工程力學雙語課件ch05-equilibrium-of-a-rigid-body

1、Chapter 5 Equilibrium of a rigid body,Wang Ming-lu,2,Specialized Vocabulary,Particle:質(zhì)點Support condition：支撐條件Roller：卷軸; 輥子, 滾軸, 滾子 Cable：纜, 索, 鋼絲繩 Pin：釘, 栓, 銷; Hinge：鉸鏈; 折[合]葉; 活頁; 轉(zhuǎn)軸 Fixed：固定的; 確定的; 不變的; 固執(zhí)的

2、Clamp：夾子， 夾鉗， 卡釘,3,Chapter Objectives.,.1.Draw complete and accurate free-body diagrams.2.Write and solve the equations of equilibrium for a rigid body.3.Describe the equilibrium for two- and three-force bodies.,4,,,,P

3、rinciples: conclusions obtained from longtime practice and from many experiments. They are repeatedly proved by practice and now are recognized without prove.,公理：是人類經(jīng)過長期實踐和經(jīng)驗而得到的結(jié)論，它被反復的實踐所驗證，是無須證明而為人們所公認的結(jié)論。,5.1 Introdu

4、ction.,5,,,,1st Principle: two force equilibrium 公理1 二力平衡公理,A rigid body subjected to the action of two forces can be in equilibrium, the necessary and sufficient conditions are:作用于剛體上的兩個力，使剛體平衡的必要與充分條件是： the two

5、 forces are equal in magnitude, | F1 | = | F2 | 這兩個力大小相等 | F1 | = | F2 | the two forces are opposite in direction, F1 = –F2 方向相反 F1 = –F2 the two forces are collinear, 作用線共線， the two

6、forces act on the same body. 作用于同一個物體上。,6,Explanation: 說明：①As far as rigid bodies are concerned, the conditions above are necessary and sufficient. ①對剛體來說，上面的條件是充要的　 　②As far as deformable bodies are concerned,

7、the conditions above are only necessary. ②對變形體來說，上面的條件只是必要條件(或多體中),1st Principle: two force equilibrium 公理1 二力平衡公理,7,③二力體：只在兩個力作用下平衡的剛體叫二力體。 (P145),③double force equivalent body: is a rigid body subjected to

8、 two forces must be in equilibrium. (P145),1st Principle: two force equilibrium 公理1 二力平衡公理,8,,9,,,10,,11,,12,,[證] ∵ 為平衡力系， ∴ 也為平衡力系。 又∵ 二力平衡必等值、反向、共線，

9、 ∴ 三力 必匯交，且共面。,13,,14,15,公理5告訴我們：處于平衡狀態(tài)的變形體，可用剛體靜力學的平衡理論。,16,,5.2 Support conditions for bodies in plane,17,,Reaction force of a constraint: The force by which a constraint acts on a body, thereby

10、restricting its displacements, is called the force of reaction of the constraint.約束反力：約束給被約束物體的力叫約束反力。,Constraint: Anything that restricts the displacement of a given body in space. (here, constraint is a nou

11、n, not a verb).約束：對非自由體的某些位移預先施加的限制條件稱為約束。 （這里，約束是名詞，而不是動詞的約束。）,非自由體：位移受限制的物體叫非自由體。,Constrained body: A body whose displacement in space is restricted by other bodies, either connected to or in contact with

12、 it, is called a constrained body.,5.2 Support conditions for bodies in plane,18,,①their magnitude is usually unknown;①大小常常是未知的；②their direction is always opposite to the direction in which the given constraint pr

13、events the displacement of the body;②方向總是與約束限制的物體的位移方向相反；,5.2 Support conditions for bodies in plane,19,,③the point of application is the contact point between the body and the constraint.③作用點在物體與約束相接觸的那一點。,G,5.2 S

14、upport conditions for bodies in plane,20,5.2 Support conditions for bodies in plane,21,5.2 Support conditions for bodies in plane,22,,5.2 Support conditions for bodies in plane,23,,5.2 Support conditions for bodies in pl

15、ane,24,,Pin,5.2 Support conditions for bodies in plane,25,,5.2 Support conditions for bodies in plane,26,A,5.2 Support conditions for bodies in plane,27,,Pin,5.2 Support conditions for bodies in plane,28,5.2 Support cond

16、itions for bodies in plane,29,5.2 Support conditions for bodies in plane,30,5.2 Support conditions for bodies in plane,31,5.2 Support conditions for bodies in plane,32,5.2 Support conditions for bodies in plane,33,5.2 Su

17、pport conditions for bodies in plane,34,5.2 Support conditions for bodies in plane,35,5.2 Support conditions for bodies in plane,36,5.2 Support conditions for bodies in plane,37,,5.2 Support conditions for bodies in plan

18、e,38,1. Force analysis To solve a problem of static. First select the body to be studied. 一、受力分析 解決力學問題時，首先要選定需要進行研究的物體，即選擇研究對象；,5.3 Construction of free-body diagrams,39,Secondly, in accordance with the given cond

19、itions and constraints, analyze the forces which act on the bodies using the basic concepts and principles. This process is called force analysis.然后根據(jù)已知條件，約束類型并結(jié)合基本概念和公理分析它的受力情況，這個過程稱為物體的受力分析。,5.3 Construction of free-b

20、ody diagrams,40,The forces acting on a body can be classified as:作用在物體上的力有：1: active forces, such as the gravitational force, the force of wind power, the pressure of gases, etc.一類是:主動力,如重力,風力,氣體壓力等。2: passive forces

21、, such as reaction force of constraints二類是：被動力，即約束反力。,5.3 Construction of free-body diagrams,41,The main steps to draw the force diagrams: ①select the object under study; ①選研究對象；

22、 ②divide the object into several parts and study the parts one by one; ②取分離體； ③draw the active forces; ③畫上主動力； ④draw the

23、 reaction force of the constraints. ④畫出約束反力。,5.3 Construction of free-body diagrams,42,5.3 Construction of free-body diagrams,43,,5.3 Construction of free-body diagrams,44,,Rod,5.3 Construction of free-

24、body diagrams,45,5.3 Construction of free-body diagrams,46,5.3 Construction of free-body diagrams,47,5.3 Construction of free-body diagrams,48,,,5.3 Construction of free-body diagrams,49,,,5.3 Construction of free-body d

25、iagrams,50,,,5.3 Construction of free-body diagrams,51,,,5.3 Construction of free-body diagrams,52,,,5.3 Construction of free-body diagrams,53,,,5.3 Construction of free-body diagrams,54,5.3 Construction of free-body dia

26、grams,55,5.3 Construction of free-body diagrams,56,The direction of constraint reaction force must be strictly according to the type of constraint, never be simply supposed by direct-view or the direction of the active f

27、orce. When analyzing the action forces and the reaction forces between two bodies,,約束反力的方向必須嚴格地按照約束的類型來畫，不能單憑直觀或根據(jù)主動力的方向來簡單推想。在分析兩物體之間的作用力與反作用力時，,5.3 Construction of free-body diagrams,57,please give attention to the dir

28、ection of the forces. If the direction of the action force has been determined, the direction of the reaction force is opposite to the action force. Don’t draw the direction of forces wrong. 要注意，作用力的方向一旦確定，反作用力的方向

29、一定要與之相反，不要把箭頭方向畫錯。,5.3 Construction of free-body diagrams,58,5.3 Construction of free-body diagrams,59,5.3 Construction of free-body diagrams,60,5.3 Construction of free-body diagrams,61,Therefore, the necessary and suff

30、icient conditions for the equilibrium of any general coplanar force system are:所以平面任意力系平衡的充要條件為：,If =0 then forces are in equilibrium.由于 =0 為力平衡 MO=0 then force couples are in equil

31、ibrium. MO=0 為力偶也平衡,,5.4 Equations for equilibrium of a rigid body,62,,①the equations of one moment①一矩式,The three independent equations above can only determine three unknown quantities.上式有三個獨立方程，只能求出三

32、個未知數(shù)。,The necessary and sufficient conditions for the equilibrium of any general coplanar force system are:,5.4 Equations for equilibrium of a rigid body,63,,③the equations of three moments ③三矩式,Condition: the points A,

33、 B and C are not collinear條件：A,B,C不在同一直線上,,②the equations of two moments②二矩式,Condition: the axis xis not line AB條件：x 軸不 AB 連線,5.4 Equations for equilibrium of a rigid body,64,[Example] P and a are giv

34、en. Determine the reaction forces of the constraints at the point A and B.[例] 已知：P, a , 求：A、B兩點的支座反力？,Solution:解：①investigate beam AB,①選AB梁研究②draw the forces diagram ②畫受力圖(If we write “fre

35、e from constraints” under the diagram,以后注明 解除約束）,free from constraints解除約束,5.4 Equations for equilibrium of a rigid body,65,（ we can directly draw the reaction forces of the constraints on the original diagram of the w

36、hole structure.可把支反力直接畫在整體結(jié)構(gòu)的原圖上）,free from constraints解除約束,5.4 Equations for equilibrium of a rigid body,66,Therefore, the equations for the equilibrium of the coplanar system of parallel forces are:平面平行力系的平衡方程為：,Co

37、ndition: line AB is not parallel to the line of the action of the force.條件：AB連線不能平行于力的作用線,5.4 Equations for equilibrium of a rigid body,67,Actually, the x projections of all the forces are zero, .Therefore,

38、there are only two independent equations, they can determine only two unknown quantities.實質(zhì)上是各力在x 軸上的投影恒等于零，即 恒成立，所以只有兩個獨立方程，只能求解兩個獨立的未知數(shù)。,5.4 Equations for equilibrium of a rigid body,68,[Example] Let P=2

39、0kN, m=16kN·m, q=20kN/m, a=0.8. Determine the reaction forces at the points A and B.[例] 已知：P=20kN, m=16kN·m, q=20kN/m, a=0.8m 求：A、B的支反力。,Solution: Study the beam AB.解：研究AB梁,5.4 Equ

40、ations for equilibrium of a rigid body,69,,解得：Solving them, we obtain:,5.4 Equations for equilibrium of a rigid body,70,5.4.2 Example.Example 5.4-5.7 (P131).5.4.3 Problems.Do the problems 5.1-4.36 by yourself, and th