哈工大机械原理大作业第27题

1、哈工大机械原理大作业 1- 第 27 题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

2、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yGEFAX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。I 级杆组 RRII 级杆组 RRRII 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 xA 0,yA 0,速度vxA 0,vyA 0,加速度 axA 0,ayA 0. 为AB=28OmmB 的位置坐标xB 280cos ,yB 280

3、sin , 速度 vxB 280 sin ,vyB 280 cos , 加速度 axB280 cos ,ayB 280 sinaxB)BCD 杆( II 级杆组 RRR)xD 0,yD 0,速度 vxD 0,vyD 0, 加速度 axD 0,ayD运动副 D 的位置坐标BC=350mm,CD=320mm0.BD由余弦定理，可求得22AD2 AB2 2AD* ABcos(2由正弦定理得ADBABsin( )arc sin 2BDCDB由余弦定理得CD 2 BD 2 BC 2 arccos2 DC DBx CD sin(ADBBDC)y AD CD cos(ADB CDB)C 的位置坐标（ X,Y

4、）,速度（ vx，vy ）和加速度 （ax,ay）,杆由此可以求出运动副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 x G 25,yG 80,速度 vxG 0,vyG 0,加速度 axG 0,ayG 0. 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

5、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 DoubleD

6、im 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

7、 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 Double Dim 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 l

8、cd 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 D

9、im 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 D

10、im 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 Double Dim Sj As DoubleDim G1 As DoubleDim G2 As DoubleDim G3 As DoubleDim val As Do

11、uble 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 = 280 lbc = 350 lcd = 320 lad = 160 lbe = 175 lef = 220wab = 10 eab = 0 delt = 0 xA = 0 yA = 0 vxA = 0 vyA = 0 axA = 0 ayA = 0 xD = 0 yD = 160 vxD = 0 vyD = 0 axD = 0 ayD = 0 xG = -25 yG

12、= 80 vxG = 0 vyG = 0 axG = 0ayG = 0 pi = 3.1415926 pa = pi / 180 fj1 = 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) YFor i = -180 To 120 Step 30X 轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 250)-(i

13、, -50)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = -60 To 240 Step 30 Y 轴坐标 Picture1.DrawStyle = 2Picture1.Line (-200, i)-(100, i) Picture1.CurrentX = -10: Picture1.CurrentY = i Picture1.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RR1Call RRRCall RR

14、2Picture1.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) YFor i = -30 To 390 Step 30X 轴坐标Picture2.DrawStyle = 2 Picture2.Line (i, 8)-(i, -2)Picture2.CurrentX = i - 10: Picture2.CurrentY

15、= 0 Picture2.Print iNext iFor i = -2 To 8 Step 1 Y 轴坐标Picture2.DrawStyle = 2Picture2.Line (-10, i)-(380, i)Picture2.CurrentX = -10: Picture2.CurrentY = i Picture2.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RR1Call RRRCall RR2Call RPRPicture2.PSet (fj1, ffg)Next fj1End SubPrivate Sub

16、 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

17、Step 5 Y 轴坐标Picture3.Line (0, i)-(380, i)Picture3.CurrentX = -20: Picture3.CurrentY = iPicture3.Print iNext iFor 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)Pi

18、cture4.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 iNext iFor i = -800 To 800 Step 80 Y 轴坐标 Picture4.Line (0, i)-(380, i)Picture4.Curre

19、ntX = -25: Picture4.CurrentY = i + 5 Picture4.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RR1Call RRRCall RR2Call RPRPicture4.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

20、 + 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

21、(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 (

22、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 =

23、 yB Then 第二象限 fbd = Atn(yD - yB) / (xD - xB) + pi fbc = 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) + pi ElseEnd 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