哈工大机械原理大作业 第25题 连杆机构设计

1、机械原理大作业一课程名称： 机械原理 设计题目： 连杆机构运动分析 院 系： 机电工程学院 指导教师： 林琳 刘福利 设计时间： 2014.06 哈尔滨工业大学题目：如图所示机构，已知机构各构件的尺寸为齿轮1，齿轮2，齿轮5的齿数分为z1=17，z2=z5=40，BC=70mm，CD=270mm，EF=80mm，DE=240mm,DG=500mm,h=160mm,齿轮1的角速度为w1=10rad/s,试求点D的轨迹及构件7上点G的位移，速度和加速度，并对计算结果进行分析。1.建立直角坐标系上述机构可简化为下图中所示机构，其中BC和EF杆的角速度相等（大小相同，方向相同），所以该机构只有一个自由

2、度。可建立如图所示的坐标系2机构结构分析该机构可认为由两个I级杆组RR（杆BC、杆EF）、II级杆组RRR（杆3、4）、II级杆组RRP（杆6及滑块G）组成。如下图所示：I级杆组RR（杆BC、杆EF）III级杆组RRR（杆3、4）II级杆组RRP（杆6及滑块G）3. 各基本杆组运动分析数学模型3.1 I级杆组BC如上图中所示，已知运动副C的位置BC=70mm, xB=0 , yB=0 , 速度vxB=0 , vyB=0 ，加速度axB=0,ayB=0.角速度=-101740 rad/s,角加速度=0. 初始转角0=0. 转角符号为.1） 位置分析 2） 速度和加速度分析 将已知参数带入便可求得

3、3.2 I级杆组EF 如上图中所示，已知运动副E的位置EF=80mm,可令BF=350mm从而=0, 初始转角0=0.转角符号为.1)位置分析 2)速度和加速度分析 将已知参数带入便可求得3.3 II级杆组RRR（杆3、4）如下图中的示意图所示，已知参数 DC=270mm,DE=240.可求出运动副D点的一系列的 参数 以及两杆的角位置3,4,角速度3，4 , 和角加速度3，4 .1） 位置方程 （1） （2） 连立（1）、（2）求得: 3 其中： 求得分3后，带入（1）、（2），便可求得4 , 2） 速度方程 其中： 3） 加速度方程 其中： 3.4 II级杆组RRP（杆6、滑块G）如下图中

4、的示意图所示已知运动副D点的 杆长DG=,滑块导路方向角f(j)=0和计算位移s的参考点K的参数可求G点的 1） 位置方程 可求得，其中带入上式即可解的xG,yG,s2） 速度方程 其中： 3） 加速度方程 其中：四、计算编程采用VB编程，其窗口代码如下： Private s7(3600) As Double Private v7(3600) As Double Private a7(3600) As Double Private x3(3600) As Double Private y3(3600) As DoublePrivate Sub Form_Load() Dim i As Sing

5、le Dim pi As Double Dim pa As Double Dim RR1 As RR Dim RR2 As RR Dim RRR1 As RRR Dim RRP1 As RRP Set RR1 = New RR Set RR2 = New RR Set RRR1 = New RRR Set RRP1 = New RRP pi = 3.1415926 pa = pi / 180 For i = 0 To 3600 RR1.xA = 0: RR1.yA = 0 RR1.vxA = 0: RR1.vyA = 0 RR1.axA = 0: RR1.ayA = 0 RR1.f = i *

6、 pa / 10: RR1.delt = 0 RR1.w = -10 * 17 / 40: RR1.e = 0 RR1.L = 70 RR1.cal RR2.xA = 350: RR2.yA = 0 RR2.vxA = 0: RR2.vyA = 0 RR2.axA = 0: RR2.ayA = 0 RR2.f = i * pa / 10: RR2.delt = pi RR2.w = -10 * 17 / 40: RR2.e = 0 RR2.L = 80 RR2.cal RRR1.xB = RR1.xB: RRR1.yB = RR1.yB RRR1.vxB = RR1.vxB: RRR1.vyB

7、 = RR1.vyB RRR1.axB = RR1.axB: RRR1.ayB = RR1.ayB RRR1.xD = RR2.xB: RRR1.yD = RR2.yB RRR1.vxD = RR2.vxB: RRR1.vyD = RR2.vyB RRR1.axD = RR2.axB: RRR1.ayD = RR2.ayB RRR1.Li = 270: RRR1.Lj = 240 RRR1.M = 1 RRR1.cal x3(i) = RRR1.xC: y3(i) = RRR1.yC RRP1.xB = RRR1.xC: RRP1.yB = RRR1.yC RRP1.vxB = RRR1.vx

8、C: RRP1.vyB = RRR1.vyC RRP1.axB = RRR1.axC: RRP1.ayB = RRR1.ayC RRP1.Li = 500: RRP1.Lj = 0 RRP1.fj = 0: RRP1.wj = 0: RRP1.ej = 0 RRP1.xK = 0: RRP1.yK = 160: RRP1.vxK = 0: RRP1.vyK = 0 RRP1.axK = 0: RRP1.ayK = 0 RRP1.M = 1 RRP1.cal s7(i) = RRP1.ss: v7(i) = RRP1.vss / 1000: a7(i) = RRP1.ass / 1000 Nex

9、t i End SubPrivate Sub Command1_Click() D点轨迹 Picture1.Scale (-10, 300)-(500, -100) Picture1.Line (-10, 0)-(500, 0) X Picture1.Line (0, 300)-(0, -100) Y For i = 0 To 500 Step 50 X轴坐标 Picture1.DrawStyle = 2 Picture1.Line (i, 300)-(i, -100) Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Pri

10、nt i Next i For i = -100 To 300 Step 50 Y轴坐标 Picture1.DrawStyle = 2 Picture1.Line (-10, i)-(500, i) Picture1.CurrentX = -10: Picture1.CurrentY = i Picture1.Print i Next i For i = 0 To 3600 Picture1.PSet (x3(i), y3(i) Next iEnd SubPrivate Sub Command2_Click() G点位移 Picture2.Scale (-10, 1000)-(380, 400

11、) Picture2.Line (-10, 500)-(380, 500) X Picture2.Line (0, 1000)-(0, 400) Y For i = -10 To 380 Step 30 X轴坐标 Picture2.DrawStyle = 2 Picture2.Line (i, 1000)-(i, 400) Picture2.CurrentX = i - 10: Picture2.CurrentY = 500 Picture2.Print i Next i For i = 400 To 1000 Step 50 Y轴坐标 Picture2.DrawStyle = 2 Pictu

12、re2.Line (-10, i)-(380, i) Picture2.CurrentX = -10: Picture2.CurrentY = i Picture2.Print i Next i For i = 0 To 3600 Picture2.PSet (i / 10, s7(i) Next iEnd SubPrivate Sub Command3_Click() G点速度 Picture3.Scale (-10, 1)-(380, -1) Picture3.Line (-10, 0)-(380, 0) X Picture3.Line (0, 1)-(0, -1) Y For i = -

13、10 To 380 Step 30 X轴坐标 Picture3.DrawStyle = 2 Picture3.Line (i, 1)-(i, -1) Picture3.CurrentX = i - 10: Picture3.CurrentY = 0 Picture3.Print i Next i For i = -5 To 5 Step 1 Y轴坐标 Picture3.DrawStyle = 2 Picture3.Line (-10, i / 5)-(380, i / 5) Picture3.CurrentX = -10: Picture3.CurrentY = i / 5 Picture3.

14、Print i / 5 Next i For i = 0 To 3600 Picture3.PSet (i / 10, v7(i) Next iEnd SubPrivate Sub Command4_Click() G点加速度 Picture4.Scale (-10, 5)-(380, -5) Picture4.Line (-10, 0)-(380, 0) X Picture4.Line (0, 5)-(0, -5) Y For i = -10 To 380 Step 30 X轴坐标 Picture4.DrawStyle = 2 Picture4.Line (i, 5)-(i, -5) Pic

15、ture4.CurrentX = i - 10: Picture4.CurrentY = 0 Picture4.Print i Next i For i = -5 To 5 Step 1 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 i = 0 To 3600 Picture4.PSet (i / 10, a7(i) Next i End Sub五、计算结果（1）D点的轨迹图：（2）G点的位