哈工大机械原理大作业1-第27题.doc

Harbin Institute of Technology机械原理大作业设计说明书（论文）课程名称： 机械原理 设计题目： 连杆机构运动分析 院 系：能源科学与工程学院 班 级： 1002101 设 计 者： 学 号： 指导教师： 赵永强 设计时间： 2012年6月10日6月24日 运动分析题目第二十七题27 如图所示机构，已知机构各构件的尺寸为AB=280mm，BC=350mm，CD=320mm, AD=160mm, BE=175mm, EF=220mm, xc=25mm, yg=80mm, 构件一的角速度为w1=10rad/s, 试求构件2上点F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。YAGCBEFyGXgDX一、建立坐标系以A点为原点建立如图所示坐标系x-y，如上图所示二、机构结构分析该机构可认为由一个I级杆组RR（杆AB）、II级杆组RRR（杆2、3）、II级杆组RPR（杆5及滑块4）组成。如下图所示。AB II级杆组RRCBDE II级杆组RRRGF II级杆组RPR三、确定已知参数和设计流程一）AB（I级杆组RR）运动副A的位置坐标为AB=28OmmB的位置坐标二）BCD杆（II级杆组RRR）运动副D的位置坐标BC=350mm, CD=320mm由余弦定理，可求得由正弦定理得由余弦定理得由此可以求出运动副C的位置坐标（X,Y）,速度（vx，vy）和加速度(ax,ay),杆BC与x轴的夹角，杆BC的角速度，杆BC的角加速度，杆CD与x轴的夹角，角速度，角加速度。三）GF （II级杆组RPR）G点的位置坐标由此可以求出构件GF的转角，角速度和角加速度四）构件BC上E点的运动BE=175mm，根据前面求出的量，可以得到E的位置坐标，速度和加速度，同理可以得到F点的运动规律。四、编程计算编程语言为VB，编程环境为visual basic6.0精简版源代码如下Dim xA As Double 点A的坐标，速度，加速度Dim yA As DoubleDim vxA As DoubleDim vyA As DoubleDim axA As DoubleDim ayA As DoubleDim xB As Double 点B的坐标，速度，加速度Dim yB As DoubleDim vxB As DoubleDim vyB As DoubleDim axB As DoubleDim ayB As DoubleDim xC As Double 点C的坐标，速度，加速度Dim yC As DoubleDim vxC As DoubleDim vyC As DoubleDim axC As DoubleDim ayC As DoubleDim xD As Double 点D的坐标，速度，加速度Dim yD As DoubleDim vxD As DoubleDim vyD As DoubleDim axD As DoubleDim ayD As DoubleDim xE As Double 点E的坐标，速度，加速度Dim yE As DoubleDim vxE As DoubleDim vyE As DoubleDim axE As DoubleDim ayE As DoubleDim xF As Double 点F的坐标，速度，加速度Dim yF As DoubleDim vxF As DoubleDim vyF As DoubleDim axF As DoubleDim ayF As DoubleDim xG As Double 点G的坐标，速度，加速度Dim yG As DoubleDim vxG As DoubleDim vyG As DoubleDim axG As DoubleDim ayG As DoubleDim delt As Double 构件1的初始角位移Dim lab As Double 杆AB的长度Dim lbc As Double 杆BC的长度Dim lcd As Double 杆CD的长度Dim lad As Double 杆AD的长度Dim lbe As Double 杆BE的长度Dim lef As Double 杆EF的长度Dim lbf As Double BF两点间的距离Dim fab As Double 杆AB的角位移Dim fbc As Double 杆BC的角位移Dim fcd As Double 杆CD的角位移Dim fef As Double 杆EF的角位移Dim ffg As Double 杆FG的角位移Dim febf As Double 角EBF的角度Dim fj1 As DoubleDim wab As Double 杆AB的角速度Dim wbc As Double 杆BC的角速度Dim wcd As Double 杆CD的角速度Dim wce As Double 杆CE的角速度Dim wef As Double 杆EF的角速度Dim wfg As Double 杆FG的角速度Dim eab As Double 杆AB的角加速度Dim ebc As Double 杆BC的角加速度Dim ecd As Double 杆CD的角加速度Dim ece As Double 杆CE的角加速度Dim eef As Double 杆EF的角加速度Dim efg As Double 杆FG的角加速度Dim LBD As Double BD的长度Dim LGF As Double GF的长度Dim JCBD As Double 角CBD的角度Dim fbd As Double 杆BD的角位移Dim Ci As Double RRR杆组的中间变量Dim Cj As DoubleDim Si As DoubleDim Sj As DoubleDim G1 As DoubleDim G2 As DoubleDim G3 As DoubleDim val As Double 角CBD的余弦值Dim pi As Double 圆周率Dim pa As Double 角度与弧度转换的系数Dim i As Double 循环变量Private Sub Form_Load() 附值lab = 280lbc = 350lcd = 320lad = 160lbe = 175lef = 220wab = 10eab = 0delt = 0xA = 0yA = 0vxA = 0vyA = 0axA = 0ayA = 0xD = 0yD = 160vxD = 0vyD = 0axD = 0ayD = 0xG = -25yG = 80vxG = 0vyG = 0axG = 0ayG = 0pi = 3.1415926pa = pi / 180fj1 = 0End SubPrivate Sub Command1_Click() 点F的轨迹Picture1.Scale (-200, 250)-(100, -50)Picture1.Line (-200, 0)-(100, 0) XPicture1.Line (0, 250)-(0, -50) Y For i = -180 To 120 Step 30 X轴坐标 Picture1.DrawStyle = 2 Picture1.Line (i, 250)-(i, -50) Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print i Next i For i = -60 To 240 Step 30 Y轴坐标 Picture1.DrawStyle = 2 Picture1.Line (-200, i)-(100, i) Picture1.CurrentX = -10: Picture1.CurrentY = i Picture1.Print i Next iFor fj1 = 0 To 360 Step 0.01 fab = fj1 * pa Call RR1 Call RRR Call RR2 Picture1.PSet (xF, yF) Next fj1End SubPrivate Sub Command2_Click() 杆5的角位移Picture2.Scale (-10, 8)-(380, -2)Picture2.Line (-10, 0)-(380, 0) XPicture2.Line (0, 8)-(0, -2) Y For i = -30 To 390 Step 30 X轴坐标 Picture2.DrawStyle = 2 Picture2.Line (i, 8)-(i, -2) Picture2.CurrentX = i - 10: Picture2.CurrentY = 0 Picture2.Print i Next i For i = -2 To 8 Step 1 Y轴坐标 Picture2.DrawStyle = 2 Picture2.Line (-10, i)-(380, i) Picture2.CurrentX = -10: Picture2.CurrentY = i Picture2.Print i Next i For fj1 = 0 To 360 Step 0.01 fab = fj1 * paCall RR1Call RRRCall RR2Call RPRPicture2.PSet (fj1, ffg) Next fj1End SubPrivate Sub Command3_Click() 杆5的角速度Picture3.Scale (-20, 90)-(380, -10)Picture3.Line (-20, 0)-(380, 0) XPicture3.Line (0, 90)-(0, -10) YFor i = 0 To 360 Step 30 X轴坐标Picture3.DrawStyle = 2Picture3.Line (i, 90)-(i, -10)Picture3.CurrentX = i - 10: Picture3.CurrentY = 0Picture3.Print iNext iFor i = -10 To 90 Step 5 Y轴坐标 Picture3.Line (0, i)-(380, i) Picture3.CurrentX = -20: Picture3.CurrentY = i Picture3.Print i Next i For fj1 = 0 To 360 Step 0.01 fab = fj1 * paCall RR1Call RRRCall RR2Call RPRPicture3.PSet (fj1, wfg) Next fj1End SubPrivate Sub Command4_Click() 杆5的角加速度Picture4.Scale (-20, 800)-(380, -800)Picture4.Line (-20, 0)-(380, 0) XPicture4.Line (0, 800)-(0, -800) YFor i = 0 To 360 Step 30 X轴坐标 Picture4.DrawStyle = 2 Picture4.Line (i, 800)-(i, -800) Picture4.CurrentX = i - 10: Picture4.CurrentY = 0 Picture4.Print i Next i For i = -800 To 800 Step 80 Y轴坐标 Picture4.Line (0, i)-(380, i) Picture4.CurrentX = -25: Picture4.CurrentY = i + 5 Picture4.Print i Next i For fj1 = 0 To 360 Step 0.01 fab = fj1 * pa Call RR1 Call RRR Call RR2 Call RPR Picture4.PSet (fj1, efg) Next fj1End 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 * Sin(fab + delt)End SubPrivate Sub RR2() 构件2上点F的运动分析lbf = Sqr(lbe 2 + lef 2)febf = Atn(lef / lbe)xF = xB + lbf * Cos(fbc + febf)yF = yB + lbf* Sin(fbc + febf)vxF = vxB - wbc * lbf * Sin(fbc + febf)vyF = vyB + wbc * lbf* Cos(fbc + febf)axF = axB - wbc 2 * lbf * Cos(fbc + febf) - ebc * lbf * Sin(fbc + febf)ayF = ayB - wbc 2 * lbf * Sin(fbc + febf) + ebc * lbf* Sin(fbc + febf)End SubPrivate Sub RRR() 级杆组RRR（杆2、杆3）xB = xA + lab * Cos(fab + delt)yB = yA + lab * Sin(fab + delt)LBD = Sqr(xD - xB) 2 + (yD - yB) 2)val = (lbc 2 + LBD 2 - lcd 2) / (2 * lbc * LBD)JCBD = Atn(-val / Sqr(-val * val + 1) + 2 * Atn(1)If xD xB And yD yB Then 第一象限fbd = Atn(yD - yB) / (xD - xB)fbc = fbd - JCBDElseEnd IfIf xD = yB Then 第二象限fbd = Atn(yD - yB) / (xD - xB) + pifbc = fbd - JCBDElseEnd IfIf xD xB And yD xB And yD yB Then y轴正向fcd = 0.5 * piElseEnd IfIf xB = xD And yD xD And yC = yD Then 第一象限fcd = Atn(yC - yD) / (xC - xD)ElseEnd IfIf xC = yD Then 第二象限fcd = Atn(yC - yD) / (xC - xD) + piElseEnd IfIf xC xD And yC xD And yC yD Then y轴正向fcd = 0.5 * piElseEnd IfIf xC = xD And yC xG And yF yD Then 第一象限ff