机械毕业设计英文外文翻译314连杆机构

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

2、 bodies for 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 l

3、ink.an arrangement based on components 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. b5examples of lower pairs include hinges rotary

4、bearings, slideways , universal couplings. examples 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.   ki

5、nematic synthesis is the process of designing a mechanism to accomplish a desired task.   here, both choosing the types as well as the dimensions of the new mechanism can be part of kinematic synthesis.p1planar, spatial and spherical mechanismsa planar mechanism is one in which all particl

6、es describe plane curves is space and all of the planes are co-planar.  the majority 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 engine

7、er requiring computer synthesis.   planar mechanisms ultilising only lower pairs are called planar linkages. 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 mechanism

8、s are sub-sets of spatial mechanisms.spatial mechanisms / linkages are not considered on this pagespherical mechanisms 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

9、 are concentric and can be repesented by their shadow on a spherical surface which is centered on the common location.spherical mechanisms /linkages are not considered on this pagedxmobilityan important factor is considering a linkage is the mobility expressed as the number of degrees of freedom.

10、60; the mobility of a linkage is the number of input parameters which must be controlled independently in order to bring the device to 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 li

11、nk 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 linkage assembly with n links = dof = 3 (n-1) rtconnecting two links using a joint which has only on degree of freedom adds two constraints.

12、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 with 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 25pexample

13、s linkages showing the mobility are shown below. jla system with a mobility of 0 is a structure. a system with a mobility of 1 can be fixed in position my positioning only 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

14、nature and there are exceptions but it can provide a very useful initial guide as the the mobility of an arrangement of links.xhgrashof's lawwhen designing a linkage where the input linkage is continuously rotated e.g. driven by a motor it is important that the input link can freely rotate throu

15、gh complete revolutions.   the arrangement would not work if the linkage locks at any point.   for the four bar linkage grashof's law provides a simple test for this conditiongrashof's law is as follows: ldfor a planar four bar linkage, the sum of the shortest and longest

16、 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 .grashof's law states that one of the links (generally the shortest link) will be able to rotate continu

17、ously if the following condition is met. zzb (shortest link ) + c(longest link) < a + dfour inversions of a typical four bar linkagedvnote: if the above condition was not met then only rocking motion would be possible for any link.rqmechanical advantage of 4 bar linkagethe mechanical advantage of

18、 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 coupler link(c) and the driven link(d), and is inversely propor

19、tional 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.emthe linkage positions shown below with an angle = 0 o and 180 o has a near infinite mechanical advantage. &

20、#160;these positions are referred to as toggle positions.   these positions allow the 4 bar linkage to be used a clamping tools.sithe angle is called the "transmission angle".   as the value sin(transmission angle) becomes small the mechanical advantage of the linkage a

21、pproaches 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

22、 the driver the system shown is in a locked position.  the system has no toggle positions and the linkage is a poor design 6efreudenstein's equationthis equation provides a simple algebraic method of determining the position of an output lever knowing the four link lengths and the posi

23、tion of the input lever.   consider the 4 -bar linkage chain as shown below. kathe position vector of the links are related as follows y6l 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 m2equating vertical distances l 1 sin 1 + l 2 s

24、in 2 + l 3 sin 3 + l 4 sin 4 = 0 0yassuming 1 = 1800 then sin 1 = 0 and cos 1 = -1 therefore eu- 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 sqmoving all terms except those containing l 3 to the rhs and squaring both sides gml 32 cos 2 3 = (l 1 - l 2 cos 2

25、 - l 4 cos 4 ) 2til 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2adding the above 2 equations and using the relationshipscos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 )    and    sin2 + cos2 = 1the following relationship results.7efreudenstein's equation results from this relationship

26、 as lzk 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 zvthis equation enables the analytic synthesis of a 4 bar linkage.    if three position of the output lever are requi

27、red 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 simultaneous equations. nrvelocity vectors for linksthe velocity of one point on a link must be perpendicular to the axis of the link,

28、 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. 1nconsidering the four bar arrangement shown below. the velocity vector diagram is built up as follows: fj· as a and d

29、are fixed then the velocity of d relative to a = 0 a and d are located at the same point tf· the velocity of b relative to a is vab = .ab perpendicular to a-b. this is drawn to scale as shown hb· the velocity of c relative to b is perpedicular to cb and passes through b v7· the veloci

30、ty of c relative to d is perpedicular to cd and passes through d 83· the velocity of p is obtained from the vector diagram by using the relationship bp/bc = bp/bc mzthe velocity vector diagram is easily drawn as shown. avvelocity of sliding block on rotating linkconsider a block b sliding on a

31、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. oracceleration vectors for linksthe accel

32、eration of a point on a link relative to another has two components: 2m· 1) the centripetal component due to the angular velocity of the link. 2.length gi· 2) the tangential component due to the angular acceleration of the link. ue· the diagram below shows how to to construct a vector

33、 diagram for the acceleration components on a single link.the centripetal acceleration ab' = 2.ab towards the centre of rotation.   the tangential component b'b = . ab in a direction perpendicular to the link. iathe diagram below shows how to construct an acceleration vector drawin

34、g for a four bar linkage. ww· for a and d are fixed relative to each other and the relative acceleration = 0 ( a,d are together ) as· the acceleration of b relative to a are drawn as for the above link oo· the centripetal acceleration of c relative to b = v 2cb and is directed towards

35、 b ( bc1 ) bk· the tangential acceleration of c relative to b is unknown but its direction is known pg· the centripetal acceleration of c relative to d = v 2cd and is directed towards d( dc2) 3c· the tangential acceleration of c relative to d is unknown but its direction is known. h8&

36、#183; the intersection of the lines through c1 and c 2 locates c v4the location of the acceleration of point p is obtained by proportion bp/bc = bp/bc and the absolute acceleration of p = ap j0the diagram below shows how to construct and acceleration vector diagram for a sliding block on a rotating

37、link. the link with the sliding block is drawn in two positions.at an angle dthe 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 sli

38、ding block relative to the coincident point on the link changes from v = a b 3 to v + dv = a b 4.the change in velocity = b3b4 which has radial components dv and tangential components v d xvthe total change in velocity in the radial direction = dv- r d brradial acceleration = dv / dt = r d / dt = a

39、- 2 r pnthe total change in velocity in the tangential direction = v d + dr + r djtangential acceleration = v d / dt + dr/dt + r d / dt = v + v + r = r + 2 v qfthe acceleration vector diagram for the block is shown below4bnote : the term 2 v representing the tangential acceleration of the block rela

40、tive to the coincident point on the link is called the coriolis component and results whenever a block slides along a rotating link and whenever a link slides through a swivelling blockix连杆机构连杆存在于车库门装置,汽车擦装置,齿轮挪动部分.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,联杆或者占据一相关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

41、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,依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制定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.wt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制定费事.平面低副机构被称作二维的连接装置.平面的连接

42、仅仅包括旋转的和一对等截面的使用.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.kp可动性3个自由度(x , y, ).由于自由度数的限制在n=3(n-1).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 2yl以下为可动的连杆机构装置的例如01必必需要一个可动的2与两个连杆来确定连接位置.

43、这是个一般的规那么,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照.ch格朗定律当制定一连接连杆时,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.假设连杆锁在任一点那么grashof定律对这个状况进展提供了简单的测验.qd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：假设非之上状况那么只有连杆滑块机构可行.四连杆机构的优点证实其正比例系数是sin( )其中是c、d sin( ).其中是b、c两杆之间的角度.这些角度不恒定,因此很显然,机构的优点是规律性的变幻.e8 如以以下列图显示当角度=0 o或那么=180

44、 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.s4角被叫做“传输角度.当传输角度的sin值趋于无限小时,机构的增距接近于0.在这样的状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,防止采纳低于400到500的传输角度.50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 程度位移方程：l 1 cos 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 jw垂直位移方程：l 1 sin 1 + l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 xs假设 1 = 1800 then sin 1 = 0 and 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 = 0lo方程两边同时消去l 3：l 3