理论力学第三章-英文(课件)

Chapter 3 Reduction and Equilibrium of Force System3.1 Changing the Line of Action of a Force 3.2 Reduction of a Force system in Space to a Given Point3.3 Discussion of the result of the reduction of a force system in space3.4 Equilibrium equations of a force system and their applicationsA force acting on a rigid body can be moved parallel to its lines of action to any point of the body,if we add a couple with a moment equal to the moment of the force about the point to which it is translated,which is called the theorem of translation of a force:force system force force system3.1 Changing the Line of Action of a Force The theorem of translation of a force gives a relationship between the force and force couple:force force+couple.The condition of translation of a force is to add a couple m,which is called the couple of transfer,where m is given by m=Fd.The theorem of translation of a force is the theoretical basis of the reduction of a force system.analysisConsider a force system in space acting on a rigid bodyReduce the force system to an arbitrary point O.3.2 Reduction of a Force system to a Given Point1 According to the Theorem of Translation of a Force we translate every force to the point O.We get a concurrent force system in space and a couple system of transfer are the moments of the original forces about point O.2 Because the couple in space is a free vector it can be always translated to the point O.3 Composing ,we get the principle vector .In other words,(the principle vector passes the point O,and it does not depend on the position of the point O)4 Composing ,we get the principle moment ,(the principle moment depends on the position of point O)Force system in space Resultant Force-couple system in space Let the center O be the origin of the coordinate system,then the magnitude of the principle vector is The magnitude of the principle force isAccording to the relationship between the moment of the force about a point(O)and the moment of the force about an axis through the point(O)The magnitude of the principle moment isFixed support(rigid embedding)Explanation:Example1Consider the forces Fi acting in the same plane;2Reduce Fi to the point A,get the force and the couple;3 the direction of RA is uncertain,but it can bedetermined by its rectangular components YA and XA;4 YA,XA and MA are the reaction forces of the fixed support;5 YA and Xa restrict the translation of the rigid body,Ma restrict the rotation.1.If the force system is in equilibrium.3.3 Discussion of the result of the reduction of a force system in spaceIf a force system in space is reduced to a given point,we get a resultant force (principle vector)and a resultant couple(principle moment).It follows:3.If the force system is equivalent to a resultant force.The principle vector which is equal to the resultant force vector of the force system,the principle vector passes the given center O(it depends on the position of the given center.If the point O is changed,the principle moment is not zero).2.If the force system is equivalent to a couple,the moment is equal to the sum of the moments MO of the forces about the given point.The principle moment does not depend on the position of the given center.4.If ,there are three kinds of situations separately:Turn MO into(R,R),where R counteracts R,so we get R.the force system is equivalent to a resultant forceA collinear force couple system,called the wrenchExample:the operation of a screwdriver R and M0 are not perpendicular to each other A collinear force couple system,called the wrench 1 resolve MO into M/and M,which are parallel and perpendicular to R 2 M and R are mutually perpendicular,They can be replaced by the single force R acting at point O3 Because M/is a free vector it can be translated to the point O.Therefore,there is a wrench about point O.After the reduction of the force system in space to the point O,we get a principle vector R and a principle moment MO.If MO R we can get a new resultant R which acts on the new given center O.The law of the resultant moment of the force system in space:1.The necessary and sufficient conditions for equilibrium of a force system in space areSo the equilibrium equations of a force system in space are They can also be expressed by 4 moment equations,5 moment equations or 6 moment equations,but each case has some limiting conditions.3.4 Equilibrium equations of a force system and their applicationsBecause all action lines of the forces are concurrent at a point and all axes are through this point the moment equations are identities.2 The equilibrium equations of a concurrent force system in space3 The equilibrium equations of a parallel force system in space,supposing all the action lines are parallel to the axis z areBecause4 The equilibrium equations of a general coplanar force system If forces act on a body,the lines of action of these forces are in the same plane,which called coplanar force systemCoplanar force system:1 coplanar system of concurrent forces2 coplanar system of parallel forces 3 general coplanar force systemTo a general coplanar force system,set the plane of action of forces is xoy plane,then The equilibrium equations of a general coplanar force system5 The equilibrium equations of coplanar system of concurrent forces6 The equilibrium equations of coplanar system of parallel forces supposing all the action lines are parallel to the axis y7 Couples Couples in space Couples in the same plane xy Summary concurrent force system in space general force system in space parallel force system in space general coplanar force systemcoplanar system of concurrent forcescoplanar system of parallel forcesApplication statically determinate and statically indeterminate problems ConceptTo a general coplanar force system,there are three independent equations which can determine three unknown quantities.When The number of equationsthe number of unknown quantities it is a statically determinate problem(which can be solved).When The number of equationsthe number of unknown quantities it is a statically indeterminate problem.ExampleStatically indeterminate problems can be solved by the conditions of compatibilityStatically indeterminate(four unknown quantities)Statically determinate (three unknown quantities)ExampleLet OA horizontal(OA=R,AB=l)and the pressure force to be P.Determine:M,the reaction forces of point O,the internal forces of the rod AB,the side pressure acting on the lead rail.Solution:Study the point BExamplethe sign of minus means the direction of the forces is opposite to the direction indicated in the diagramThen study the wheel:Solution:All elements are connected with each other by joints,the bottom B is inserted into the earth,P=1000N,AE=BE=CE=DE=1m,the weight of the elements are negligible.Determine the forces of the member AC and the reaction of the support B.1 Study the whole rigid structure;Draw the force diagram;Example2 Study the element CD;Draw the free-body diagram;ExampleP=100N,AC=1.6m,BC=0.9m,CD=EC=1.2m,AD=2m AB is horizontal,ED is vertical,BD is perpendicular to the incline.Determine and the reaction forces of the constraints.1 Study the whole rigid structure;Draw the force diagram;Solution:3 study the element AB,draw the free-body diagram as above;Consider a continuous beam,P=10kN,Q=50kN,CE is vertical,neglect the weight of the beam.Determine the reaction forces of the points A,B and D.1 study the craneExampleSolution:3 Study the whole rigid structure:2 Then study the beam CDExample1 The steps solving problems:Summary 2 The skills of solving problems:1)Select a body to study2)Draw the free-body diagram 3)Select the coordinates,a center,and write down equilibrium equations;4)Solve the equations,determine the unknown quantities.1)Choose the coordinate axes unknown quantities;2)Choose the given center of the forces at the point of crossing;A truss is a structure that is made of straight,slender bars that are joined Together to form a pattern of triangles.plane trussesStructure of truss in engineeringStructure of truss in engineeringStructure of truss in engineeringStructure of truss in engineering1 the weights of the members are negligible2 all joints are pin 3 the applied forces act at joints.jointjointjointelementelementelementThree assumptions:According to three assumptions,we can get conclusion:each member of a truss is a two-force bodyCalculation of the forces in the members of a truss:1 Method of joint2 Method of sectionP=10kN,determine the internal forces on every elements.Solution:study the whole structure,determine the reaction forces:Study the joints A,C and D one by one,determine the internal forces of every member.We get:(denote that the rod is compressed)ExampleThe another equation of the joint D can be used to verify the result of calculation.It equals to ,the calculation is accurate.Substituting intoWe obtain:Substituting into we get:The solution of which isExample Knowing h,a and P,please determine the internal forces on the element 4,5 and 6.Solution:1 Study the whole structure,determine the reaction forces.2 Imagine section I-I to cut the truss,study the left-part of the truss.AAII2 If no load acts on a three-element joint and two elements of them are collinear,the third is a zero element.3 If no load acts on the four-elementjoint and two pairs of them are collinear,the internal forces of eachpair are equal in magnitude and of the same character.1 If no load acts on a two-element joint and the two elements are not collinear,the internal forces of them are zero,they are called zero elements.3.The judgments of the internal forces of special elementsP d are known,determine the internal forces of the rods a,b,c and d.Solution:1 from the judgments of the zero element2 Study the point A:Exampleanalysis Method of joint:This is convenient when the stresses in all elements of the truss have to be determined.Method of section:This is convenient in determining the stresses in individual members,notably for verifying results.Before the calculation,assume first all elements to be under tension,if the result is negative it means the element concerned is compressed,opposite to the direction assumed.