静力学第3章.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,Chapter 3 Coplanar Equilibrium Analysis,3.1 Definition of Equilibrium,A body is said to be in equilibrium if the resultant of the force system that acts on the body,vanishes.,Equilibrium means that,both the resultant force and the resultant couple are zero.,For,a coplanar force system,its resultant can always be represented as:,R,=,F,C,R,=M,o,In scalar form:,R,x,=,F,x,R,y,=,F,y,C,R,=M,o,The equations of equilibrium are:,F,x,=0,F,y,=0,M,o,=0,F,x,=0,F,y,=0,M,o,=0,-one moment equations,(1).,Two moment equations,If M,A,=0 and R,0,R,must act through point A,If M,B,=0 and,R,acting through point B,Action line of,R,is line AB.,If,F,x,=0 and,x,not,AB,then,R,=0,the force system in equilibrium.,M,A,=0,M,B,=0,F,x,=0,when,x,is not perpendicular to the line AB.,3.2 Coplanar Equilibrium Equation,a.General case,(2).Three,moment equations,If M,A,=0 M,B,=0 R,0,R,acting through point A and point B,M,A,=0 M,B,=0 M,C,=0,when,A,B,C,are any three,noncolinear,points.,If,M,C,=0 and,C,is not on the,AB,line,then,R,=0,the force system in equilibrium.,b.Parallel,Coplanar force systems,(,F,i,y,),Obviously,F,x,0,then equilibrium equations are,F,y,=0,M,o,=0,or M,A,=,0,M,B,=,0,AB,line is not parallel to the y-axis,.,思考,:,如果各力共线,结果如何,?,F,y,=0 Why?,M,o,=,d,F,y,=0,与上式不独立,c.Concurrent,Coplanar force systems,M,o,0,F,x,=0,F,y,=0,x,y,are any two nonparallel directions.,or,M,A,=,0,M,B,=,0,A,B,O,are not on same line.,or,M,A,=,0,F,x,=0,x,is not perpendicular to line,OA,.,Sample 3-1,The weight of the hoist is P,1,and the weight of the body is P,2,.,P,1,=,10kN,，,P,2,=,40kN,。Determine the reactions at thrust bearing A and slider bearing B.,Solution：,1.FBD (Coplanar force system),2.Write and solve equilibrium equations,Sample problem 3-2,The weight of the structure is P,and applied forces are shown in the Fig.,P,=100kN,M,=20kN,M,F,=400kN,q,=20kN/m,l,=1m,.Determine the reactions at A.,Solution:1.FBD,2.Write and solve equilibrium equations:,F,x,=0,F,Ax,+,F,1,-,F,sin60,=0,F,y,=0,F,Ay,-,P,+,F,cos60,=0,M,A,=,0,M,A,-,M,-,F,1,l,-,F,cos60,l,+,F,sin60,3,l,=0,The negative sign for,F,Ay,and,M,A,indicate their correct senses are opposite to those assumed in the FBD.,m,Sample 3-3,A beam and applied forces is shown in Fig.Determine the reactions of support A and support B.,Solution,：,1.FBD,2.Writing and solving equations,The negative sign for,F,Ay,indicate its correct sense is opposite to that assumed in the FBD.,Example 3-4,塔式起重机,(tower crane),机架重,P,1,=700KN,，作用线通过塔架的中心。最大起重量,P,2,=200KN,，最大悬臂长,12m,，轨道,AB,间距,4m,。平衡重,(equalizer)P,3,到机身中心线距离为,6m,。,(1),保证起重机在满载和空载时都不致翻倒，平衡重应为多少,?(2),当,P,3,=180KN,，满载时轨道,A,、,B,的约束力。,Solution,：,（,1,）,FBD,满载时起重机即将绕,B,点翻倒，有,F,A,=0,。,P,3min,。,3.4 Statically determinate problem and statically indeterminate problem,The number of unknowns on the FBD equals the number of independent equations-,statically determinate problem,.,The number of unknowns exceeds the number of independent equations-,statically indeterminate problem,.,For example:,解决之道：补充方程（变形协调条件）属材料力学。,3.5 Equilibrium Analysis of Composite Bodies,Composite bodies:,物体系是指由几个物体通过约束组成的系统,(,包括,Frames,Machines),。,3-3,中分析平衡问题的方法适用于此。但其特点有：,（,1,）整体系统平衡，每个物体也平衡。因此，可取整体或部分系统或单个物体为研究对象。,（,2,）分清内力和外力。在受力图上不考虑内力。,（,3,）灵活选取平衡对象和列写平衡方程。尽量减少未知量，最好是一个方程解一个未知量，简捷求解。,（,4,）如系统由,n,个物体组成，而每个物体在平面力系作用下平衡，则有,3,n,个独立的平衡方程，可解,3,n,个未知量。可用不独立的方程校核计算结果。,作为练习，大家可分析下述例题中独立的平衡方程数。,Example 3-5,The beams,AB,and,BC,are connected by a pin at,B,.Assuming that q=15KN/m,M=20KNm，find the reactions at points,A,、,B,、,C.,Solution:,For the entire structure,we have,F,Cx,=0,For the beam,AB,FBD as Fig.b,F=2q.,Example 3-6,The bridge consists of two arches,each of weight,P,1,=40kN,acting at,D,or,E,.The arches are pointed together at,A,and attached to the piers with pins at,B,and C.The load at,H,is,P=20kN,.Find the reactions at,A,B,and,C,.,Solution:,For the bridge,FBD as Fig.(b).,in equilibrium,For the right arch,FBD as Fig.(c).,in equilibrium,For the entire bridge,we have,X,=0,F,Bx,-F,Cx,=0 ,F,Bx,=20kN,Solution:,Analysis:,From the FBD of the entire frame:4 unknowns 3 independent equations,So we begin by considering the equilibrium of ABD.,1.ABD:,M,D,=0 Q,(40+60+80)-N,AB,(60+80)=0 N,AB,=5400 N,2.Entire frame:,M,C,=0 Q,100-N,AB,60-P,400=0,Q,100-5400,60-400P=0,P=(100Q-60,5400),400=240 N Answer,N,AB,Y,C,X,C,Sample 3-7,Determine the force,P,required for the equilibrium of the compound lever if,Q,=4200N.,Solution,:,1.Analysis entire frame FBD shown in the Fig.,M,A,=0 N,E,4-W(2-0.4)=0,N,E,=2.4,kN,2.For the BD with the weight,M,B,=0 -W 0.6-W 0.4+Y,D,2=0,Y,D,=3,kN,3.CE,M,C,=0,X,D,1.5+N,E,2-W(1.5-0.4)-Y,D,1=0,X,D,=3.2,kN,Y,A,X,A,N,E,N,E,X,C,Y,C,Y,D,X,D,W,C,W,Sample 3-8,W=6kN.Neglecting the weights of the bars and the pulley,find the magnitude of the pin reaction at,D.,X,B,Y,B,W,X,D,Y,D,W,3-6 Analysis of Plane Trusses,平面简单桁架,a.D,escription of a Truss,A truss is a structure that is made of straight,slender bars that are joined together to form a pattern of triangles.,Trusses are usually designed to transmit forces over relatively long spans.,c.Method of Sections,The three assumptions,of analysis of trusses,1.The weight of the members are negligible.,2.All joints are smooth pins.,3.The applied forces are at the joints.,Therefore each member of a truss is a,two-force body,.,b.Method of Joints,Since the members are two-force bodies,the forces in the FBD of a joint are concurrent.Consequently,two independent equilibrium equations are available for each joint.,In the method of sections,a part of the truss is isolated on an FBD so that it exposes the force,or forces,to be computed.When applying the method of sections,the force systems will generally be non-concurrent,coplanar(three independent equilibrium equations).,Sample problem3-9,A truss is shown in Fig.a.P=10KN.Determine the force in each member.,Solution:,(1)The FBD of entire truss is shown in Fig.a.For the symmetry,we get:,F,Bx,=0,F,By,=,F,Ay,=P/2=5KN,(2)Method of joint:assuming the force in each member to be tensile.,Joint,A,：,FBD in the Fig(b),Sample problem 3-10,Determine the forces in member 1,2,3.The length of each member is 1 meter.,P,1,=10kN，,P,2,=7kN.,Solution:,(1)The FBD of entire truss is shown in Fig.(a),P,1,(2)Isolating the truss with an assumed plane m-n and studying the left of it:The FBD is shown in Fig.(b),P,1,