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机械毕业设计英文翻译
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英文翻译外文文献翻译231连杆机构,机械毕业设计英文翻译
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1 Link mechanism Linkages 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 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 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. Examples of lower pairs include hinges rotary 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. Kinematic 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. Planar, Spatial and Spherical Mechanisms A planar mechanism is one in which all particles 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 engineer requiring computer synthesis. Planar mechanisms ultilising only lower pairs are called planar linkages. Planar linkages only involve the use of revolute and prismatic pairs A 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 page Spherical mechanisms has one point on each linkage which is stationary and the stationary point nts 2 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.Spherical mechanisms /linkages are not considered on this page Mobility An 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 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 linkage 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 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 2 Examples 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 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 nature and there are exceptions but it can provide a very useful initial guide as the the mobility of an arrangement of links. Grashofs Law When 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 through complete revolutions. The nts 3 arrangement would not work if the linkage locks at any point. For the four bar linkage Grashofs law provides a simple test for this condition Grashofs 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 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 + d Four Inversions of a typical Four Bar Linkage Note: If the above condition was not met then only rocking motion would be possible for any link. Mechanical Advantage of 4 bar linkage The 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 coupler link(c) and the driven link(d), and is inversely proportional to sin( ) the angle between the driver link (b) and the coupler nts 4 (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 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 approaches 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 is in a locked position. The system has no toggle positions and the linkage is a poor design Freudensteins Equation This 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 linkage chain as shown below. nts 5 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 and 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 ) 2 l 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2 Adding the above 2 equations and using the relationships cos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1 nts 6 the 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 ) / 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 simultaneous equations. Velocity Vectors for Links The 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 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 nts 7 The velocity of 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 sliding Block on Rotating Link Consider 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. Acceleration Vectors for Links The 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 nts 8 The diagram below shows how to to construct 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 four 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 acceleration 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 location of the acceleration of point p is obtained by proportion bp/bc = BP/BC and the absolute acceleration of P = ap nts 9 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 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 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 Tangential acceleration = v d / dt + dr/dt + r d / dt = v + v + r = r + 2 v The acceleration vector diagram for the block is shown below nts 10 Note : 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 whenever 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nts 11 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，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nts 12 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。 nts 13 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 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 nts 14 - l 4 sin 4) 2 由以上两式可得如下关系 cos ( 2 - 4 ) = cos
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