連桿彈簧復(fù)位自動調(diào)偏裝置設(shè)計

購買設(shè)計請充值后下載，資源目錄下的文件所見即所得，都可以點(diǎn)開預(yù)覽，資料完整，充值下載可得到資源目錄里的所有文件。【注】：dwg后綴為CAD圖紙，doc,docx為WORD文檔，原稿無水印，可編輯。具體請見文件預(yù)覽，有不明白之處，可咨詢QQ:12401814 連桿機(jī)構(gòu)連桿存在于車庫門裝置，汽車擦裝置，齒輪移動裝置中。它是一給予很少關(guān)注的機(jī)械工程學(xué)的組成部分。聯(lián)桿是具有兩個或更多運(yùn)動副元件的剛性機(jī)構(gòu)，用它的連接是為了傳遞力或運(yùn)動。在每個機(jī)器中，在運(yùn)動期間，聯(lián)桿或者占據(jù)一相對于地面的固定位置或者作為一個整體來承載機(jī)床。這些連桿是機(jī)器主體被稱為固定連桿?；谟裳h(huán)的或滑動的分界面的元件連接的布局被稱作連接。這類旋轉(zhuǎn)的和菱形的連接機(jī)構(gòu)被稱作低副。高副基于接觸點(diǎn)或彎曲分界面的。低副的例子包括鉸鏈循環(huán)的軸承和滑道以及萬向接頭。高副的例子包括通信區(qū)主站和齒輪。動力分析得到，根據(jù)機(jī)件幾何學(xué)有利條件研究是一特別的機(jī)構(gòu),它是識別輸入角速度和角加速度等等的運(yùn)動。運(yùn)動合成作用是處理機(jī)構(gòu)設(shè)計到完成完成要求的任務(wù)。這里, 兩者的選擇類型和新的機(jī)制尺寸可能是運(yùn)動學(xué)的綜合部份。平面的、空間性的和球面運(yùn)動機(jī)構(gòu)平面的機(jī)構(gòu)是其中全部的點(diǎn)描述平面曲線是間隔和全部平面是共面的， 大多數(shù)連桿和機(jī)構(gòu)被設(shè)計成這樣，例如刨床體系。主要的理由是這個平面的體系對工程師來說更方便。計算機(jī)綜合法對工程師來說空間性的裝置會有更多的麻煩。平面低副機(jī)構(gòu)被稱作二維的連接裝置。平面的連接僅僅包括旋轉(zhuǎn)的和一對等截面的使用?？臻g機(jī)構(gòu)沒有對相對運(yùn)動的點(diǎn)的限制。平面的和球面運(yùn)動機(jī)構(gòu)是亞墊鐵等鍛工工具的空間機(jī)構(gòu)?？臻g機(jī)構(gòu)的連接不是被認(rèn)為這時候被記錄。球面運(yùn)動機(jī)構(gòu)有一接觸點(diǎn)接通各個連桿，它是不動的并且平穩(wěn)點(diǎn)在所有當(dāng)中聯(lián)桿場中工作。在所有機(jī)件當(dāng)中，運(yùn)動是同心并且由他們的盲區(qū)接通球面表現(xiàn)出來，它是集中于普遍的定位?？臻g機(jī)構(gòu)的連接認(rèn)為不是這時候被記錄。可動性連桿在運(yùn)動中所表現(xiàn)的自由度數(shù)是一個很重要的問題。為了使裝置被送到指定位置應(yīng)控制獨(dú)立的活動自由度。它可能是由桿的數(shù)量和連接方式?jīng)Q定的。一自由連桿通常有3個自由度(x , y, )。由于自由度數(shù)的限制在n連桿裝置中，通常把一個桿固定。自由度數(shù)=3(n-1).連接二連桿的機(jī)構(gòu)有兩個自由度約束的增加。有兩個約束的二連桿連接，其中一個自由度是來約束這個系統(tǒng)的。有一個約束的連桿機(jī)構(gòu)的自由度是j1，有兩個約束的連桿機(jī)構(gòu)的自由度是j2。這個系統(tǒng)的自由度數(shù)可表示為 m = 3 (n-1) - 2 j 1 - j 2以下為可動的連桿機(jī)構(gòu)裝置的示例0是這個體系中可動的機(jī)構(gòu)。系統(tǒng)中僅僅由一連桿的位置固定可以將可動1安裝在固定位置。系統(tǒng)中需要一個可動的2與兩個連桿來確定連接位置。這是個一般的規(guī)則，但也存在例外，它可以作為一個可動性連桿布局的很有用的參考。格朗定律當(dāng)設(shè)計一連接連桿時，在連續(xù)地旋轉(zhuǎn)連桿處，例如由一馬達(dá)輸入時，連線可以自由地旋轉(zhuǎn)完全運(yùn)行驅(qū)動是很重要的。如果連桿鎖在任一點(diǎn)則方案不會工作。四桿聯(lián)動機(jī)構(gòu)和grashof定律對這個情況進(jìn)行提供了簡單的測驗(yàn)。格朗的定律如下：b(短的鏈環(huán))+c(長的鏈環(huán))a+d四個典型的四連桿機(jī)構(gòu)注意：如果非之上情況則只有連桿滑塊機(jī)構(gòu)可行。四連桿機(jī)構(gòu)的優(yōu)點(diǎn)四連桿機(jī)構(gòu)按比例增大了施加在主動桿上的輸入扭矩?？梢宰C明其正比例系數(shù)是Sin( )其中是c、d 兩桿之間的角度。反比例于sin( )。其中是b、c兩桿之間的角度。這些角度不恒定，因此很明顯，機(jī)構(gòu)的優(yōu)點(diǎn)是規(guī)律性的變幻。 如下圖顯示當(dāng)角度=0 o或則=180 o時接近于無限增矩機(jī)構(gòu)。這些位置是極限位置， 這些位置使四連桿機(jī)構(gòu)可以用于夾具機(jī)構(gòu)。角被叫做“傳輸角度”。當(dāng)傳輸角度的sin值趨于無限小時，機(jī)構(gòu)的增距接近于0。在這樣的情況下連桿容易因?yàn)楹苄〉哪Σ炼a(chǎn)生自鎖。一般來說，當(dāng)使用四連桿機(jī)構(gòu)時，避免采用低于400到500的傳輸角度。弗洛伊德方程這些方程提供了確定內(nèi)外連桿位置及連桿長度的簡單代數(shù)學(xué)方法。假設(shè)四連桿機(jī)構(gòu)如下所示：四連桿的位置矢量如下：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 假設(shè) 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方程兩邊同時消去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由以上兩式可得如下關(guān)系cos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1結(jié)果如下所示弗洛伊德方程得出這樣的參數(shù)關(guān)系結(jié)論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 這個方程符合四連桿機(jī)構(gòu)的有限元分析。如果外連桿機(jī)構(gòu)中的三個參量已知，那么可以由公式得出其他連桿的位置與長度參數(shù)。連桿的速度矢量桿上一點(diǎn)的速度必須與桿的軸向垂直否則連桿的長度將產(chǎn)生變化。在B下的連桿速度為vAB = .AB，方向垂直于AB桿，速度矢量圖如下： 考慮到下面四連桿機(jī)構(gòu)的實(shí)例，速度矢量圖表示如下：1）A和D相連并固定，相對加速度=0，A和D位于同一點(diǎn)2）B點(diǎn)相對A點(diǎn)加速是vAB = .AB垂直于AB桿。3）C點(diǎn)相對D點(diǎn)速度通過D點(diǎn)垂直于CD桿。4）P店讀速度由速度矢量圖和比例bp/bc = BP/BC獲得。速度矢量簡圖如下所示：連桿上滑塊的速度認(rèn)為B滑塊繞著A在連桿上滑動，滑塊瞬間位移到B點(diǎn)。B點(diǎn)的速度為A = .AB并垂直于線的方向。其鏈接滑塊和速度矢量圖如下所示： 連桿的加速矢量桿上一點(diǎn)相對另一點(diǎn)的加速矢量由兩部分組成：1）向心加速度由其角速度和連桿長度決定為 2.L2）角加速度由連桿角加速度度決定以下圖表顯示如何到構(gòu)造一矢量圖表下圖顯示如何構(gòu)造單連桿機(jī)構(gòu)的加速矢量向心加速度ab = 2.AB方向指向圓心，角加速度為bb = . AB方向垂直于桿。下圖顯示如何構(gòu)造四連桿機(jī)構(gòu)的加速矢量畫法1）. A和D相連并固定，相對加速度=0(a,d同)2）. B點(diǎn)相對A點(diǎn)加速在上面的桿上畫出3）. B點(diǎn)相對C點(diǎn)向心加速度為：B = v 2CB，方向指向B。4）. B點(diǎn)相對C點(diǎn)角加速度未知但是方向已知5）. C點(diǎn)相對D點(diǎn)向心加速度為：D = v 2CD， 與d( dc2)方向相同。6）. C點(diǎn)相對D點(diǎn)角加速度未知但是方向已知7）. 通過線c1 和c 2的交叉點(diǎn)找出cP點(diǎn)的速度由比例bp/bc=bp/bc獲得，且其絕對加速度為P = ap。下面的圖表顯示其構(gòu)造方式和轉(zhuǎn)桿上滑塊的加速矢量圖。兩個滑塊之間呈dw角。連桿上點(diǎn)的速度與B點(diǎn)變化一致，變化范圍為.r =a b 1 到 ( + d) (r +dr) = a b 2b1b2速度的變化分為沿桿方向的r d 及沿其切線方向的dr + r d。滑塊上B點(diǎn)的速度與連桿上相關(guān)點(diǎn)的變化有關(guān)v = a b 3 to v + dv = a b 4.沿著dv與v d 方向速度的變化= b3b4 。在速度切線方向總變化= dv- r d 加速度 = dv / dt = r d / dt = a - 2 r 速度在正切方向總變化= v d + dr + r 正切加速= v d / dt + dr/dt + r d / dt = v + v + r = r + 2 v 加速矢量圖表顯示如下：注: 其中2 v代表塊的正切加速度。每當(dāng)鏈接滑通過一個旋轉(zhuǎn)的塊，相對一致點(diǎn)沿著一旋轉(zhuǎn)鏈環(huán)一塊滑動。- 7 -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 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 MechanismsA 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 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 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 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 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 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 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 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 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 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 conditionGrashofs 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 + dFour Inversions of a typical Four Bar LinkageNote: If the above condition was not met then only rocking motion would be possible for any 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 coupler 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 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 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 linkage 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 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 ) 2l 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.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 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 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 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 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. 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 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 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 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 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 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 whenever a block slides along a rotating link and whenever a link slides through a swivelling block- 9 -遼寧工程技術(shù)大學(xué)課程設(shè)計遼寧工程技術(shù)大學(xué)課 程 設(shè) 計題 目：連桿彈簧復(fù)位自動調(diào)偏裝置班級：姓名：指導(dǎo)教師：完成日期：一、設(shè)計題目二、設(shè)計要求三、完成后應(yīng)上交的材料四、進(jìn)度安排五、指導(dǎo)教師評語成 績： 指導(dǎo)教師日期摘要在帶式輸送機(jī)運(yùn)轉(zhuǎn)過程中，輸送帶的縱向中心線偏離輸送機(jī)理論中心線的現(xiàn)象稱輸送帶跑偏。它的表現(xiàn)是輸送帶邊緣至托輥或滾筒邊緣的距離與理論值相比或大或小。輸送帶的跑偏會使輸送帶與機(jī)架、托輥支架相摩擦，造成邊膠磨損。嚴(yán)重的跑偏會使輸送帶翻邊，若在滾筒表面邊緣有凸起的螺絲頭、覆蓋膠層局部剝離、劃傷等事故。跑偏會導(dǎo)致輸送機(jī)的事故停機(jī)次數(shù)增多，影響生產(chǎn)；跑偏還可能引起物料外撒，使輸送機(jī)系統(tǒng)的運(yùn)營經(jīng)濟(jì)性下降。為此，曲柄連桿式自動調(diào)偏裝置最突出的特色是曲柄連桿機(jī)構(gòu)。調(diào)偏架通過滾動軸承安裝在底座上，兩端對稱連接制作的連桿，連桿末端固定立輥。安裝在曲柄上的兩側(cè)立輥靠近輸送帶邊緣，一有跑偏出現(xiàn)，輸送帶就壓向該側(cè)立輥，使立輥隨著曲柄向外轉(zhuǎn)動，曲柄和連桿拉動調(diào)偏架按輸送帶運(yùn)行方向旋轉(zhuǎn)一定角度，從而產(chǎn)生調(diào)偏。SummaryWhile the bringing type conveyer operates, the phenomenon that the longitudinal centre line of the conveyer belt deviates from the mechanism of transporting and talks about the centre line claims the conveyer belt runs partially. Its behavior is either large or small on compared with theory value to holding the distance on the roller or edge of cylinder the edge of conveyer belt. The running and leaning towards and will enable conveyer belt and framework, ask the roller support to rub of conveyer belt, cause the glue is worn and torn. A serious one run, lean towards, can enable conveyer belt turn-ups, if surface edge have protruding screw head, cover glue layers of part strip, accident of scratching etc. in cylinder. Run, lean towards, can lead to the fact accident of conveyer shut down number of times increase, influence and produce; Run, may cause supplies let go also simply, make conveyer economic decline of operation of system. For this reason, it is the crank connecting rod organization that the crank connecting rod type adjusts and leans towards the most outstanding characteristic of the device automatically. Adjust and lean towards the shelf and install the connecting rod made of the symmetrical connection of both ends on the base through the rolling bearing, the connecting rod sets up the roller terminal fixedly. Both sides that are installed on the crank set up the roller close to the edge of conveyer belt, run and simply appear, conveyer belt press to should incline, set up roller, make, set up roller rotate outwards with the crank, the crank and connecting rod spur and adjust and lean towards the shelf to rotate certain angle according to the conveyer belt operation direction, thus produced and adjusted partially. 目 錄1引言12系統(tǒng)分析22.1工作原理2 2.2初始條件22.3計算過程與分析22.3.1阻力計算22.3.2載荷計算32.3.3總阻力計算42.3.4彈簧彈力計算與選擇42.3.5調(diào)偏架的固定52.3.6兩連桿的固定及計算62.3.7其它桿件的安裝62.3.8立輥的規(guī)格63系統(tǒng)說明74. 心得體會8參考文獻(xiàn)8