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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 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, θ ). ?? 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...
Grashof's 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 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 condition
Grashof'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 ..Grashof's 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( β ) the angle between the coupler link(c) and the driven link(d), and is inversely proportional to sin( α ) 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 α = 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 β 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
Freudenstein's 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..
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 θ 1 + l 2 cos θ 2 + l 3 cos θ 3 + l 4 cos θ 4 = 0
Equating Vertical distances
l 1 sin θ 1 + l 2 sin θ 2 + l 3 sin θ 3 + l 4 sin θ 4 = 0
Assuming θ 1 = 1800 then sin θ 1 = 0 and cosθ 1 = -1 Therefore
- l 1 + l 2 cosθ 2 + l 3 cosθ 3 + l 4 cos θ 4 = 0
and .. l 2 sin θ 2 + l 3 sin θ 3 + l 4 sin θ 4 = 0
Moving all terms except those containing l 3 to the RHS and Squaring both sides
l 32 cos 2 θ 3 = (l 1 - l 2 cos θ 2 - l 4 cos θ 4 ) 2
l 32 sin 2 θ 3 = ( - l 2 sin θ 2 - l 4 sin θ 4) 2
Adding the above 2 equations and using the relationships
cos ( θ 2 - θ 4 ) = cos θ 2 cos θ 4 + sin θ 2sin θ 4 ) ?? and ?? sin2θ + cos2θ = 1
the following relationship results..
Freudenstein's Equation results from this relationship as
K 1 cos θ 2 + K2 cos θ 4 + K 3 = cos ( θ 2 - θ 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 = ω.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 = ω.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 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 = ω.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.ω 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' = ω 2.AB towards the centre of rotation.?? The tangential component b'b = α. 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 dω
The velocity of the point on the link coincident with B changes from ω.r =a b 1 to ( ω + dω) (r +dr) = a b 2
The change in velocity b1b2has a radial component ωr d θ and a tangential component ωdr + r dω
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 θ
The total change in velocity in the radial direction = dv- ω r d θ
Radial acceleration = dv / dt = ω r d θ / dt = a - ω2 r
The total change in velocity in the tangential direction = v dθ + ω dr + r α
Tangential acceleration = v dθ / dt + ω dr/dt + r d ω / dt
= v ω + ω v + r α = α r + 2 v ω
The acceleration vector diagram for the block is shown below
Note : The term 2 v ω 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
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湘潭大學(xué)興湘學(xué)院
畢業(yè)論文(設(shè)計(jì))任務(wù)書(shū)
論文(設(shè)計(jì))題目: 小型谷類(lèi)干燥機(jī)的設(shè)計(jì)-振動(dòng)篩分部分
學(xué)號(hào): 2006183906 姓名: 譚理想 專(zhuān)業(yè): 機(jī)械設(shè)計(jì)制造及其自動(dòng)化
指導(dǎo)教師: 周后明 系主任: 周友行
一、主要內(nèi)容及基本要求
小型谷類(lèi)干燥機(jī)的設(shè)計(jì)包括干燥前的振動(dòng)篩分、干燥本體二部分組成。本設(shè)計(jì)為振動(dòng)篩分部分的設(shè)計(jì),其主要技術(shù)指標(biāo)與要求如下:
1、最大凈谷處理量:600-800kg/小時(shí);
2、篩面層數(shù):2層;
3、振動(dòng)頻率:600-700HZ
4、振動(dòng)類(lèi)型:機(jī)械式振動(dòng),要求構(gòu)造簡(jiǎn)單、成本低
5、整機(jī)形式:立式
設(shè)計(jì)要求:
1、完成振動(dòng)篩的方案設(shè)計(jì)和選型論證
2、振動(dòng)篩的結(jié)構(gòu)設(shè)計(jì),繪制部件裝配圖和主要零件圖,圖紙總量折合成A0,不少于2張
3、撰寫(xiě)設(shè)計(jì)說(shuō)明書(shū),關(guān)鍵零件應(yīng)進(jìn)行強(qiáng)度和剛度計(jì)算,說(shuō)明書(shū)字?jǐn)?shù)不少于1~5萬(wàn)
4、完成資料查閱和3000字的文獻(xiàn)翻譯
二、重點(diǎn)研究的問(wèn)題
振動(dòng)篩分機(jī)的結(jié)構(gòu)設(shè)計(jì)及相關(guān)強(qiáng)度校核。
三、進(jìn)度安排
序號(hào)
各階段完成的內(nèi)容
完成時(shí)間
1
查閱資料、調(diào)研
第1,2周
2
制訂設(shè)計(jì)方案
第3,4周
3
分析與計(jì)算
第5,6周
4
繪部件裝配圖
第7,8、9周
5
繪零件圖
第10,11周
6
撰寫(xiě)設(shè)計(jì)說(shuō)明書(shū)
第12,13周
7
準(zhǔn)備答辯材料
第14周
8
畢業(yè)答辯
第15周
四、應(yīng)收集的資料及主要參考文獻(xiàn)
1、機(jī)械設(shè)計(jì)手冊(cè)
2、機(jī)械傳動(dòng)設(shè)計(jì)手冊(cè)
3、王峰,王皓. 篩分機(jī)械[M] . 北京:機(jī)械工業(yè)出版社,1998.
4、任德樹(shù). 粉碎篩分原理與設(shè)備[M]. 北京:冶金工業(yè)出版社,1984.472~580.
5、張露霞,李云峰. 振動(dòng)篩篩分效率的影響因素分析[J]. 煤礦機(jī)械,2008,(11):
74~76.
6、石秀東,錢(qián)林方,徐亞棟等. 離散式振動(dòng)篩篩分性能指標(biāo)及篩面配置研究[J]. 金屬礦山,2008,(12):113~115.
7、劉家平,王世杰. 篩分機(jī)驅(qū)動(dòng)功率計(jì)算[J]. 機(jī)械傳動(dòng),2002,26(4):31~32
8、杜皖寧. 篩分機(jī)械處理量的合理選擇[J]. 金屬礦山,2001,(4):55~56
9、夏玉林,嚴(yán)志全,朱滿平. 一種新型篩分機(jī)的開(kāi)發(fā)與設(shè)計(jì)[J]. 機(jī)械研究與應(yīng)用,2005,18(5):114~115
湘潭大學(xué)興湘學(xué)院
畢業(yè)論文(設(shè)計(jì))鑒定意見(jiàn)
學(xué)號(hào): 2006183906 姓名: 譚理想 專(zhuān)業(yè): 機(jī)械設(shè)計(jì)制造及其自動(dòng)化
畢業(yè)論文(設(shè)計(jì)說(shuō)明書(shū)) 42 頁(yè) 圖 表 7 張
論文(設(shè)計(jì))題目: 小型谷類(lèi)干燥機(jī)——直線振動(dòng)篩選部分的設(shè)計(jì)
內(nèi)容提要:直線振動(dòng)篩選的運(yùn)送部分的作用十分重要,經(jīng)過(guò)這樣的反復(fù)處理最終將物料
全部做成符合要求的產(chǎn)品。本課題的振動(dòng)篩為自同步雙振動(dòng)電機(jī)驅(qū)動(dòng)的,其特點(diǎn)是結(jié)構(gòu)
簡(jiǎn)單、安裝方便、成本低、容易操作及維護(hù)等。其篩箱為板梁鉚焊組合結(jié)構(gòu),由主副側(cè)
板、管梁、入料擋板出料板、篩板等組成,側(cè)板選用低合金壓力容器鋼板,強(qiáng)度高、可
焊性好,周邊折彎并在振動(dòng)方向及沿縱向連接多根角鋼,使側(cè)板剛度大大增強(qiáng),有利于
強(qiáng)度的提高和噪音的降低。管梁由法蘭盤(pán)、無(wú)縫鋼管、加強(qiáng)槽鋼等組成,重量輕、強(qiáng)度大,便于制造安裝,具有互換性。加強(qiáng)槽鋼上有T形孔,使用T形螺栓,便于篩板的安裝維護(hù),消除U形螺栓對(duì)管梁的磨損。工作原理:兩臺(tái)振動(dòng)電機(jī)的型號(hào)相同,可以產(chǎn)生一種組合的直線振動(dòng)。
指導(dǎo)教師評(píng)語(yǔ)
譚理想同學(xué)在畢業(yè)設(shè)計(jì)過(guò)程中,態(tài)度認(rèn)真,并對(duì)畢業(yè)設(shè)計(jì)任務(wù)作了認(rèn)真的理解與分析。通過(guò)查閱有關(guān)干燥機(jī)和振動(dòng)篩分機(jī)的文獻(xiàn)資料,提出了一可行的篩分機(jī)械的結(jié)構(gòu)設(shè)計(jì)方案。完成了振動(dòng)篩主要設(shè)計(jì)參數(shù)的計(jì)算,繪制了設(shè)計(jì)二維裝配圖主要零部件圖紙,其工作量適當(dāng),圖紙、說(shuō)明書(shū)符合要求,達(dá)到了本科生畢業(yè)設(shè)計(jì)的目的和要求。同意答辯,推薦等級(jí)為中。
指導(dǎo)教師:
年 月 日
答辯簡(jiǎn)要情況及評(píng)語(yǔ)
答辯小組組長(zhǎng):
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答辯委員會(huì)意見(jiàn)
答辯委員會(huì)主任:
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