哈工大机械原理大作业-连杆(16页).doc

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1、-Harbin Institute of Technology机械原理大作业一课程名称： 机械原理 设计题目： 连杆运动分析 院 系： 机电工程学院 班 级： 设 计 者： 学 号： 指导教师： 陈明 设计时间： 2013年6月25日 1、 运动分析题目在图1-10中所示的干草压缩机中，已知LAB=150mm,LBC=600mm,LCE=120mm,LCD=500mm,LEF=600mm,XD=400mm,YD=500mm,YF=600mm,曲柄1作等速转动，其转速n1=50r/min。求在一个运动循环中活塞5的位移、速度和加速度的变化曲线。 图1-102、 机构的结构分析（1） 基本杆组的划

2、分 AB即杆件1为原动件 DECB即杆件2、3为RRR型II级杆组，其中CE为同一构件上点。 EF和滑块即4、5为RRP型II级杆组（2）、建立以点A为原点的固定平面直角系3、确定已知参数和求解流程(1)原动件1（I级杆组RR） 如图所示，已知原动件1的转角 原动件杆1的角速度 原动件1的角加速度运动副A的位置坐标运动副A的速度运动副A的加速度原动件杆I的长度 可求出B的位置B的速度B的加速度 (2) 构件2、3（II级杆组RRR）D的位置D的速度D的加速度杆长，由关系其中可解得由上面两个式子可以得到两杆的角速度其中，可得E的位置E的速度E的加速度（3)、构件4、5杆组（II级杆组RRP）在建

3、立的坐标系中取一参考点K则速度加速度杆长,设F位移为s由由上面两个式子可以得到所以：F点位移速度加速度四、编程计算并输出结果（VB编程）主程序：Private Sub Command1_Click()Dim s5(3600) As DoubleDim v5(3600) As DoubleDim a5(3600) As DoubleDim pi As DoubleDim pa As Doublepi = 3.1415926pa = pi / 180Dim i As LongDim f1(3600) As DoubleDim RR1 As RRDim RR2 As RRDim RRR1 As RR

4、RDim RRP1 As RRPSet RR1 = New RRSet RR2 = New RRSet RRR1 = New RRRSet RRP1 = New RRPFor i = 0 To 3600 Step 1f1(i) = i * pa / 10RR1.delt = 0RR1.f = f1(i)RR1.w = 5.24RR1.e = 0RR1.L = 150RR1.xA = 0RR1.yA = 0RR1.vxA = 0RR1.vyA = 0RR1.axA = 0RR1.ayA = 0RR1.calRRR1.Li = 600RRR1.Lj = 500RRR1.xB = RR1.xBRRR

5、1.yB = RR1.yBRRR1.vxB = RR1.vxBRRR1.vyB = RR1.vyBRRR1.axB = RR1.axBRRR1.ayB = RR1.ayBRRR1.xD = 400RRR1.yD = 500RRR1.vxD = 0RRR1.vyD = 0RRR1.axD = 0RRR1.ayD = 0RRR1.M = 1RRR1.calRRRRR2.delt = 0RR2.f = RRR1.fiRR2.w = RRR1.wiRR2.e = RRR1.eiRR2.L = 480RR2.xA = RR1.xBRR2.yA = RR1.yBRR2.vxA = RR1.vxBRR2.v

6、yA = RR1.vyBRR2.axA = RR1.axBRR2.ayA = RR1.ayBRR2.calRRP1.Li = 600RRP1.Lj = 0RRP1.fj = piRRP1.wj = 0RRP1.ej = 0RRP1.xB = RR2.xBRRP1.yB = RR2.yBRRP1.vxB = RR2.vxBRRP1.vyB = RR2.vyBRRP1.axB = RR2.axBRRP1.ayB = RR2.ayBRRP1.xK = 0RRP1.yK = 600RRP1.vxK = 0RRP1.vyK = 0RRP1.axK = 0RRP1.ayK = 0RRP1.M = 1RRP

7、1.cals5(i) = RRP1.ssv5(i) = RRP1.vssa5(i) = RRP1.assNext iPicture1.Scale (-30, 700)-(360, 580)Picture1.Line (0, 0)-(360, 0) XPicture1.Line (0, 580)-(0, 700) Y For i = 0 To 360 Step 10 X轴坐标 Picture1.DrawStyle = 2 Picture1.Line (i, 700)-(i, 580) Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Pictur

8、e1.Print i Next i For i = 580 To 700 Step 10 Y轴坐标 Picture1.DrawStyle = 2 Picture1.Line (0, i)-(360, i) Picture1.CurrentX = -10: Picture1.CurrentY = i Picture1.Print i Next iFor i = 0 To 3600 Step 1 Picture1.PSet (i / 10, s5(i) Next iEnd SubRR:Public L As DoublePublic f As DoublePublic delt As Double

9、Public w As DoublePublic e As DoublePublic xA As DoublePublic yA As DoublePublic vxA As DoublePublic vyA As DoublePublic axA As DoublePublic ayA As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic Sub cal()xB = xA + L

10、* Cos(f + delt)yB = yA + L * Sin(f + delt)vxB = vxA - w * L * Sin(f + delt)vyB = vyA + w * L * Cos(f + delt)axB = axA - w 2 * L * Cos(f + delt) - e * L * Sin(f + delt)ayB = ayA - w 2 * L * Sin(f + delt) + e * L * Cos(f + delt)End SubRRR:Public Li As DoublePublic Lj As DoublePublic fi As DoublePublic

11、 fj As DoublePublic wi As DoublePublic wj As DoublePublic ei As DoublePublic ej As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic xC As DoublePublic yC As DoublePublic vxC As DoublePublic vyC As DoublePublic axC As D

12、oublePublic ayC As DoublePublic xD As DoublePublic yD As DoublePublic vxD As DoublePublic vyD As DoublePublic axD As DoublePublic ayD As DoublePublic M As DoublePublic Sub calRRR()Dim fDB As DoubleDim Ci As DoubleDim Cj As DoubleDim Si As DoubleDim Sj As DoubleDim G1 As DoubleDim G2 As DoubleDim G3

13、As DoubleDim LBD As DoubleDim JCBD As DoubleDim val As Doublepi = 3.1415926LBD = Sqr(xB - xD) 2 + (yD - yB) 2)If LBD Abs(Li - Lj) Thenval = (Li 2 + LBD 2 - Lj 2) / (2 * Li * LBD)JCBD = Atn(-val / Sqr(-val * val + 1) + 2 * Atn(1)End IfRRP:Public Li As DoublePublic Lj As DoublePublic fi As DoublePubli

14、c fj As DoublePublic wi As DoublePublic wj As DoublePublic ei As DoublePublic ej As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic xK As DoublePublic yK As DoublePublic vxK As DoublePublic vyK As DoublePublic axK As

15、DoublePublic ayK As DoublePublic xC As DoublePublic yC As DoublePublic vxC As DoublePublic vyC As DoublePublic axC As DoublePublic ayC As DoublePublic xD As DoublePublic yD As DoublePublic vxD As DoublePublic vyD As DoublePublic axD As DoublePublic ayD As DoublePublic M As SinglePublic ss As DoubleP

16、ublic vss As DoublePublic ass As DoublePublic Sub cal()Dim A0 As DoubleDim Q1 As DoubleDim Q2 As DoubleDim Q3 As DoubleDim Q4 As DoubleDim Q5 As DoubleDim val As DoubleDim pi As Doublepi = 3.14159216A0 = Lj + (yK - yB) * Cos(fj) - (xK - xB) * Sin(fj)val = A0 / Lifi = M * Atn(val / Sqr(-val * val + 1

17、) + fjxC = xB + Li * Cos(fi)yC = yB + Li * Sin(fi)ss = (xC - xK) * Cos(fj) + (yC - yK) * Sin(fj)xD = xK + ss * Cos(fj)yD = yK + ss * Sin(fj)Q1 = vxK - vxB - wj * (ss * Sin(fj) + Lj * Cos(fj)Q2 = vyK - vyB + wj * (ss * Cos(fj) - Lj * Sin(fj)Q3 = Li * Sin(fi) * Sin(fj) + Li * Cos(fi) * Cos(fj)wi = (-Q

18、1 * Sin(fj) + Q2 * Cos(fj) / Q3vss = -(Q1 * Li * Cos(fi) + Q2 * Li * Sin(fi) / Q3vxC = vxB - wi * Li * Sin(fi)vyC = vyB + wi * Li * Cos(fi)vxD = vxK + vss * Cos(fj) - ss * wj * Sin(fj)vyD = vyK + vss * Sin(fj) + ss * wj * Cos(fj)Q4 = axK - axB + wi 2 * Li * Cos(fi) - ej * (ss * Sin(fj) + Lj * Cos(fj

19、) - wj 2 * (ss * Cos(fj) - Lj * Sin(fj) - 2 * vss * wj * Sin(fj)Q5 = ayK - ayB + wi 2 * Li * Sin(fi) + ej * (ss * Cos(fj) - Lj * Sin(fj) - wj 2 * (ss * Sin(fj) + Lj * Cos(fj) + 2 * vss * wj * Cos(fj)ei = (-Q4 * Sin(fj) + Q5 * Cos(fj) / Q3ass = (-Q4 * Li * Cos(fi) - Q5 * Li * Sin(fi) / Q3axC = axB -

20、ei * Li * Sin(fi) - wi 2 * Li * Cos(fi)ayC = ayB + ei * Li * Cos(fi) - wi 2 * Li * Sin(fi)axD = axK + ass * Cos(fj) - ss * ej * Sin(fj) - ss * wj 2 * Cos(fj) - 2 * vss * wj * Sin(fj)ayD = ayK + ass * Sin(fj) + ss * ej * Cos(fj) - ss * wj 2 * Sin(fj) + 2 * vss * wj * Cos(fj)End Sub五、计算结果数据如图：位移曲线：速度曲线：加速度曲线：六、计算结果分析 主动件转角为0时,滑块的位移为628mm,随着转角的匀速增加,滑块位移先上升,速度为负快速下降,加速度为负且开始值较小并逐渐下降,到达最低点-2200左右,此时速度为0, 然后位移开始继续下降,速度继续下降,然后到达最小值-100左右,此时加速度为0,又开始上升(向左运动),速度正向增大,在后面一段时期速度继续增大,加速度也正向增大,然后速度下降,加速度下降.滑块分别在4.2s,11.4s速度达到正向最大，1.7s,7.7s速度达到负向最大，0.8s,2.7s,5.8s,9.6s加速度达到极值，滑块就是这样周期性的左右运动-第 16 页-