机械专业毕业设计(论文)外文翻译-连杆机构.doc

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1、Link mechanismLinkages include garage door mechanisms, car wiper mechanisms, gear shift mechanisms.They are a very important part of mechanical engineering which is given very little attention.A link is defined as a rigid body having two or more pairing elements which connect it to other bodies for

2、the purpose of transmitting force or motion . In every machine, at least one link either occupies a fixed position relative to the earth or carries the machine as a whole along with it during motion. This link is the frame of the machine and is called the fixed link.An arrangement based on component

3、s connected by rotary or sliding interfaces only is called a linkage. These type of connections, revolute and prismatic, are called lower pairs. Higher pairs are based on point line or curve interfaces. Examples of lower pairs include hinges rotary bearings, slideways , universal couplings. Examples

4、 of higher pairs include cams and gears.Kinematic analysis, a particular given mechanism is investigated based on the mechanism geometry plus factors which identify the motion such as input angular velocity, angular acceleration, etc. Kinematic synthesis is the process of designing a mechanism to ac

5、complish a desired task. Here, both choosing the types as well as the dimensions of the new mechanism can be part of kinematic synthesis.Planar, Spatial and Spherical MechanismsA planar mechanism is one in which all particles describe plane curves is space and all of the planes are co-planar.The maj

6、ority of linkages and mechanisms are designed as planer systems. The main reason for this is that planar systems are more convenient to engineer. Spatial mechanisma are far more complicated to engineer requiring computer synthesis. Planar mechanisms ultilising only lower pairs are called planar link

7、ages. Planar linkages only involve the use of revolute and prismatic pairsA spatial mechanism has no restrictions on the relative movement of the particles. Planar and spherical mechanisms are sub-sets of spatial mechanisms.Spatial mechanisms / linkages are not considered on this pageSpherical mecha

8、nisms has one point on each linkage which is stationary and the stationary point of all the links is at the same location. The motions of all of the particles in the mechanism are concentric and can be repesented by their shadow on a spherical surface which is centered on the common location.Spheric

9、al mechanisms /linkages are not considered on this pageMobilityAn important factor is considering a linkage is the mobility expressed as the number of degrees of freedom.The mobility of a linkage is the number of input parameters which must be controlled independently in order to bring the device to

10、 a set position.It is possible to determine this from the number of links and the number and types of joints which connect the links.A free planar link generally has 3 degrees of freedom (x , y, ). One link is always fixed so before any joints are attached the number of degrees of freedom of a linka

11、ge assembly with n links = DOF = 3 (n-1) Connecting two links using a joint which has only on degree of freedom adds two constraints. Connecting two links with a joint which has two degrees of freedom include 1 restraint to the systems. The number of 1 DOF joints = say j 1 and the number of joints w

12、ith two degrees of freedom = say j 2. The Mobility of a system is therefore expressed as mobility = m = 3 (n-1) - 2 j 1 - j 2Examples linkages showing the mobility are shown below. A system with a mobility of 0 is a structure. A system with a mobility of 1 can be fixed in position my positioning onl

13、y one link. A system with a mobility of 2 requires two links to be positioned to fix the linkage position.This rule is general in nature and there are exceptions but it can provide a very useful initial guide as the the mobility of an arrangement of links.Grashofs LawWhen designing a linkage where t

14、he input linkage is continuously rotated e.g. driven by a motor it is important that the input link can freely rotate through complete revolutions. The arrangement would not work if the linkage locks at any point. For the four bar linkage Grashofs law provides a simple test for this conditionGrashof

15、s law is as follows: For a planar four bar linkage, the sum of the shortest and longest links cannot be greater than the sum of the remaining links if there is to be continuous relative rotation between two members.Referring to the 4 inversions of a four bar linkage shown below .Grashofs law states

16、that one of the links (generally the shortest link) will be able to rotate continuously if the following condition is met. b (shortest link ) + c(longest link) a + dFour Inversions of a typical Four Bar LinkageNote: If the above condition was not met then only rocking motion would be possible for an

17、y link.Mechanical Advantage of 4 bar linkageThe mechanical advantage of a linkage is the ratio of the output torque exerted by the driven link to the required input torque at the driver link. It can be proved that the mechanical advantage is directly proportional to Sin( ) the angle between the coup

18、ler link(c) and the driven link(d), and is inversely proportional to sin( ) the angle between the driver link (b) and the coupler (c) .These angles are not constant so it is clear that the mechanical advantage is constantly changing.The linkage positions shown below with an angle = 0 o and 180 o has

19、 a near infinite mechanical advantage.These positions are referred to as toggle positions. These positions allow the 4 bar linkage to be used a clamping tools.The angle is called the transmission angle. As the value sin(transmission angle) becomes small the mechanical advantage of the linkage approa

20、ches zero. In these region the linkage is very liable to lock up with very small amounts of friction.When using four bar linkages to transfer torque it is generally considered prudent to avoid transmission angles below 450 and 500.In the figure above if link (d) is made the driver the system shown i

21、s in a locked position.The system has no toggle positions and the linkage is a poor design Freudensteins EquationThis equation provides a simple algebraic method of determining the position of an output lever knowing the four link lengths and the position of the input lever. Consider the 4 -bar link

22、age chain as shown below. The position vector of the links are related as follows l 1 + l 2 + l 3 + l 4 = 0 Equating horizontal distances l 1 cos 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 Equating Vertical distances l 1 sin 1 + l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 Assuming 1 = 1800 then sin 1 = 0 a

23、nd cos 1 = -1 Therefore - l 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 and . l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 Moving all terms except those containing l 3 to the RHS and Squaring both sides l 32 cos 2 3 = (l 1 - l 2 cos 2 - l 4 cos 4 ) 2l 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2Adding the above

24、 2 equations and using the relationshipscos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1the following relationship results.Freudensteins Equation results from this relationship as K 1 cos 2 + K2 cos 4 + K 3 = cos ( 2 - 4 )K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 )

25、/ 2 l 2 l 4 This equation enables the analytic synthesis of a 4 bar linkage. If three position of the output lever are required corresponding to the angular position of the input lever at three positions then this equation can be used to determine the appropriate lever lengths using three simultaneo

26、us equations. Velocity Vectors for LinksThe velocity of one point on a link must be perpendicular to the axis of the link, otherwise there would be a change in length of the link.On the link shown below B has a velocity of vAB = .AB perpendicular to A-B. The velocity vector is shown. Considering the

27、 four bar arrangement shown below. The velocity vector diagram is built up as follows: As A and D are fixed then the velocity of D relative to A = 0 a and d are located at the same point The velocity of B relative to a is vAB = .AB perpendicular to A-B. This is drawn to scale as shown The velocity o

28、f C relative to B is perpedicular to CB and passes through b The velocity of C relative to D is perpedicular to CD and passes through d The velocity of P is obtained from the vector diagram by using the relationship bp/bc = BP/BC The velocity vector diagram is easily drawn as shown. Velocity of slid

29、ing Block on Rotating LinkConsider a block B sliding on a link rotating about A. The block is instantaneously located at B on the link.The velocity of B relative to A = .AB perpendicular to the line. The velocity of B relative to B = v. The link block and the associated vector diagram is shown below

30、. Acceleration Vectors for LinksThe acceleration of a point on a link relative to another has two components: 1) the centripetal component due to the angular velocity of the link. 2.Length 2) the tangential component due to the angular acceleration of the link. The diagram below shows how to to cons

31、truct a vector diagram for the acceleration components on a single link.The centripetal acceleration ab = 2.AB towards the centre of rotation. The tangential component bb = . AB in a direction perpendicular to the link. The diagram below shows how to construct an acceleration vector drawing for a fo

32、ur bar linkage. For A and D are fixed relative to each other and the relative acceleration = 0 ( a,d are together ) The acceleration of B relative to A are drawn as for the above link The centripetal acceleration of C relative to B = v 2CB and is directed towards B ( bc1 ) The tangential acceleratio

33、n of C relative to B is unknown but its direction is known The centripetal acceleration of C relative to D = v 2CD and is directed towards d( dc2) The tangential acceleration of C relative to D is unknown but its direction is known. The intersection of the lines through c1 and c 2 locates c The loca

34、tion of the acceleration of point p is obtained by proportion bp/bc = BP/BC and the absolute acceleration of P = ap The diagram below shows how to construct and acceleration vector diagram for a sliding block on a rotating link. The link with the sliding block is drawn in two positions.at an angle d

35、The velocity of the point on the link coincident with B changes from .r =a b 1 to ( + d) (r +dr) = a b 2 The change in velocity b1b2has a radial component r d and a tangential component dr + r d The velocity of B on the sliding block relative to the coincident point on the link changes from v = a b

36、3 to v + dv = a b 4.The change in velocity = b3b4 which has radial components dv and tangential components v d The total change in velocity in the radial direction = dv- r d Radial acceleration = dv / dt = r d / dt = a - 2 r The total change in velocity in the tangential direction = v d + dr + r Tan

37、gential acceleration = v d / dt + dr/dt + r d / dt = v + v + r = r + 2 v The acceleration vector diagram for the block is shown belowNote : The term 2 v representing the tangential acceleration of the block relative to the coincident point on the link is called the coriolis component and results whe

38、never a block slides along a rotating link and whenever a link slides through a swivelling block连杆机构连杆存在于车库门装置，汽车擦装置，齿轮移动装置中。它是一给予很少关注的机械工程学的组成部分。联杆是具有两个或更多运动副元件的刚性机构，用它的连接是为了传递力或运动。在每个机器中，在运动期间，联杆或者占据一相对于地面的固定位置或者作为一个整体来承载机床。这些连杆是机器主体被称为固定连杆。基于由循环的或滑动的分界面的元件连接的布局被称作连接。这类旋转的和菱形的连接机构被称作低副。高副基于接触点或弯曲分

39、界面的。低副的例子包括铰链循环的轴承和滑道以及万向接头。高副的例子包括通信区主站和齿轮。动力分析得到，根据机件几何学有利条件研究是一特别的机构,它是识别输入角速度和角加速度等等的运动。运动合成作用是处理机构设计到完成完成要求的任务。这里, 两者的选择类型和新的机制尺寸可能是运动学的综合部份。平面的、空间性的和球面运动机构平面的机构是其中全部的点描述平面曲线是间隔和全部平面是共面的， 大多数连杆和机构被设计成这样，例如刨床体系。主要的理由是这个平面的体系对工程师来说更方便。计算机综合法对工程师来说空间性的装置会有更多的麻烦。平面低副机构被称作二维的连接装置。平面的连接仅仅包括旋转的和一对等截面的

40、使用。空间机构没有对相对运动的点的限制。平面的和球面运动机构是亚垫铁等锻工工具的空间机构。空间机构的连接不是被认为这时候被记录。球面运动机构有一接触点接通各个连杆，它是不动的并且平稳点在所有当中联杆场中工作。在所有机件当中，运动是同心并且由他们的盲区接通球面表现出来，它是集中于普遍的定位。空间机构的连接认为不是这时候被记录。可动性连杆在运动中所表现的自由度数是一个很重要的问题。为了使装置被送到指定位置应控制独立的活动自由度。它可能是由杆的数量和连接方式决定的。一自由连杆通常有3个自由度(x , y, )。由于自由度数的限制在n连杆装置中，通常把一个杆固定。自由度数=3(n-1).连接二连杆的机

41、构有两个自由度约束的增加。有两个约束的二连杆连接，其中一个自由度是来约束这个系统的。有一个约束的连杆机构的自由度是j1，有两个约束的连杆机构的自由度是j2。这个系统的自由度数可表示为 m = 3 (n-1) - 2 j 1 - j 2以下为可动的连杆机构装置的示例0是这个体系中可动的机构。系统中仅仅由一连杆的位置固定可以将可动1安装在固定位置。系统中需要一个可动的2与两个连杆来确定连接位置。这是个一般的规则，但也存在例外，它可以作为一个可动性连杆布局的很有用的参考。格朗定律当设计一连接连杆时，在连续地旋转连杆处，例如由一马达输入时，连线可以自由地旋转完全运行驱动是很重要的。如果连杆锁在任一点则

42、方案不会工作。四杆联动机构和grashof定律对这个情况进行提供了简单的测验。格朗的定律如下：b(短的链环)+c(长的链环)a+d四个典型的四连杆机构注意：如果非之上情况则只有连杆滑块机构可行。四连杆机构的优点四连杆机构按比例增大了施加在主动杆上的输入扭矩。可以证明其正比例系数是Sin( )其中是c、d 两杆之间的角度。反比例于sin( )。其中是b、c两杆之间的角度。这些角度不恒定，因此很明显，机构的优点是规律性的变幻。 如下图显示当角度=0 o或则=180 o时接近于无限增矩机构。这些位置是极限位置， 这些位置使四连杆机构可以用于夹具机构。角被叫做“传输角度”。当传输角度的sin值趋于无限

43、小时，机构的增距接近于0。在这样的情况下连杆容易因为很小的摩擦而产生自锁。一般来说，当使用四连杆机构时，避免采用低于400到500的传输角度。弗洛伊德方程这些方程提供了确定内外连杆位置及连杆长度的简单代数学方法。假设四连杆机构如下所示：四连杆的位置矢量如下：l 1 + l 2 + l 3 + l 4 = 0 水平位移方程：l 1 cos 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 垂直位移方程：l 1 sin 1 + l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 假设 1 = 1800 then sin 1 = 0 and

44、cos 1 = -1 Therefore 而l 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0方程两边同时消去l 3：l 32 cos 2 3 = (l 1 - l 2 cos 2 - l 4 cos 4 ) 2 l 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2由以上两式可得如下关系cos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1结果如下所示弗洛伊德方程得出这样的参数关系

45、结论K 1 cos 2 + K2 cos 4 + K 3 = cos ( 2 - 4 )K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4 这个方程符合四连杆机构的有限元分析。如果外连杆机构中的三个参量已知，那么可以由公式得出其他连杆的位置与长度参数。连杆的速度矢量杆上一点的速度必须与杆的轴向垂直否则连杆的长度将产生变化。在B下的连杆速度为vAB = .AB，方向垂直于AB杆，速度矢量图如下： 考虑到下面四连杆机构的实例，速度矢量图表示如下：1）A和D相连并固定，相对加速度=0，A和D位于同一点

46、2）B点相对A点加速是vAB = .AB垂直于AB杆。3）C点相对D点速度通过D点垂直于CD杆。4）P店读速度由速度矢量图和比例bp/bc = BP/BC获得。速度矢量简图如下所示：连杆上滑块的速度认为B滑块绕着A在连杆上滑动，滑块瞬间位移到B点。B点的速度为A = .AB并垂直于线的方向。其链接滑块和速度矢量图如下所示： 连杆的加速矢量杆上一点相对另一点的加速矢量由两部分组成：1）向心加速度由其角速度和连杆长度决定为 2.L2）角加速度由连杆角加速度度决定以下图表显示如何到构造一矢量图表下图显示如何构造单连杆机构的加速矢量向心加速度ab = 2.AB方向指向圆心，角加速度为bb = . AB

47、方向垂直于杆。下图显示如何构造四连杆机构的加速矢量画法1）. A和D相连并固定，相对加速度=0(a,d同)2）. B点相对A点加速在上面的杆上画出3）. B点相对C点向心加速度为：B = v 2CB，方向指向B。4）. B点相对C点角加速度未知但是方向已知5）. C点相对D点向心加速度为：D = v 2CD， 与d( dc2)方向相同。6）. C点相对D点角加速度未知但是方向已知7）. 通过线c1 和c 2的交叉点找出cP点的速度由比例bp/bc=bp/bc获得，且其绝对加速度为P = ap。下面的图表显示其构造方式和转杆上滑块的加速矢量图。两个滑块之间呈dw角。连杆上点的速度与B点变化一致，变化范围为.r =a b 1 到 ( + d) (r +dr) =