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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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遼寧工程技術(shù)大學(xué)課程設(shè)計(jì)
遼寧工程技術(shù)大學(xué)
課 程 設(shè) 計(jì)
題 目：連桿彈簧復(fù)位自動(dòng)調(diào)偏裝置
班 級(jí)：
姓 名：
指導(dǎo)教師：
完成日期：
一、設(shè)計(jì)題目
二、設(shè)計(jì)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(píng)語(yǔ)
成 績(jī)：
指導(dǎo)教師
日 期
摘 要
在帶式輸送機(jī)運(yùn)轉(zhuǎn)過(guò)程中，輸送帶的縱向中心線(xiàn)偏離輸送機(jī)理論中心線(xiàn)的現(xiàn)象稱(chēng)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(huì)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(huì)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(huì)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(yíng)經(jīng)濟(jì)性下降。
為此，曲柄連桿式自動(dòng)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(guò)滾動(dòng)軸承安裝在底座上，兩端對(duì)稱(chēng)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(dòng)，曲柄和連桿拉動(dòng)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。
Summary
While 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…………………………………………………………1
2系統(tǒng)分析……………………………………………………2
2.1工作原理……………………………………………2
2.2初始條件……………………………………………2
2.3計(jì)算過(guò)程與分析……………………………………2
2.3.1阻力計(jì)算……………………………………2
2.3.2載荷計(jì)算……………………………………3
2.3.3總阻力計(jì)算…………………………………4
2.3.4彈簧彈力計(jì)算與選擇………………………4
2.3.5調(diào)偏架的固定………………………………5
2.3.6兩連桿的固定及計(jì)算………………………6
2.3.7其它桿件的安裝……………………………6
2.3.8立輥的規(guī)格…………………………………6
3．系統(tǒng)說(shuō)明…………………………………………………7
4. 心得體會(huì)…………………………………………………8
參考文獻(xiàn)………………………………………………………8
遼寧工程技術(shù)大學(xué)課程設(shè)計(jì)
1引言
皮帶運(yùn)輸機(jī)在運(yùn)輸?shù)倪^(guò)程中由于運(yùn)輸?shù)臇|西沉重，和貨物在往運(yùn)輸帶上放的時(shí)候的方向方法不同，所以會(huì)導(dǎo)致運(yùn)輸帶的跑偏，我們這里研究的就是防止運(yùn)輸帶的跑偏過(guò)程。
在帶式輸送機(jī)運(yùn)轉(zhuǎn)過(guò)程中，輸送帶的縱向中心線(xiàn)偏離輸送機(jī)理論中心線(xiàn)的現(xiàn)象稱(chēng)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(huì)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(huì)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(huì)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(yíng)經(jīng)濟(jì)性下降。
為此，曲柄連桿式自動(dòng)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(guò)滾動(dòng)軸承安裝在底座上，兩端對(duì)稱(chēng)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(dòng)，曲柄和連桿拉動(dòng)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。
特點(diǎn)：
1 這個(gè)裝置是無(wú)源裝置，結(jié)構(gòu)簡(jiǎn)單，經(jīng)濟(jì)實(shí)惠。
2 通過(guò)連桿作用，比不使用連桿的調(diào)偏方式大，作用快。
3 通過(guò)復(fù)位彈簧的作用，可以減少?zèng)]有復(fù)位彈簧時(shí)，皮帶在托輥上左右蛇形的運(yùn)動(dòng)狀態(tài)。
1—立輥 2—調(diào)偏架 3—支架 4—底座
5—中間轉(zhuǎn)動(dòng)部分 6—連桿1 7—連桿2
整體調(diào)偏工作原理圖
2系統(tǒng)分析
2.1工作原理
連桿彈簧復(fù)位式自動(dòng)調(diào)偏裝置最突出的特色是連桿彈簧復(fù)位機(jī)構(gòu)。由上圖分析，調(diào)偏架通過(guò)滑動(dòng)軸承固定在軌道上，兩端對(duì)稱(chēng)連接連桿，連桿2末端固定立輥。安裝在連桿上的兩側(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，使立輥立即隨著連桿2向外轉(zhuǎn)動(dòng)，連桿拉動(dòng)支架按輸送帶的運(yùn)行方向旋轉(zhuǎn)一定角度，從而通過(guò)受力分析，產(chǎn)生調(diào)偏作用，當(dāng)皮帶回到中心時(shí)，在一側(cè)彈簧的作用下，支架又被拉回到了不偏轉(zhuǎn)的位置上。從而也就完成了整個(gè)的調(diào)偏過(guò)程
2.2初始條件：
皮帶帶寬B＝1200mm 輸送量Q＝2000t/h 帶速V＝2.5m/s 托輥間距a=1.5m
2.3計(jì)算過(guò)程與分析
2.3.1阻力計(jì)算
（1）輸送機(jī)所受的主要阻力（ISO5048）：
FH＝fLg[qR0+qRu+(2qB+qG)cosδ
f 模擬摩擦系數(shù)
f＝0.020 工作環(huán)境良好、制造、安裝良好，帶速低，物料內(nèi)摩擦系數(shù)小。
L 托輥長(zhǎng)度
g 重力加速度
qR0輸送機(jī)承載分支每米托輥旋轉(zhuǎn)部分質(zhì)量 查表取20kg/m
qRu輸送機(jī)回程分支每米托輥旋轉(zhuǎn)部分質(zhì)量
qB 每米輸送帶質(zhì)量
查表為23kg/m
qG 物料每米質(zhì)量
=222.22kg/m
δ 機(jī)身傾角 δ取3°
最后求得
FH＝26.57N
(2)輸送機(jī)所受的特殊阻力：
FT=μN(yùn)=
μ皮帶與托輥的摩擦系數(shù)
μ取0.4
b1 前傾托輥與輸送帶接觸長(zhǎng)度
當(dāng)跑偏0.075m時(shí)通過(guò)計(jì)算知
b1＝0.443m
最后求得
FT=359.66N
2.3.2載荷計(jì)算
（1）由CEMA所給出的托輥載荷的計(jì)算方法去計(jì)算承載托輥：
由公式：
K1物料系數(shù)
等長(zhǎng)三輥槽形托輥組
K1＝0.7
K2輸送帶系數(shù)
等長(zhǎng)三輥槽形托輥組
K2＝0.4
通過(guò)計(jì)算得
P中＝1977.05N
P側(cè)＝＝423.65N
CEMA方法中：
作用在中間托輥上的力就70%，作用在側(cè)托輥上的力為15%.
2.3.3總阻力計(jì)算
帶在未跑偏時(shí)，所受的總的阻力為：
F總＝FH+FT+(P中+ P側(cè)*2)μ=2590.9N
當(dāng)帶跑偏75mm時(shí)，額外增加的對(duì)側(cè)托輥的載荷
F額= =318.88N
F摩側(cè)＝127.55N
F總半＝＝1295.45N這里取1300N
2.3.4彈簧彈力計(jì)算與選擇
故彈簧的最小應(yīng)提供的力為1300N，才能保證彈簧能夠?qū)⒄{(diào)心托輥拉回或推回到原位，這里取
F彈＝1300N
彈簧的選擇：
最小工作載荷：
P1＝1300N/m
由表查得：
彈簧截面圓直徑 d=8mm
中徑 D=80mm
工作極限載荷 Pj=1504N
工作極限載荷下的變形量 fj=19.03N/mm
具體計(jì)算：
工作行程先取 h=50mm
最大工作載荷 Pn=1300N
初算彈簧剛度 P==10.84N/mm
由表通過(guò)彈簧截面直徑和中徑得到
單圈剛度 Pd=79N/mm，
通過(guò)計(jì)算有效圈數(shù)
取標(biāo)準(zhǔn)值n=10
總?cè)?shù)
彈簧剛度
工作極限載荷下的變形量
節(jié)距
自由高度
彈簧外徑
彈簧內(nèi)經(jīng)
螺旋角
展開(kāi)長(zhǎng)度
最小載荷時(shí)的高度
最大載荷時(shí)的高度
極限載荷時(shí)的高度
實(shí)際工作行程
故彈簧的總技術(shù)要求如下：
1.總?cè)?shù)： n1=12
2.旋向?yàn)橛倚?
3.展開(kāi)長(zhǎng)度 L=3031.79mm
4.硬度為 HRC=4550
5.彈簧固定在兩桿中間，起到當(dāng)支架跑偏時(shí)復(fù)位的作用。
2.3.5調(diào)偏架的固定
調(diào)偏架通過(guò)滑動(dòng)軸承，固定在離支架500mm的位置上。調(diào)偏架的高度由余弦定理求出，調(diào)偏架的高度為424mm.通過(guò)滑動(dòng)軸承固定在底座上。
2.3.6兩連桿的固定及計(jì)算
由于支架只能旋轉(zhuǎn)15°，所以需保證兩桿之間的夾角為105°，由余弦定理
c=500 θ＝105° a=b=314.98mm
故桿應(yīng)為
314.98mm
當(dāng)連桿拉動(dòng)支架旋轉(zhuǎn)15°時(shí)，支架擺過(guò)的距離為mm
通過(guò)上面的計(jì)算，知彈簧被壓縮了50mm，提供了1300N的力，這個(gè)力恰好能使偏轉(zhuǎn)的支架回到平衡位置。
由于彈簧的自由長(zhǎng)度為 H0＝282.3mm
由余弦定理算出固定位置為距離兩連桿連接處的178.48mm
2.3.7其它桿件的安裝
兩立輥與皮帶接觸就留用10㎜間距。
2.3.8立輥的規(guī)格
立輥依據(jù)經(jīng)驗(yàn)及所受的載荷和帶寬確定為60mm.直徑為25mm
3．系統(tǒng)說(shuō)明
安裝調(diào)試完畢后，自動(dòng)糾偏裝置的維護(hù)與保養(yǎng)
一、 緊固螺栓要每周檢查一次，防止松動(dòng)變位。
二、 每周應(yīng)將傳動(dòng)部分的灰塵進(jìn)行一次清除。
三、 每當(dāng)設(shè)備檢修時(shí)，應(yīng)對(duì)自動(dòng)糾偏裝置進(jìn)行除銹，并涂刷面漆。
四、 使用過(guò)程中要經(jīng)常檢查，如發(fā)現(xiàn)有損壞部件，要及時(shí)更換。
4. 心得體會(huì)
經(jīng)過(guò)兩個(gè)星期的努力,我終于將機(jī)械設(shè)計(jì)課程設(shè)計(jì)做完了.在這次作業(yè)過(guò)程中,我遇到了許多困難,一遍又一遍的計(jì)算,一次又一次的設(shè)計(jì)方案修改這都暴露出了前期我在這方面的知識(shí)欠缺和經(jīng)驗(yàn)不足。至于畫(huà)裝配圖和零件圖,由于前期計(jì)算比較充分,整個(gè)過(guò)程用時(shí)不到一周,在此期間,我還得到了許多同學(xué)和老師的幫助.在此我要向他們表示最誠(chéng)摯的謝意.
盡管這次作業(yè)的時(shí)間是漫長(zhǎng)的,過(guò)程是曲折的,但我的收獲還是很大的.不僅對(duì)制圖有了更進(jìn)一步的掌握;Auto CAD ,Word這些僅僅是工具軟件,熟練掌握也是必需的.對(duì)我來(lái)說(shuō),收獲最大的是方法和能力.那些分析和解決問(wèn)題的方法與能力.在整個(gè)過(guò)程中,我發(fā)現(xiàn)像我們這些學(xué)生最最缺少的是經(jīng)驗(yàn),沒(méi)有感性的認(rèn)識(shí),空有理論知識(shí),有些東西很可能與實(shí)際脫節(jié).總體來(lái)說(shuō),我覺(jué)得做這種類(lèi)型的作業(yè)對(duì)我們的幫助還是很大的,它需要我們將學(xué)過(guò)的相關(guān)知識(shí)都系統(tǒng)地聯(lián)系起來(lái),從中暴露出自身的不足,以待改進(jìn).有時(shí)候,一個(gè)人的力量是有限的,合眾人智慧,我相信我們的作品會(huì)更完美!
參考文獻(xiàn)
[1]宋偉剛，通用帶式輸送機(jī)設(shè)計(jì)，機(jī)械工業(yè)出版社，2006.5
[2]張鉞 ，新型帶式輸送機(jī)設(shè)計(jì)手冊(cè)，冶金工業(yè)出版社，2001.2
[3]成大先，機(jī)械設(shè)計(jì)手冊(cè)單行本彈簧起重運(yùn)輸件五金件，化學(xué)工業(yè)出版社，2004.1
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