ME 101: Engineering Mechanics - IIT Guwahati

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ME 101: Engineering MechanicsRajib Kumar BhattacharjyaDepartment of Civil EngineeringIndian Institute of Technology GuwahatiM Block : Room No 005 : Tel:

ME101: Division II &IV (3 1 0 8)Lecture Schedule: Venue L2 (Div. II & IV)DAYDIV IIDIV IVMONDAY3.00-3.55 (PM)10.00-10.55 (AM)TUESDAY2.00-2.55 (PM)11.00-11.55 (AM)FRIDAY4.00-4.55 (PM)09.00-09.55 (AM)Tutorial Schedule: Thurs: 8:00-8:55 (AM)2

ME101: SyllabusRigid body static: Equivalent force system. Equations of equilibrium, Free body diagram, Reaction,Static indeterminacy and partial constraints, Two and three force systems.Structures: 2D truss, Method of joints, Method of section. Frame, Beam, types of loading andsupports, Shear Force and Bending Moment diagram, relation among load-shear force-bendingmoment.Friction: Dry friction (static and kinematics), wedge friction, disk friction (thrust bearing), belt friction,square threaded screw, journal bearings (Axle friction), Wheel friction, Rolling resistance.Center of Gravity and Moment of Inertia: First and second moment of area and mass, radius ofgyration, parallel axis theorem, product of inertia, rotation of axes and principal M. I., Thin plates,M.I. by direct method (integration), composite bodies.Virtual work and Energy method: Virtual Displacement, principle of virtual work, mechanicalefficiency, work of a force/couple (springs etc.), Potential Energy and equilibrium, stability.UP TO MID SEMKinematics of Particles: Rectilinear motion, curvilinear motion rectangular, normal tangential, polar,cylindrical, spherical (coordinates), relative and constrained motion, space curvilinear motion.Kinetics of Particles: Force, mass and acceleration, work and energy, impulse and momentum, impact.Kinetics of Rigid Bodies: Translation, fixed axis rotation, general planner motion, work-energy, power,potential energy, impulse-momentum and associated conservation principles, Euler equations ofmotion and its application.

Course web: 1 Basic principles: Equivalent force system; Equations of equilibrium; Freebody diagram; Reaction; Static indeterminacy.2 Structures: Difference between trusses, frames and beams, Assumptionsfollowed in the analysis of structures; 2D truss; Method of joints; Methodof section3 Frame; Simple beam; types of loading and supports; Shear Force andbending Moment diagram in beams; Relation among load, shear force andbending moment.4 Friction: Dry friction; Description and applications of friction in wedges,thrust bearing (disk friction), belt, screw, journal bearing (Axle friction);Rolling resistance.5 Virtual work and Energy method: Virtual Displacement; Principle of virtualwork; Applications of virtual work principle to machines; Mechanicalefficiency; Work of a force/couple (springs etc.);6 Potential energy and equilibrium; stability. Center of Gravity and Momentof Inertia: First and second moment of area; Radius of gyration;7 Parallel axis theorem; Product of inertia, Rotation of axes and principalmoment of inertia; Moment of inertia of simple and composite bodies.Mass moment of inertia.Department of Civil Engineering: IIT GuwahatiTutorial123QUIZ45Assignment

ME101: Text/Reference BooksI. H. Shames, Engineering Mechanics: Statics and dynamics, 4th Ed, PHI, 2002.F. P. Beer and E. R. Johnston, Vector Mechanics for Engineers, Vol I - Statics, Vol II– Dynamics, 9th Ed, Tata McGraw Hill, 2011.J. L. Meriam and L. G. Kraige, Engineering Mechanics, Vol I – Statics, Vol II –Dynamics, 6th Ed, John Wiley, 2008.R. C. Hibbler, Engineering Mechanics: Principles of Statics and Dynamics, PearsonPress, 2006.Andy Ruina and Rudra Pratap, Introduction to Statics and Dynamics, OxfordUniversity Press, 2011

Marks DistributionEnd SemesterMid SemesterQuizTutorialsAssignmentClassroom Participation40201015051075% Attendance MandatoryTutorials: Solve and submit on each ThursdayAssignments: Solve later and submit it in the next classDepartment of Civil Engineering: IIT Guwahati

ME101: Tutorial 10061G11G21207210121023202Name of the TutorDr. Karuna KalitaDr. Satyajit PandaDr. Deepak SharmaDr. M Ravi SankarDr. Ganesh NatrajanDr. Sachin S GautamDr. Swarup BagProf. Sudip TalukdarDr. Arbind SinghProf. Anjan DuttaDr. Kaustubh DasguptaT124001Dr. Bishnupada MandalT13T144G34G4Prof. V. S. MoholkarDr. A. K. GolderTutorial sheet has three sectionsSection I: Discuss by the tutor(2 questions)Section II: Solve by the students inthe class (4 questions)Section II: Solve by the studentsAs assignment(4 questions)

ME101: Engineering MechanicsMechanics: Oldest of the Physical SciencesArchimedes (287-212 BC): Principles of Lever and Buoyancy!Mechanics is a branch of the physical sciences that isconcerned with the state of rest or motion of bodies subjectedto the action of forces.Rigid-body Mechanics ME101StaticsDynamicsDeformable-Body Mechanics, andFluid Mechanics

Engineering MechanicsRigid-body Mechanics a basic requirement for the study of themechanics of deformable bodies and themechanics of fluids (advanced courses). essential for the design and analysis of manytypes of structural members, mechanicalcomponents, electrical devices, etc, encounteredin engineering.A rigid body does not deform under load!

Engineering MechanicsRigid-body MechanicsStatics: deals with equilibrium of bodies underaction of forces (bodies may be either at rest ormove with a constant velocity).

Engineering MechanicsRigid-body Mechanics Dynamics: deals with motion of bodies(accelerated motion)

Mechanics: Fundamental ConceptsLength (Space): needed to locate position of a point in space, &describe size of the physical system Distances, GeometricPropertiesTime: measure of succession of events basic quantity inDynamicsMass: quantity of matter in a body measure of inertia of abody (its resistance to change in velocity)Force: represents the action of one body on another characterized by its magnitude, direction of its action, and itspoint of application Force is a Vector quantity.

Mechanics: Fundamental ConceptsNewtonian MechanicsLength, Time, and Mass are absolute conceptsindependent of each otherForce is a derived conceptnot independent of the other fundamental concepts.Force acting on a body is related to the mass of the bodyand the variation of its velocity with time.Force can also occur between bodies that are physicallyseparated (Ex: gravitational, electrical, and magnetic forces)

Mechanics: Fundamental ConceptsRemember: Mass is a property of matter that does notchange from one location to another. Weight refers to the gravitational attraction ofthe earth on a body or quantity of mass. Itsmagnitude depends upon the elevation atwhich the mass is located Weight of a body is the gravitational force acting on it.

Mechanics: IdealizationsTo simplify application of the theoryParticle: A body with mass but with dimensionsthat can be neglectedSize of earth is insignificantcompared to the size of itsorbit. Earth can be modeledas a particle when studying itsorbital motion

Mechanics: IdealizationsRigid Body: A combination of large number of particles inwhich all particles remain at a fixed distance (practically)from one another before and after applying a load.Material properties of a rigid body are not required to beconsidered when analyzing the forces acting on thebody.In most cases, actual deformations occurring in structures,machines, mechanisms, etc. are relatively small, and rigidbody assumption is suitable for analysis

Mechanics: IdealizationsConcentrated Force: Effect of a loading which isassumed to act at a point (CG) on a body. Provided the area over which the load is appliedis very small compared to the overall size of thebody.Ex: Contact Forcebetween a wheeland ground.40 kN160 kN

Mechanics: Newton’s Three Laws of MotionBasis of formulation of rigid body mechanics.First Law: A particle originally at rest, or moving in a straight linewith constant velocity, tends to remain in this state provided theparticle is not subjected to an unbalanced force.First law contains the principle ofthe equilibrium of forces maintopic of concern in Statics

Mechanics: Newton’s Three Laws of MotionSecond Law: A particle of mass “m” acted upon by anunbalanced force “F” experiences an acceleration “a” thathas the same direction as the force and a magnitude that isdirectly proportional to the force.mSecond Law forms the basis for most ofthe analysis in DynamicsF ma

Mechanics: Newton’s Three Laws of MotionThird Law: The mutual forces of action and reaction betweentwo particles are equal, opposite, and collinear.Third law is basic to our understanding of Force Forces alwaysoccur in pairs of equal and opposite forces.

Mechanics: Newton’s Law of Gravitational AttractionWeight of a body (gravitational force acting on a body) is required to becomputed in Statics as well as Dynamics.This law governs the gravitational attraction between any two particles.m1m2F G 2rF mutual force of attraction between two particlesG universal constant of gravitationExperiments G 6.673x10-11 m3/(kg.s2)Rotation of Earth is not taken into accountm1, m2 masses of two particlesr distance between two particles

Gravitational Attraction of the EarthWeight of a Body: If a particle is located at or near the surface ofthe earth, the only significant gravitational force is that betweenthe earth and the particleWeight of a particle having mass m1 m :Assuming earth to be a nonrotating sphere of constant densityand having mass m2 MemM eW G 2rr distance between the earth’scenter and the particleW mgLet g G Me /r2 acceleration due to gravity(9.81m/s2)

Mechanics: UnitsFour Fundamental QuantitiesQuantityDimensionalSymbolSI ondsForceFNewtonNF ma N kg.m/s2W mg N kg.m/s2Basic Unit1 Newton is the forcerequired to give a mass of 1kg an acceleration of 1 m/s2

Mechanics: Units Prefixes

Scalars and VectorsScalars: only magnitude is associated.Ex: time, volume, density, speed, energy, massVectors: possess direction as well as magnitude, and must obey theparallelogram law of addition (and the triangle law).Ex: displacement, velocity, acceleration,force, moment, momentumEquivalent Vector: V V1 V2 (Vector Sum)Speed is the magnitude of velocity.

VectorsA Vector V can be written as: V VnV magnitude of Vn unit vector whose magnitude is one and whose direction coincides withthat of VUnit vector can be formed by dividing any vector, such as the geometricposition vector, by its length or magnitudeVectors represented by Bold and Non-Italic letters (V)Magnitude of vectors represented by Non-Bold, Italic letters (V)yjxziki, j, k – unit vectors

VectorsFree Vector: whose action is not confined to orassociated with a unique line in spaceEx: Movement of a body without rotation.Sliding Vector: has a unique line ofaction in space but not a uniquepoint of applicationEx: External force on a rigid body Principle of Transmissibility Imp in Rigid Body MechanicsFixed Vector: for which a unique point ofapplication is specifiedEx: Action of a force on deformable body

Vector Addition: Procedure for AnalysisParallelogram Law (Graphical)Resultant Force (diagonal)Components (sides ofparallelogram)Algebraic SolutionUsing the coordinate systemTrigonometry (Geometry)Resultant Force and Componentsfrom Law of Cosines and Law ofSines

Force SystemsForce: Magnitude (P), direction (arrow) and point of application (point A) isimportantChange in any of the three specifications will alter the effect on the bracket.Force is a Fixed VectorIn case of rigid bodies, line of action of force is important (not its point ofapplication if we are interested in only the resultant external effects of theforce), we will treat most forces asExternal effect: Forces applied (appliedforce); Forces exerted by bracket, bolts,Foundation (reactive force)Cable Tension PInternal effect: Deformation, strainpattern – permanent strain; depends onmaterial properties of bracket, bolts, etc.

Force SystemsConcurrent force:Forces are said to be concurrent at a point if their lines of actionintersect at that pointF1, F2 are concurrent forces; R will be on same plane; R F1 F2F2F2RF2AF1F2RRAR F1 F2F1F1AF1PlaneForces act at same point Forces act at different pointTriangle Law(Apply Principle of Transmissibility)

Components and Projections of ForceComponents of a Force are not necessarily equal to the Projectionsof the Force unless the axes on which the forces are projected areorthogonal (perpendicular to each other).F1 and F2 are components of R.R F1 F2Fa and Fb are perpendicular projections onaxes a and b, respectively.R Fa Fb unless a and b are perpendicular toeach other

Components of ForceExamples


Components of ForceExample 1:Determine the x and yscalar components ofF1, F2, and F3 actingat point A of the bracket

Components of ForceSolution:

Components of ForceAlternative Solution

Components of ForceAlternative Solution

Components of ForceExample 2: The two forces act on a bolt at A. Determine theirresultant.Graphical solution - construct aparallelogram with sides in the samedirection as P and Q and lengths inproportion. Graphically evaluate theresultant which is equivalent indirection and proportional in magnitudeto the diagonal.Trigonometric solution - use thetriangle rule for vector addition inconjunction with the law of cosinesand law of sines to find the resultant.

Components of ForceSolution: Graphical solution - A parallelogram with sidesequal to P and Q is drawn to scale. Themagnitude and direction of the resultant or ofthe diagonal to the parallelogram aremeasured,R 98 N α 35 Graphical solution - A triangle is drawn with Pand Q head-to-tail and to scale. Themagnitude and direction of the resultant or ofthe third side of the triangle are measured,R 98 Nα 35

Components of ForceTrigonometric Solution: Apply the triangle rule.From the Law of Cosines,R 2 P 2 Q 2 2 PQ cos B (40 N )2 (60 N )2 2 (40 N )(60 N ) cos 155 R 97 .73 NFrom the Law of Sines,sin A sin B QRsin A sin BQR sin 155 A 15 .04 α 20 Aα 35 .04 60 N97 .73 N

Components of Force

Components of ForceExample 3:Tension in cable BC is 725-N, determine the resultant of the threeforces exerted at point B of beam AB.Solution: Resolve each force into rectangularcomponents. Determine the components of theresultant by adding thecorresponding force components. Calculate the magnitude anddirection of the resultant.

Components of ForceResolve each force into rectangular componentsMagnitude (N)X-component (N)Y-component (N)725-525500500-300-400780720-300Calculate the magnitude and direction

Components of ForceAlternate solutionCalculate the magnitude and direction

Rectangular Components in Spacer The vector F iscontained in theplane OBAC.r Resolve F intohorizontal and verticalcomponents.Fy F cosθ yFh F sin θ y Resolve Fh intorectangularcomponentsFx Fh cos φ F sin θ y cos φFz Fh sin φ F sin θ y sin φ

Rectangular Components in Space

Rectangular Components in SpaceDirection of the force is defined by the location of two points

Rectangular Components in SpaceExample: The tension in the guywire is 2500 N. Determine:SOLUTION:a) components Fx, Fy, Fz of theforce acting on the bolt at A, Based on the relative locations of thepoints A and B, determine the unitvector pointing from A towards B.b) the angles qx, qy, qz defining the Apply the unit vector to determinedirection of the forcethe components of the force actingon A. Noting that the components of theunit vector are the direction cosinesfor the vector, calculate thecorresponding angles.

Rectangular Components in SpaceSolutionDetermine the unit vector pointing from Atowards B.Determine the components of the force.

Rectangular Components in SpaceSolutionNoting that the components of the unitvector are the direction cosines for thevector, calculate the corresponding angles.θ x 115.1oθ y 32.0oθ z 71.5o

Vector ProductsDot ProductApplications:to determine the angle between two vectorsto determine the projection of a vector in a specified directionA.B B.A (commutative)A.(B C) A.B A.C (distributive operation)

Vector ProductsCross Product:Cartesian Vector

Moment of a Force (Torque)Sense of the moment may be determined bythe right-hand rule

Moment of a ForcePrinciple of TransmissibilityAny force that has the samemagnitude and direction as F, isequivalent if it also has the sameline of action and therefore,produces the same moment.Varignon’s Theorem(Principle of Moments)Moment of a Force about a point is equal tothe sum of the moments of the force’scomponents about the point.

Rectangular Components of a MomentThe moment of F about O,

Rectangular Components of the MomentThe moment of F about B,

Moment of a Force About a Given AxisMoment MO of a force F applied atthe point A about a point OScalar moment MOL about an axisOL is the projection of the momentvector MO onto the axis,Moments of F about the coordinateaxes (using previous slide)

Moment of a Force About a Given AxisMoment of a force about an arbitrary axis and If we take point in place of point . . . are in the same line0

Moment: ExampleCalculate the magnitude of the moment about thebase point O of the 600 N force in different waysSolution 1.Moment about O isSolution 2.

Moment: ExampleSolution 3.Solution 4.Solution 5.The minus sign indicates that thevector is in the negative z-direction

Moment of a CoupleMoment produced by two equal, opposite andnon-collinear forces is called a couple.Magnitude of the combined moment ofthe two forces about O: The moment vector of the couple is independentof the choice of the origin of the coordinate axes,i.e., it is a free vector that can be applied at anypoint with the same effect.

Moment of a CoupleTwo couples will have equal moments if The two couples lie in parallel planesThe two couples have the same sense or thetendency to cause rotation in the same direction.Examples:

Addition of CouplesConsider two intersecting planes P1and P2 with each containing a couple in plane in plane Resultants of the vectors also form a couple By Varigon’s theorem Sum of two couples is also a couple that is equal tothe vector sum of the two couples

Couples VectorsA couple can be represented by a vector with magnitude anddirection equal to the moment of the couple.Couple vectors obey the law of addition of vectors.Couple vectors are free vectors, i.e., the point of application is notsignificant.Couple vectors may be resolved into component vectors.

Couple: ExampleCase IMoment required to turn the shaft connected atcenter of the wheel 12 NmCase I: Couple Moment produced by 40 Nforces 12 NmCase II: Couple Moment produced by 30 Nforces 12 NmIf only one hand is used?Force required for case I is 80NForce required for case II is 60NWhat if the shaft is not connected at the centerof the wheel?Is it a Free Vector?Case II

Equivalent SystemsAt support O! "

Equivalent Systems: Resultants # What is the value of d?Moment of the Resultant force about the grip must be equal to themoment of the forces about the grip # Equilibrium Conditions

Equivalent Systems: ResultantsEquilibriumEquilibrium of a body is a condition in which theresultants of all forces acting on the body is zero.Condition studied in Statics

Equivalent Systems: ResultantsVector Approach: Principle of Transmissibility can be usedMagnitude and dir

Engineering Mechanics Rigid-body Mechanics a basic requirement for the study of the mechanics of deformable bodies and the mechanics of fluids (advanced courses). essential for the design and analysis of many types of structural members, mechanical components, electrical devices, etc, encountered in engineering.

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