东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A03 Equivalent Systems of Forces

Equivalent Systems of Forces

Equivalent Systems of Forces

ContentsIntroduction（绪论）VectorProducts of TwoVectors（矢量积）Moment ofaForceAbouta Point（力对点的矩）Moment of a Force About a Given Axis（力对轴的矩）MomentofaCouple（力偶矩）CouplesCanBeRepresentedByVectors（力偶矩矢量）Resolution ofaForce Into aForce and a Couple（力的平移与分解)System ofForces: Reduction to a Force and a Couple （力 系 :简化为单个力与单个矩）2

Contents • Introduction（绪论） • Vector Products of Two Vectors（矢量积） • Moment of a Force About a Point（力对点的矩） • Moment of a Force About a Given Axis（力对轴的矩） • Moment of a Couple（力偶矩） • Couples Can Be Represented By Vectors（力偶矩矢量） • Resolution of a Force Into a Force and a Couple（力的平移与 分解） • System of Forces: Reduction to a Force and a Couple（力系： 简化为单个力与单个矩） 2

Introduction: Treatment of a body as a single particle is not always possible. Ingeneral, the size of the body and the specific points of application of theforces must be considered.: Most bodies in elementary mechanics are assumed to be rigid, i.e., theactual deformations are small and do not affect the conditions ofequilibrium or motion of the body: Current chapter describes the effect of forces exerted on a rigid body andhow to replace a given system of forces with a simpler equivalent system. moment ofa force about a pointmomentofaforceaboutanaxis: moment due to a couple: Any system of forces acting on a rigid body can be replaced by anequivalent system consisting of one force acting at a given point and onecouple.3

Introduction • Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. • Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. • Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. • moment of a force about a point • moment of a force about an axis • moment due to a couple • Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. 3

Vector Product of Two Vectors: Concept of the moment of a force about a point isV=PxQmore easily understood through applications ofthe vector product or cross product.Vector product of two vectors P and Q is definedas the vector V which satisfies the following(a)conditions:1. Line of action of V is perpendicular to planecontaining P and Q2.Magnitude of Vis V = POsin0(b)3.Direction of V is obtained from the right-handrule.. Vector products:- are not commutative, Q× P=-(P×Q)P×(Q1 +02)= P×Q1 + P×Q2- are distributive,(P×Q)xS+ P×(Q×S)arenotassociative,4

Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1.Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is 3. Direction of V is obtained from the right-hand rule. V  PQsin • Vector products: - are not commutative, - are distributive, - are not associative, Q P  PQ   P Q1 Q2  PQ1  PQ2 PQ S  PQ S 4

Vector Products: Rectangular ComponentsVectorproductsofCartesianunitvectors,ixi=0jxi=-k kxi=jixi--ixj=kjxj=0kxj=-ikxk=0ixk=-j jxk=iixj=kVectorproductsintermsofrectangularcoordinates = (Pi + P,j + P,k)x(Qxi +O,j+Q.k)(P,Q, - P.0,) +(P.x - PxO.))+(PxOy - PyOx)kjkiP2PsPyOxy Q5

Vector Products: Rectangular Components • Vector products of Cartesian unit vectors, 0 0 0                      i k j j k i k k i j k j j k j i i i j i k k i j                         • Vector products in terms of rectangular coordinates V P i P j P k  Q i Q j Q k  x y z x y z                  P Q P Q k P Q P Q i P Q P Q j x y y x y z z y z x x z          x y z x y z Q Q Q P P P i j k     5

Moment of a Force About a Point.A force vector is defined by its magnitude anddirection. Its effect on the rigid body also dependsMoon its point of application.: The moment of F about O is defined asMo=rxF: The moment vector Mo is perpendicular to theplane containing O and the force F.(a): Magnitude of Mo measures the tendency of the forceto cause rotation of the body about an axis along MoMoMo =rFsin0=FdThe sense of the moment may be determined by theright-hand rule(b):Anyforce F'that has the same magnitude anddirection as F, is equivalent if it also has the same lineof action and therefore,produces the same moment6

Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on its point of application. • The moment of F about O is defined as MO  r F • The moment vector MO is perpendicular to the plane containing O and the force F. • Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment. • Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO . The sense of the moment may be determined by the right-hand rule. MO  rF sin  Fd 6

Sample ProblemA 100-lb vertical force is applied to the end of aleverwhichisattachedtoashaftatODetermine:24 in.a) moment about O,1001bb) horizontal force at A which creates the samemoment,60°c) smallest force at A which produces the samemoment,d) location for a 240-lb vertical force to producethe same moment,)whetheranyoftheforcesfromb,c,anddisequivalenttotheoriginalforce.7

Sample Problem A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force. 7

a) Moment about O is equal to the product of theforce and the perpendicular distance between theline of action of theforce and O.Since the forcetendstorotatetheleverclockwise.themoment24 in1001hvectoris intotheplaneofthepaper.Mo = Fd60°d =(24in.)cos60°=12 inMMo = (100 1b)(12 in.)Mo =12001b·in8

a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.   100 lb12 in. 24in. cos60 12 in.      O O M d M Fd MO 1200 lbin 8

c) Horizontal force at A that produces the samemoment,d = (24 in.)sin60° = 20.8 in24indMo = Fd1200 lb in. = F(20.8 in)60°1200lbinHF =57.71b20.8in.9

c) Horizontal force at A that produces the same moment,     20.8 in. 1200 lb in. 1200 lb in. 20.8 in. 24 in. sin 60 20.8 in.         F F M Fd d O F  57.7 lb9

c)The smallestforceAto producethe samemomentoccurs when the perpendicular distance is amaximum or when F is perpendicular to OA24 in.Mo = Fd1200 1b in. = F(24 in.)60°1200lb·inFF = 501b24 in.M10

c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.   24 in. 1200 lb in. 1200 lb in. 24 in.      F F MO Fd F  50 lb10