi

bj

bi di φcos=

bj dj φcos=

where,

bi bj, diameters of base circles=

φ pressure angle=

di

dj

φ

3. Using the base circle calculate the gear tooth profile for both gears using the involute equations described before.

4. Import the geometry into the cad system as a line. convert this to a complete tooth (this may involve a sweep, or mirror and join depending on the CAD software). Trim-ming, fillets, etc will be done later.

5. Calculate the addendum and dedendum radii for both gears - using the standards given earlier. Create dedendum circles for each gear body with the diameters of the pitch diameter minus the dedendum. (In one case this will be below the base, in the other this will be above). In the final application the centers between these gears will be d1+d2 apart.

ai aj1P---= = ci cj

1.25P

----------= =

Bi di 2ci–= Bj dj 2cj–=

where,

ai aj, addendums for both gears=

ci cj, dedendums for both gears=

Bi Bj, dedendum diameter - size of circle for the gear bodies=

Ai di 2ai+= Aj dj 2aj+=

Ai Aj, Addendum circles for gears=

6. Calculate the angular spacing of the teeth using the number of teeth. Align the teeth at the right distance from the center of rotation. The addendum, dedendum and pitch circles must cross the tooth at the right location. Then use the cad system to copy the teeth about the center of the gear. If required do a join step and/or cut the base circle for the gear.

∆θi360°

Ni-----------= ∆θj

360°Nj

-----------=

where,

∆θi ∆θj, angles between gear teeth on gear i and j=

• If we are dealing with a rack, it is effectively a circular gear with an infinite radius.

• When we have internal gears (one gear inside another) we need to adjust the methods to reflect that both gears are on the same side of the pitch line.

10.1.3 Design Issues

• During motion the gear teeth undergo a combination of sliding and rolling. The direction of slid-ing reverses at the pitch point, where the motion is pure rolling.

10.1.3.1 - Undercutting and Contact Ratios

• Undercutting occurs on some gears. This is a gouging of teeth that occurs when teeth contact below the base circle of the gear during motion.

• During manufacturing some processes (generation) can remove the excess material that lead to undercutting. But this can reduce the base width of the teeth and weakening of the gear.

• Undercutting problems can be reduced by increasing the radius of the gear, and increasing the number of teeth.

• The gear teeth are in contact along the pressure line between the points where it intersects the addendum lines.

7. Add other features as required, such as holes, keyways, holes, spokes, etc. Note: fil-lets are needed at the base of the teeth to prevent stress concentrations, and small rounds are needed at the top of the teeth to reduce wear. The maximum radii of the fillets can be determined using the clearance.

rifilletci aj–<

rjfilletcj ai–<

where,

rjfilletrifillet

, maximum fillet radii at the bottom of the gear teeth=

• The contact ratio is as defined below. A value of 1 means that at any time only one tooth is engaged. A value greater than 1 means that only one tooth is engaged. A value of 2 would mean that at any time 2 teeth are engage. A value less than 1 means that at times the teeth are not in contact (bad).

• Undercutting will not occur during production of the profiles if the following addendum values

pressure line

pitch point ‘P’addendum

circle

dedendum+clearan

motion

addendum circle

pitch circle

qa

qr

qa arc of approach=

qr arc of recess=

A P

B

Note: my artistry is not so good. Here the dis-tances between the circle should be equal.

mc

qt

p----

ua ur+

p φcos----------------= =

where,

mc the contact ratio=

ua approach contact path length (from A to P)=

ur recess contact path length (from P to B)=

ua dj aj+( )2bj

2– dj φsin–=

ur di ai+( )2bi

2– di φsin–=

observed,

10.1.3.2 - Changing the Center Distance

• If the center distance between gears is changed, then the pitch circles on both gears will move away from the center.

• The result of the enlarging of pitch circles will be a reduction in the contact ratio. This will lead to a smaller contact ratio.

• This condition also allows some play in the gears (backlash). This play means that a reversing of direction can lead to a small reversing rotation before the other tooth is impacted. This leads to errors and the impact forces can shorten the life of gears significantly.


10.2 HELICAL GEARS

• Helical gears are essentially spur gears, but with a bit of a twist in the normally straight profile of the teeth.

egr352a1.jpg

• The helical arrangement means that the teeth engage at one point (as opposed to a line of con-tact), and then slowly mesh along the face of the tooth. This also means that the number of engaged teeth (contact ratio) can be higher.

• If we contrast spur and helical gears. The spur gear teeth contact fully all at once.

• These gears mesh very smoothly, so they find application for,- high speeds- heavy loads

aj bj2

di dj+( )2 φsin( )2+ dj–<

ai bi2

di dj+( )2 φsin( )2+ di–<

gear noise must be reduced- center lines of shafts do not intersect

• These gears,- require more effort during fabrication- need additional bearings to resist axial thrust.

• helical gears can be used to transmit torques between parallel, and non-parallel (often perpendic-ular) shafts.

• To eliminate the need for an axial thrust bearing we can use a herringbone (double helical) gear. This is effectively two helical gears with opposite twists on the helix, and joined down a center line.

10.2.1 Design of Helical Gears

• We can describe the helix with a single angle,

• This helix allows three different ways to measure the pitch and pressure angles.

ψ

ψ helix angle=

• We can now develop an equivalent pitch radius,

pn

px

pt

pn normal diametral pitch=

pt tranverse diametral pitch=

px axial pitch=

pt

pn

ψcos------------- px ψcos= =

φn

φt

φn normal pressure angle=

φt tranverse pressure angle=

ψcosφntan

φttan-------------=

• Typical design parameters include,- produced for specialized applications, and custom designed- typical pressure angle is 20°- for helix angles from 0 to 30° use the normal diametral pitch to calculate tooth propor-

tions- helix angles greater than 45° are not recommended- in mating parallel gears - one must have a left hand helix, and the other must have a right

hand helix. Use the sign of the helix angle to indicate left or right handed.

• The line of contact in spur gears is straight across the teeth. In helical gears, the line of contact is diagonal.

• To measure the contact ratio we need to use three values to be effective,

ded

ψcos( )2--------------------=

where,

d the pitch diameter of the helical gear=de the diameter of an equivalent spur gear=

Ne dePnN

ψcos( )3--------------------= =

where,

N number of teeth on the helical gear=

Pn diametral pitch of gear=

Ne equivalent number of teeth on a spur gear=

10.2.2 Perpendicular Helical Gears

• If two shafts intersect at an angle (typically 90°) we can link them using helical gears. The angle between the shafts can be,

• The example below shows one configuration for helical gears, including the location of thrust bearings.

ψb ψtan φcos( )atan=

mn

mt

ψbcos( )2----------------------=

mxFpx----- F ψtan

pt----------------= =

where,

ψb base helix angle=

mt transverse contact angle=

mn normal contact ratio (as found for spur gears)=mx axial contact ratio=

F face width=

mT total contact ratio=

mT mn mx+=

Aside: Recall, a contact ratio below 1 is unacceptable, and the gears will come out of contact. For helical gears this ratio should be greater than 2.

Σ ψi ψj±=

where,

ψi ψj, helix angle on gears i and j=

Σ angle between shafts=

• When the shaft intersection angle is large (90°) you may use same handed gears to intersect. In the example above the gears are both right handed.

• The pitch diameter for these gears can be found using,

• A minimum contact ratio of 2 is recommended for these gears.

• Typical parameters for cross-axis helical gears are given below,


Driver Helical Gears

Thrust Bearings

dN

Pn ψcos-------------------=

Helix Angle

45°60°75°86°

Driver

Minimum# teeth

20941

Helix angle

45°30°15°4°

Driven

Pressure angle

14.5°17.5°19.5°20°

10.3 BEVEL GEARS

• These gears are like normal spur gears, except that they have a conical form.egr352a2.jpg

• Their applications are characterized by,- to couple shafts with intersecting axes-

• Bevel gears are meshed so that the points of their cones are coincident.

• As we move towards the point of the cones, the number of teeth remains the same, but the diam-eter reduces towards zero. This changes the size of the teeth, and the pitch diameter.

• The form of the gears is like that of spur gears, but each has a cone angle, and when added together this gives the angle between the shafts.

• We can apply some of the basic ratios to bevel gears,

γi γj, pitch angles=

γi

γj

Σ γi γj+= Σ shaft angle=

di

dj

P

O

10.3.1 Design of Bevel Gears

• To determine the pitch angles for the gears we can write the following expressions,

• An approximate methods for creating bevel gears is called ‘Tredgold’s approximation’

• Tredgold’s technique requires that a cone on the bottom of the bear be found. This cone is then

ωi

ωj-----

dj

di----

Nj

Ni-----= =

where,

ωi ωj, input/output angular velocities=

di dj, pitch diameters of gears (normally taken at wide base)=

Ni Nj, number of teeth on gears=

OPdi

2 γisin---------------

dj

2 γjsin---------------= =

where,

OP the distance from the mesh point to the tips of the cones=

γisindi

dj---- γjsin

di

dj---- Σ γi–( )sin

di

dj---- Σ γisinsin γi Σsinsin–( )= = =

by trigonometry,

γi∴ Σsindj

di---- Σcos+

-----------------------atan ΣsinNj

Ni----- Σcos+

------------------------atan= =

γjsindj

di---- γisin

dj

di---- Σ γji–( )sin

dj

di---- Σ γjsinsin γj Σsinsin–( )= = =

γj∴ ΣsinNi

Nj----- Σcos+

------------------------atan=

flattened out, and normal gear design is done. Finally the cone is mapped back onto the bottom of the gear. The profiles are then projected up to the point of the cone.

• Typical design parameters include,- 20° pressure angle for straight bevel gears- bevel gears are always custom made, and are not interchanged- deflections mean that the wider base tends to take most of the load, so the teeth are

designed with a short length (commonly less tan 1/3 of the total cone length)-

10.3.2 Other Bevelled Gears

• Crown and Face Gear - this gear is much like a rack for spur gears. To get this, one of the gears is given a pitch angle of 90°.

• Spiral Bevel Gears - to reduce noise in beveled gears, a spiral can be added to the teeth.

• Hypoid gears - the centers of the bevelled gears are not coincident - the shaft is offset.


10.4 WORM GEARS

• Worm gears use a long helical screw that drives a larger helical gear.egr352a3.jpg

dei

di

γicos------------= dej

dj

γjcos------------=

Nei

πdei

pi---------= Nej

πdej

pj---------=

where,

deidej

, equivalent diameter for spur gear design=

NeiNej

, equivalent number of teeth for spur gear design=

• Basically we use a screw like gear (the worm) and a large cylindrical gear (worm gear) that is driven by the worm.

• The worm gear curves to the shape of the worm to increase contact. Also note that worm gear is a helical gear.

• The worm acts very much like a rack, except that it is threaded onto a cylindrical surface.

worm

worm gear

pitch

centerdistance

circle

dg

dg pitch diameter=

dg

Ngp

π---------=

where,

Ng number of worm gear teeth=

• These gears find their best applications when a large gear ratio is needed in a compact space. The shafts typically intersect at a 90° angle - when this is the case the helix angle on both gears is the same.

• The worm gear can be single enveloping or double enveloping.- The single envelope is the parallel sided gear. This type of gear is more forgiving

for position and alignment tolerance problems.- The double enveloping gear has a curvature that increases the surface on contact

between the gears. This can be very useful for power applications.

• The following values are reasonable for finding the profile dimensions of the teeth,

ψλ

λ lead angle=

where,

ψ helix angle=

p

p = axial pitchdw pitch diameter=

dw

dw

dw dg+

2------------------

0.875

2.2----------------------------------=

This relationship is suggested by the AGMA for a good power capacity.

λ 90° ψ–pNw

πdw----------

atan= =

Nw number of teeth (threads) on the worm=

addendum 0.3183p=dedendum 0.3183p=clearance 0.050p=

• Suggested pressure angles for given lead angles are listed below,

10.4.1 Harmonic Drives

• These are actually normal servo motors, but with an integrated harmonic gear. The harmonic gears are very compact, and as a result the overall size is reduced. These gears also allow very high gear ratios (eg, 100:1)

• Harmonic drives are also gaining popularity with smaller manipulators. They use a rotating elliptical core that deforms a flexible section. The flexible section is in contact with an outer section for short periods of time, and as the ellipse rotates, there is a geared down rotation gen-erated. This allows integral gears in motors


λ φ

0-16°16-25°25-35°35-45°

14.5°20°25°30°

Note that as the inner elliptical spline rotates, the flexible spline counter-rotates. The surface between the wave generator, and flexible spline is smooth, and the surfacer between the flexible spline, and the outer spline is geared.

10.5 REFERENCES



11. DESIGN OF MECHANISMS

• When we design gears, cams and mechanisms we are free to set and vary parameters. But, above this we often must select these components to start with.

• The selection of components can be aided by using techniques such as,- schematics- ratios- common approaches

11.1 SIMPLE GEAR TRAINS

• When we want to increase/reduce angular displacements/velocities/etc. we can use simple gear trains.

• As we have seen before, gears typically have an input to output ratio. The relationship below is for simple gear trains - only one gear on each axis.

• We can deal with compound gear trains (multiple gears on each axis) by using product of driven

eωn

ω2------

Ni

Ni 1+------------

i 2=

n 1–

∏di

di 1+-----------

i 2=

n 1–

∏= = =

where,

ωi The angular velocity from the first gear 2 to the last gear n=

Ni The number of teeth on gear i=

di The pitch diameter of gear i=

e The speed ratio of a gear train (can be negative)=

and driving teeth.

11.1.1 Examples - Fixed Axis Gears

• A simple gear train has only one gear on each axis.

• A compound gear train has multiple gears on the same axis. Consider the truck transmission example from Shigley and Uicker,

eωn

ω2------

Ni

i 1=

m

∏

Nj

j 1=

n

∏---------------= =

where,

Ni The number of driving teeth on gear i of m driving gears=

Nj The number of driving teeth on gear j of n driving gears=

• We may also consider an in-line gear train. These can be used for items such divide by twelve and sixty in clocks,

Motor shaft

clutch

reverse idler

2

3 4

56

7

8

9

1011

stem gear

Speed (gear)

1234reverse

Gear Train

2-3-6-92-3-5-82-3-4-7bypass gear train2-3-6-10-11-9

In this manual transmission the gear shifter will move the gears in and out of con-tact. At this point all of the needed gears will be meshed and turning. The final step is to engage the last gear in the gear train with the clutch (plate) and this couples the gears to the wheels.

• Quite often we will have a particular speed ratio in mind. We can convert this to teeth numbers by finding a suitable fractional value,

• The gear train in the previous example might look something like,

N2 60=

N3 15=

N4 60=

N5 15= eN2N4

N3N5------------- 16= =

Input shaft Output shaftIn this case the output shaft

turns 16 times faster than the input shaft. If we reversed directions the output (former input) would now turn 1/16 of the input (former out-put) shaft speed.

e 0.1561561000------------ 78

500--------- 39

250---------= = = =

assume we are given a value of e=0.156, we can begin by putting this in fraction form with the lowest integer values,

In this case the ratio is very high. We could decide to try to make a gear, or we could split the numerator and denominator into multiples. We can then put the multiples into smaller fractions. In the case below there are no common numerators and denominators so all of the gears will need to be compound.

e39250--------- 3( ) 13( )

10( ) 25( )---------------------- 3

10------

1325------

= = =

Looking at the numerators and denominators there are a few integers that are small. We should set a minimum number of teeth for practical design purposes. In this case we can set the minimum at 20.

e 310------

1325------

2170------

2650------

2150------

2670------

= = =

• Try the design below,

N2 21=

N3 50=

N4 26=

N5 70=

Output shaftInput shaft

Design a gear train for the value e=-0.2. Next design for e=0.2.

11.1.2 Examples - Moving Axis Gears

• In some cases gears move relative to each other. This can be used to generate some interesting alternatives.

11.1.2.1 - Epicyclic Gear Trains

• Epicyclic gears have many applications, such as automatic transmissions in automobiles.

• In these trains the gears typically orbit each other.

• Consider the basic epicyclic gear train,

• We can represent these gears using a notation developed by Levai. Consider the basic epicyclic gear,

Sun gear

Planet Gear

Planet

2

3

4

CarrierOR KINEMATICEQUIVALENT

• When designing with these gears, we can consider different control modes possible. In the case above we could connect the gear ‘4’ to a ring gear (internal gear) and make a simple multi-speed transmission.

• Some of the other possible gear train types include variations on the number of planets.

• Consider the compact planetary gear shown below,

4

3

2

gear 2 and 3 move relative to ground. This is different that ‘4’ because it is not directly connected to ground.

teeth

collar

4

3

2

If we ground ‘3’ then we get a speed ratio between ‘2’ (clutch stem gear) and ‘5’ (the output shaft) of,

5

eN2

N4------–

N4

N5------

N2

N5------–= =

If we connect ‘3’ to ‘5’ then we get a speed ratio of,

e 1=

• We can also construct an epicyclic gear using bevel gears. This is called Humpage’s reduction gear.


double

sun gear

planet carrier

fixed gear

planet gear

input shaft

output shaft

double planetbevel gear

planet carrier

fixed gear

output gearand shaft

11.1.2.2 - Differentials

• Differential mechanisms allow us to effectively do subtraction or averaging.

• If we want to determine the difference between two linear motions we could a mechanism like the one shown below,

2

3

4

5

1 Draw the Levai representation for the epicyclic gear to the left. Assuming that the num-ber of teeth are 20, 30, 15, 50 from gear 2 through to gear 5, find the speed ratio between the input at 2 and the output at 3. What would the speed ratio be if 4 was the output?

• An angular differential is shown below using bevel gears,

************** include image here

• In one type of automotive differential the housing above is driven, and both the output shafts turn. This allows a small difference in wheel speed. Without this simple action like turning cor-ners would exert large forces on the tires and drive train. This is also why one of the drive wheels can spin while the other is fixed.

• Various vehicles can disengage the differential for offroad conditions (where tires can slip),

aa

∆x1

∆x2

∆x12 2∆x1 ∆x2–=

θ2 θ4

θ6

θ5

θ5N2

N5------θ

1

N4

N5------θ

2–=

5

2

3

4

6

θ6 θ2 θ4+=

while others have mechanisms to balance torque to the wheels when an excessive difference in speed is detected.

• Worm gears have also been used in automotive differentials.

11.2 LINKAGES

• We have seen analysis techniques for linkages before, but now we will consider design of mech-anisms to take advantage of the basic properties.


1. We have been asked to design a coarse pitch (P=3) gear pair with stub teeth and a speed ratio of e=0.24. Find the radii and dimensions needed to draw the gears. Find at least one point on a gear tooth profile. Roughly sketch the gears showing the dimensions calculated.

11.4 REFERENCES



12. STATIC ANALYSIS OF GEARS

• As normal, we should do a static analysis of our machines as a prelude to analysis of suitability, and strength of components.

12.1 INTRODUCTION

12.2 ANALYSIS OF GEARS

• The involute profile of the gear means that the force applied at the gear teeth is not tangent to the pitch line, but actually tangent to the base circle. At the pitch point the force between the teeth acts at the pressure angle.

• For spur gears the following values and equations can be used for the applied forces. Clearly there would be a reaction force that is not shown here. (Note: this applies even if multiple teeth are in contact).

• If we are considering helical gears the helical spiral of the teeth adds a second angle to the con-tact force. This means that the contact force requires 3D analysis.

• Straight bevel gears are not so easy to calculate because the force is applied at a variable dis-tance from the center of the rotational axis.

pitchT

F

φ

d

circle

Ft F φcos=

Fn F φsin=

T d2---F φcos=

Given the gear geometry,

φt transverse pressure angle=

ψ helix angle=

we can calculate,

Ft F ψ φtcoscos=

Fn F ψ φtsincos=

Fa F ψ φtcossin=

T Ftd2--- d

2---F ψ φtcoscos= = F∴ 2T

d ψ φtcoscos------------------------------=

where,

Fa the axial component of the force (need axial bearings)=

• In these gears we compromise by assuming the force is applied at the center of the tooth. In actu-ality this force will be further from the center of the gear.

• We can calculate force components using the relationships below,

• For all of these gears we need to use the calculated forces to design bearings and supports. Most notably the axial thrusts require thrust bearings be included in the design.


12.4 REFERENCES



Tdavg

2----------Ft

davg

2----------F φcos= =

where,

davg the average pitch diameter of the teeth=

Ft F φcos=

γ angle of bevel=

Fr F φ γcossin=

F∴ 2Tdavg φcos----------------------=

Fr the force component normal to the axis of rotation=

Fa F φ γsinsin=

13. MECHANICAL COOKBOOK

• This book should be used to generate ideas.

• When designing you may have a general idea of some function in the final mechanism, try to find a reasonable match for the desired functions here.

13.1 TRANFORMING DEVICES

• Typical inputs and outputs can be characterized using the following terms- motion type (rotation, translation, other)- intermittent/continuous/variable- limited/unlimited range- uni- or bidirectional- axis of rotation/line of action- force/velocity ratios

• The conversions between these can typically convert some input to an output

• Use tables, like those on pgs 105-108 in Erdman and Sandor.


Consider the mechanism in an automobile between the engine and the rear wheels. At the engine there is a torque shaft that turns effectively over a limited range of angular veloci-ties. We need to transmit the torque to the back tires of an automobile. a) Describe the overall inputs and outputs. b) Breaks the differences down into smaller increments. c) Select components from the text that would serve the various functions.

13.2 REFERENCES


14. A MECHANICAL COOKBOOK

14.1 CONNECTORS

14.1.1 Bearings

• The interface between moving parts that should minimize friction and wear.

14.1.1.1 - Plain Bearings

• Generally used in low speed machines.

• The main bearing action comes from the lubricant.

14.1.1.1.1 - Solid Bearings

• This looks like a section of tube that is placed in a hole, and the shaft rotates inside.

• Typical materials are,- bronze- sintered bronze (with graphite)- cast iron

• Made for slowly rotating equipment

• lubrication is required, and problems will arise when not properly maintained.

• Available in standard sizes.

14.1.1.1.2 - Split Bearings

• Used on large machines at low speeds.

• The two halves of the bearings are adjusted in position using shims.

• Typical materials include,- bronze- bronze with babbitt- babbitt lined metal

• Oil grooves are used for lubrication.

14.1.1.1.3 - Thrust Bearings

• Opposses axial thrusts of rotating shafts.

• Uses shoes of a variety of shapes,- flat- kidney shaped

• An oil wedge approach is used to support the bearing.

14.1.1.2 - Rolling Bearings

• Advantages,- low friction at all times- compact- high accuracy- low wear- come in standard sizes

14.1.1.2.1 - Ball Bearings

• Low friction, high speeds, low loads.

• ball bearings are packed between two rotating rings.

• The grooves that contain the ball bearings is given different shapes for different loading condi-tions.

14.1.1.2.2 - Roller Bearings

• For heavy loads at medium or high speeds.

• The various roller bearings are designed for loads (radial and axial) and packing space.

F F

F F

F

F

for radial loads for radial and axial loads

14.1.1.2.3 - Thrust Bearings

• A set of rollers or balls are held between two washers.

• Designed mainly for lower speed axial loads and occasionally light radial loads.

F F F

F

single row cylindricalroller bearing

double row cylindricalroller bearing double row self

aligning roller bearing

F

F

14.1.2 Threads

• One of the classic forms of mechanical connector. Also used to magnify motion and force, and to convert rotation to linear motion.

• The basic temrinology is,

F

cylindrical rollerthrust bearings

F

F

spherical rollerthrust bearings

• Right hand threads are turned clockwise to tighten, left hand threads are turned the other way.

• Threads Per Inch (TPI) are the number of turns of the thread per inch of length.

• There are a number of standard threads, as outlined in the following subsections.

helix angle

ticknessof thread

flank

minor diameter

pitch diameter

major diameter

root

crest

pitch

thread angle

14.1.2.1 - Metric

FORM: P

D

60°

F

2F

RELATIONS:F = 0.125 PD = 0.54127 P (External)D = 0.6134 P (Internal)

NOTATION:

M 10 x 1.5 - 5g 6goutside diameter tolerance

pitch diameter tolerance

pitch

nominal size

metric thread

External Thread Tolerancese - large allowanceg- small allowanceh - no allowance

Internal Thread TolerancesG - small allowanceH - no allowance

EXAMPLE:

M20x2-GH/5g 6g

14.1.2.2 - American National Standard

FORM: P

D

60°

F

RELATIONS:F = 0.125 PD = 0.6495 PP = 1 / N

NOTATION:

UNC-3/4”-8 outside diameter tolerance

pitch diameter tolerance

TPI

nominal size

Unified N

ational Course

NC - National CourseNF - National FineNS - National SpecialNPT - National Pipe Thread

EXAMPLE:

UNC-

F

14.1.2.3 - British Standard Whitworth (BSW)

FORM: P

D

55°

RELATIONS:R = 0.1373 PD = 0.6403 PP = 1 / N

NOTATION:

EXAMPLE:

R

R

14.1.2.4 - The Unified Thread

FORM: P

H 60°

RELATIONS:D = 0.6134 P (external) = 0.5413 P (internal)F = 0.125 P (external) = 0.250 P (internal)

NOTATION:

EXAMPLE:

17H24

----------

H8----

H6----

14.1.2.5 - American National ACME Thread

FORM:

P

D

29°

RELATIONS:F = 0.3707 PC = 0.3707 P - 0.0052

NOTATION:

EXAMPLE:

F

C

D = 0.500 P (+0.010/-0.000)

14.1.2.6 - Brown and Sharpe Worm Thread

FORM:

P

D

29°

RELATIONS:F = 0.335 PC = 0.310 P

NOTATION:

EXAMPLE:

F

C

D = 0.6866 P

14.1.2.7 - Square Thread

14.1.3 Tapers

FORM:

P

D

RELATIONS:F = 0.500 PC = 0.500 P + 0.002

NOTATION:

EXAMPLE:

w

D = 0.500 P

w

14.2 MOTION/FORCE TRANSMISSIONS

14.2.1 Gears

• Gears are generally round or linear sets of teeth for transmitting forces or motions.

• Different combinations of gears will allow conversions of forces, motions and directions.

• Different types of gears are,- spur- internal- helical- herringbone- bevel- hypoid- worm- rack and pinion

14.2.1.1 - Spur Gears

• transmit power between parallel shafts

• have straight teeth parallel to axis of rotation

• used for slow/moderate speeds

• when two of these gears are meshed, the larger is the gear, the smaller is the pinion

14.2.1.2 - Internal Gears

• transmission between parallel shafts

• require less space

• better meshing/more contact between gears.

14.2.1.3 - Helical Gears

• Gears with teeth on an angle.

• can convert rotary motion to rotary motion on a non-parallel shaft.

• a tooth does not suddenly engage/disengage fully, so noise and vibration are reduced.

• also, more than one tooth is typically in contact, so the strength is increased.

• these gears often generate longitudinal forces that require thrust bearings.

14.2.1.4 - Herringbone Gears

• Looks like a helical gear, but it looks as if a second helical gear with a reverse helix has been attached.

• Similar to helical gears, except that thrust bearings are not required.

OR

14.2.1.5 - Bevel Gears

• Transmit rotations to another axis perpendicular to the first.

• These gears look like spur gears, but with a taper.

• When the two gears are the same size they are miter gears.

• for non-90° intersections, the gears are called angular bevel gears.

• the rotational axes of these gears intersect.

14.2.1.6 - Hypoid Gears

• Like bevel gears, but with helical teeth

• the gears are often offset also (the ases of rotation do not intersect)

• these gears are commonly used in the auto industry.

14.2.1.7 - Worm Gears

• The worm is a helical gear with one or more threads.

• The worm gear is typically a straight tooth gear that is turned by the worm.

• This combination is used to convert rotation to a perpendicular rotation, and reduce the speed.

14.2.1.8 - Rack and Pinion

• A rotating spur gear drives a linear rack of teeth.

• this combination converts a rotation to a linear motion.

14.3 POWER TRANSMISSION

14.3.1 Hydraulics

• Incompressible fluids are used to transmit volume and pressure changes throughout a system.

• Pascal’s law basically describes these systems,

• Hydrostatic force/motion multiplier,

P FA= If a pressure is applied to a non-moving (static) fluid, the pressure (hydrostatic pressure) the fluid exerts on all surfaces that it touches are the same.

• The Hydrodynamic Effect - when fluid is moving quickly, it has high levels of kinetic energy. If the fluid impacts a surface, it transmits a high quantity of energy in a short period of time.

• Hydraulic Circuits typically contain,1. Hydraulic Fluid2. An Oil Resevoir3. A Pump to Move Oil, and Apply Pressure4. Pressure Lines5. Control Valuves - to regulate fluid flow6. Piston and Cylinder - to actuate external mechanisms

• Oil Resevoirs

∆x1 ∆x2

A1A2

P V1,

P V2,F1

F2

For Force:

For Motion:

PF1

A1------

F2

A2------= =

F1

F2------∴

A1

A2------=

∆V1 ∆– x1A1 ∆– V2 ∆x2A2= = =

∆x2

∆– x1------------∴

A1

A2------=

• Pump types used include,- Reciprocating pumps - have intermitent pressures with a single piston

• Valves

fluid return

air filter

outlet tube

level

refill oil filter

access hatch

gauge

for cleaning

baffle - isolates theoutlet fluid fromturbulence in the inlet

Mechanic imp

Design

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