机械原理大作业连杆19.doc

哈工大机械原理大作业连杆机构说明书连杆机构运动分析1.运动分析题目（19）如图1所示机构，已知机构各构件的尺寸为AB=112mm，=90，AD=264mm，DG=278mm，EF=116mm，FG=24mm，BC=CE=CD=200mm，构件1的角速度1=10rad/s，试求构件2上点E的轨迹及构件5的角位移、角速度和角加速度，并对计算结果进行分析。图12.机构的结构分析，组成机构的基本杆组划分该构件由级杆组RR（机架、原动件1）、级杆组RRR（杆2、杆3）和级杆组RRR（杆4、杆5）组成。杆组如图2所示：图23.各基本杆组的运动分析数学模型3.1. 原动件1已知原动件1的转角1=0360，初始角度=0，原动件1的角速度1=10rad/s，角加速度=0. 所以运动副A的位置坐标为：XA=264mm, YA=0mm运动副A的速度：vXA=0, vYA=0，运动副A的加速度：aXA=0, aYA=0，原动件杆1长度：lAB=112mm。则可得B点位置方程：XB=XA+lABcos1, YB=XA+lABsin1。对上述位移公式求导可得B点速度方程：VXB=VXA-1lsin1, VYB=VYA+1lcos1再求导得B点加速度方程：aXB=XA-12 lAB cos1-lAB sin1， aYB=YA-12 lAB sin1+lAB cos13.2级杆组RRR（杆2、杆3）运动副D点位置坐标XD=0，YD=0，该点的速度、加速度均为0，杆BC长lBC=200mm，杆CD长lCD=200mm。可求得BC杆相对于X轴正方向转角：2=2arctanB0+A02+B02-C02A0+B0以及CD杆相对于X轴正方向转角：3=arctanyC-yDxC-xD式中，A0=2lBC（XD-XB）,B0=2 lBC（YD-YB）,C0=lBC2+lBD2-lCD2,lBD2=(xD-xB)2+(yD-yB)2求导可得BC杆2，2和CD杆3，3。则运动副C的位置坐标为：XC=XB+lBCcos2， YC=YB+lBCsin2求导得到vXc、vYC 以及aXC 、aYC。3.3构件2上点E的运动可以认为BE为1级机构，则点E的位置方程为：XE=XB+lBEcos2， YE=YB+lBEsin2点E的速度方程：vxE= vxB 2 lBE sin2, vyE= vyB+2 lBE cos2点E的加速度方程：axE= axB-22 lBE cos2-2lBE sin2, ayE=aYB-22 lBE sin2+2lBE cos23.4级杆组RRR（杆4、杆5）运动副E点位置方程、速度、加速度由3.3可知，则杆EF长LEF=116mm，杆FG长LGF=28mm。可求得EF杆相对于X轴正方向转角4=2arctanB1+A12+B12-C12A1+B1FG杆相对于X轴正方向转角5=arctanyF-yGxF-G其中A1=2lEF（XG-XE）,B1=2 lEF（YG-YE）,C1=lEF2+lEG2-lFG2, lEG2=(xG-xE)2+(yG-yE）2求导可得EF杆5，5和CD杆5，5，则运动副F的位置坐标为XF=xE+lEFcos4， YF=YE+lEFsin4最后求导得到vXF、vYF 以及aXF 、aYF。题目所要求的，构件5的角位移、角速度和角加速度，即为运动副F的角位移5，5和5。4.建立坐标系建立以点D为原点、AD为x轴的固定平面直角坐标系D-xy，如图2所示。图25.计算编程Option Explicit 定义自变量Dim xA , yA ,vxA ,vyA ,axA,ayA As DoubleDim xB , yB,vxB , vyB ,axB ,ayB As DoubleDim xC , yC, vxC , vyC ,axC,ayC As DoubleDim xD , yD , vxD, vyD , axD , ayD As DoubleDim xE,yE , vxE,vyE,axE ,ayE As DoubleDim xF,yF , vxF , vyF , axF,ayF As DoubleDim xG ,yG,vxG,vyG , axG , ayG As DoubleDim lab, lbc,lcd , lce , lef , lfg As Double 杆的长度Dim lbd , lde , leg As Double 距离Dim delt As Double 构件1的初始角位移Dim fab , fbc , fcd , fef , ffg , fbd , feg As Double 角位移Dim wab , wbc , wcd, wef, wfg As Double 杆的角速度Dim eab , ebc ,ecd ,eef , efg As Double 杆的角加速度Dim cos_cdb As Double 角CDB的余弦值Dim fcdb As Double 角CDBDim cos_fge As Double 角FGE的余弦值Dim ffge As Double 角FGEDim Ci, Cj , Si , Sj ,G1, G2 , G3 As DoubleDim Cii , Cjj , Sii , Sjj , G11, G22 ,G33 As DoubleDim pi As Double 圆周率Dim pa As Double 角度和弧度相互转换的系数Dim i As Double 循环变量Dim fj As Double 角度循环变量Private Sub Command1_Click()Picture1.ClsPicture1.Scale (-200, 500)-(200, -100)Picture1.Line (-190, 0)-(190, 0) X轴Picture1.Line (0, 500)-(0, -100) Y轴 For i = -160 To 160 Step 40 X轴坐标 Picture1.DrawStyle = 2 Picture1.Line (i, 500)-(i, -100) Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print i Next i For i = 0 To 500 Step 50 Y轴坐标 Picture1.DrawStyle = 2 Picture1.Line (-190, i)-(190, i) Picture1.CurrentX = -10: Picture1.CurrentY = i Picture1.Print i Next i For fj = 0 To 360 Step 0.01 fab = fj * pa Call RR1 Call RRR1 Call RR2 Picture1.PSet (xE, yE), vbRed Next fjEnd SubPrivate Sub Command2_Click()Picture2.Scale (-10, 390)-(380, -70)Picture2.Line (-10, 0)-(380, 0) X轴Picture2.Line (0, 380)-(0, -80) Y轴 For i = -30 To 390 Step 30 X轴坐标 Picture2.DrawStyle = 2 Picture2.Line (i, 1000)-(i, -100) Picture2.CurrentX = i - 10: Picture2.CurrentY = 0 Picture2.Print i Next i For i = -30 To 390 Step 30 Y轴坐标 Picture2.DrawStyle = 2 Picture2.Line (-10, i)-(380, i) Picture2.CurrentX = -10: Picture2.CurrentY = i Picture2.Print iNext i For fj = 0 To 360 Step 0.01 fab = fj * pa Call RR1 Call RRR1 Call RR2 Call RRR2 ffg = ffg / (2 * pi) * 360Picture2.PSet (fj, ffg), vbRed Next fjEnd SubPrivate Sub Command3_Click()Picture3.Scale (-10, 60)-(380, -80)Picture3.Line (-10, 0)-(380, 0) X轴Picture3.Line (0, 100)-(0, -100) Y轴For i = 0 To 360 Step 30 X轴坐标Picture3.DrawStyle = 2Picture3.Line (i, 60)-(i, -80)Picture3.CurrentX = i - 10: Picture3.CurrentY = 0Picture3.Print iNext iFor i = -80 To 60 Step 10 Y轴坐标 Picture3.Line (0, i)-(380, i) Picture3.CurrentX = -10: Picture3.CurrentY = i Picture3.Print i Next i For fj = 0 To 360 Step 0.01 fab = fj * pa Call RR1 Call RRR1 Call RR2 Call RRR2Picture3.PSet (fj, wfg), vbRed Next fjEnd SubPrivate Sub Command4_Click()Picture4.Scale (-10, 200)-(380, -200)Picture4.Line (-10, 0)-(380, 0) X轴Picture4.Line (0, 200)-(0, -200) Y轴 For i = 0 To 360 Step 30 X轴坐标 Picture4.DrawStyle = 2 Picture4.Line (i, 200)-(i, -200) Picture4.CurrentX = i - 10: Picture4.CurrentY = 0 Picture4.Print i Next i For i = -200 To 200 Step 40 Y轴坐标 Picture4.DrawStyle = 2 Picture4.Line (-10, i)-(380, i) Picture4.CurrentX = -10: Picture4.CurrentY = i Picture4.Print i Next i For fj = 0 To 360 Step 0.01 fab = fj * pa Call RR1 Call RRR1 Call RR2 Call RRR2Picture4.PSet (fj, efg), vbRed Next fjEnd SubPrivate Sub Form_Load() 给变量赋初始值lab = 112:lbc = 200:lce = 200:lcd = 200:lef = 116:lfg = 28wab = 10:eab = 0:delt = 0xA = 264:yA = 0:vxA = 0:vyA = 0:axA = 0:ayA = 0xD = 0:yD = 0:vxD = 0:vyD = 0:axD = 0:ayD = 0xE = 0:yE = 136:vxE = 0:vyE = 0:axE = 0:ayE = 0xG = 0:yG = 278:vxG = 0:vyG = 0:axG = 0:ayG = 0pi = 4 * Atn(1):pa = pi / 180:fj = 0End SubPrivate Sub RR1() 级杆组RR1（原动件1）xB = xA + lab * Cos(fab + delt)yB = yA + lab * Sin(fab + delt)vxB = vxA - wab * lab * Sin(fab + delt)vyB = vyA + wab * lab * Cos(fab + delt)axB = axA - wab 2 * lab * Cos(fab + delt) - eab * lab * Sin(fab + delt)ayB = ayA - wab 2 * lab * Sin(fab + delt) + eab * lab * Cos(fab + delt)End SubPrivate Sub RRR1() 级杆组RRR1(杆2,3)lbd = Sqr(xD - xB) 2 + (yD - yB) 2)If xB xD And yB yD Then fbd = Atn(yB - yD) / (xB - xD)ElseEnd IfIf xB xD And yB = yD Then fbd = 0ElseEnd IfIf xB xD And yB xG And yE yG Then feg = Atn(yE - yG) / (xE - xG)ElseEnd IfIf xE = xG And yE yG Then feg = pi / 2ElseEnd IfIf xE xG Then feg = Atn(yE - yG) / (xE - xG) + piElseEnd IfIf xE = xG And yE xG And yE yG Then feg = Atn(yE - yG) / (xE - xG) + 2 * piElseEnd Ifcos_fge = (leg 2 + lfg 2 - lef 2) / (2 * leg * lfg)ffge = Atn(-cos_fge / Sqr(1 - cos_fge 2) + pi / 2ffg = (feg - ffge)If ffg = 0 Thenffg = ffgElseEnd IfxF = xG + lfg * Cos(ffg)yF = yG + lfg * Sin(ffg)If xF xE And yF = yE Then fef = Atn(yF - yE) / (xF - xE)ElseEnd IfIf xF = xE And yF yE Thenfef = pi / 2ElseEnd IfI