1、哈工大机械原理大作业连杆黄建青Harbin Institute of Technology机械原理大作业一课程名称: 机械原理 设计题目: 连杆机构运动分析 院 系: 能源学院 班 级: 1302402 设 计 者: 黄建青 学 号: 1130240222 指导教师: 焦映厚 陈照波 设计时间: 2015年06月24日 求解步骤如下:1. 建立坐标系 建立以点A为原点的固定平面直角坐标系xAy。2. 机构的结构分析,基本干组的划分 该机构由级杆组RR(原动件AB)、级杆组RRP(杆2,3)、级杆组RRP(杆4,5)。如下图所示。3. 建立各基本杆组的运动分析的数学模型计算流程框图如下:3.1级
2、杆组RR(原动件AB)设原动件杆1的转角为,则角速度 角加速度 支座A的位置坐标为 速度为 加速度为 原动件杆1的长度 所以,运动副B的位置、速度和加速度分析如下:位置 速度 加速度分析 3.2级杆组RRP(杆2、3)杆长,设点C位移为s则 C点位移速度加速度3.3级杆组RRP(杆4、5)杆长,则D点轨迹:E点坐标:速度加速度4. 用VB编程求解源代码如下:Dim XA As Double, YA As DoubleDim 2(10) As DoubleDim XD(10) As Double, YD(10) As DoubleDim XB(10) As Double, YB(10) As D
3、oubleDim XC(10) As Double, YC(10) As Double, 1(10) As DoubleDim XE(10) As Double, YE(10) As DoubleDim 2(10) As DoubleDim t As DoubleDim w1 As DoubleDim LAB As Double, LBC As Double, LBD As Double, LDE As DoubleDim i As DoubleFunction Arcsin(X) As DoubleArcsin = Atn(X / Sqr(-X * X + 1)End FunctionPri
4、vate Sub Command1_Click()Picture1.FillColor = vbBlackPicture1.Circle (XA, YA), 6, vbBlacki = 0For t = 0 To 10 Step 0.0001If i = 1 ThenPicture1.Line (XA, YA)-(XB(t), YB(t), vbWhitePicture1.Line (XC(t), YC(t)-(XD(t), YD(t), vbWhitePicture1.Line (XE(t), YE(t)-(XD(t), YD(t), vbWhitePicture1.Circle (XA,
5、YA), 6, vbWhitePicture1.Circle (XD(t), YD(t), 6, vbWhitePicture1.Circle (XB(t), YB(t), 6, vbWhitePicture1.Circle (XC(t), YC(t), 6, vbWhitePicture1.Circle (XE(t), YE(t), 6, vbWhiteEnd If2(t) = w1 * tXB(t) = LAB * Cos(2(t)YB(t) = LAB * Sin(2(t)Picture1.FillColor = vbBlackPicture1.Circle (XB(t), YB(t),
6、 6, vbBlackPicture1.Line (XA, YA)-(XB(t), YB(t), vbBlackIf 2(t) (3.14159 / 2) And 2(t) (3.14159 / 2) And 2(t) (3 * 3.14159 / 2) ThenXD(t) = XB(t) - LBD * (Sin(1(t)2(t) = Arcsin(-(XD(t) / LDE)ElseIf 2(t) (3 * 3.14159 / 2) ThenXD(t) = XB(t) + LBD * (Sin(1(t)2(t) = Arcsin(XD(t) / LDE)End IfYD(t) = YB(t
7、) + LBD * Abs(Cos(1(t)Picture1.FillColor = vbBlackAPicture1.Circle (XC(t), YC(t), 6, vbBlackPicture1.Line (XC(t), YC(t)-(XB(t), YB(t), vbBlackPicture1.Line (XC(t), YC(t)-(XD(t), YD(t), vbBlackXE(t) = 0YE(t) = YD(t) + LDE * (Cos(2(t)Picture1.FillColor = vbBlackPicture1.Circle (XE(t), YE(t), 6, vbBlac
8、kPicture1.Line (XD(t), YD(t)-(XE(t), YE(t), vbBlackPicture1.Line (XE(t) - 100, YE(t)-(XE(t) + 100, YE(t), vbBlackPicture1.Line (XE(t) - 100, YE(t) - 450)-(XE(t) + 100, YE(t) - 450), vbBlacki = 1Next tEnd Sub得到动画图:图1、动画Dim XA As Double, YA As DoubleDim 2(10) As DoubleDim XB(10) As Double, YB(10) As D
9、oubleDim XC(10) As Double, YC(10) As Double, 1(10) As DoubleDim XD(10) As Double, YD(10) As Double, XE(10) As Double, YE(10) As DoubleDim 2(10) As DoubleDim t As DoubleDim w1 As DoubleDim LAB As Double, LBC As Double, LBD As Double, LDE As DoubleDim i As DoublePrivate Sub Command2_Click()Picture1.Cl
10、sEnd SubPrivate Sub Command3_Click()EndEnd Sub Private Sub Form_Load()XA = 0YA = 0w1 = 10LAB = 100LBC = 200LBD = 100LDE = 210End SubFunction Arcsin(X) As DoubleArcsin = Atn(X / Sqr(-X * X + 1)End FunctionPrivate Sub Picture1_Click()Picture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-3
11、60, 360)-(360, -360)Picture1.Line (-360, 0)-(360, 0)Picture1.Line (0, -360)-(0, 360)For i = 0 To 24Picture1.CurrentX = -10Picture1.CurrentY = -30 * i + 360Picture1.Print & -30 * i + 360 & Next iFor i = -18 To 18Picture1.CurrentX = i * 20Picture1.CurrentY = -10Picture1.Print & i * 20 & Next iEnd SubP
12、rivate Sub Command1_Click()For t = 0 To 10 Step 0.00012(t) = w1 * tXB(t) = LAB * Cos(2(t)YB(t) = LAB * Sin(2(t)1(t) = Arcsin(Abs(XB(t) / LBC)XC(t) = 0YC(t) = YB(t) - LBC * Abs(Cos(1(t)XD(t) = XB(t) + LBD * (Sin(1(t)YD(t) = YB(t) + LBD * Abs(Cos(1(t)2(t) = Arcsin(Abs(XD(t) / LDE)XE(t) = 0YE(t) = YD(t
13、) + LDE * (Cos(2(t)Picture1.Circle (2(t) * 180 / 3.14159, YC(t), 0.08, vbBlackPicture1.Circle (2(t) * 180 / 3.14159, YE(t) - 460), 0.08, vbBlack Picture1.Circle (XD(t), YD(t), 0.08, vbBlackNext tEnd Sub得到构件2上的D点轨迹,如图2所示:图2、D点轨迹图得到构件3、5的位移图:图3、 位移图Dim XA As Double, YA As DoubleDim 2(10) As DoubleDim
14、XB(10) As Double, YB(10) As Double, XB1(10) As Double, YB1(10) As DoubleDim XC(10) As Double, YC(10) As Double, 1(10) As Double, A1(10) As Double, vyC(10) As DoubleDim XD(10) As Double, YD(10) As Double, XD1(10) As Double, YD1(10) As DoubleDim XE(10) As Double, YE(10) As Double, vyE(10) As Double, 2
15、(10) As Double, A2(10) As DoubleDim t As DoubleDim w1 As DoubleDim LAB As Double, LBC As Double, LBD As Double, LDE As DoubleDim i As DoublePrivate Sub Command2_Click()Picture1.ClsEnd SubPrivate Sub Command3_Click()EndEnd SubPrivate Sub Form_Load()XA = 0YA = 0w1 = 10LAB = 100LBC = 200LBD = 100LDE =
16、190End SubFunction Arcsin(X) As DoubleArcsin = Atn(X / Sqr(-X * X + 1)End FunctionPrivate Sub Picture1_Click()Picture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-30, 1900)-(360, -1900)Picture1.Line (0, 0)-(360, 0)Picture1.Line (0, -1900)-(0, 1900)For i = 0 To 135Picture1.CurrentX = -2
17、0Picture1.CurrentY = -500 * i + 1900Picture1.Print & -500 * i + 1900 & Next iFor i = 0 To 18Picture1.CurrentX = i * 20Picture1.CurrentY = -10Picture1.Print & i * 20 & Next iEnd SubPrivate Sub Command1_Click()For t = 0 To 10 Step 0.00012(t) = w1 * tXB(t) = LAB * Cos(2(t)YB(t) = LAB * Sin(2(t)XB1(t) =
18、 -LAB * w1 * Sin(2(t)YB1(t) = LAB * w1 * Cos(2(t)If 2(t) (3.14159 / 2) And 2(t) (3.14159 / 2) And 2(t) (3.14159 / 2) And 2(t) (3.14159 / 2) And 2(t) (3 * 3.14159 / 2) ThenXD(t) = XB(t) - LBD * (Sin(1(t)XD1(t) = XB1(t) - LBD * A1(t) * (Cos(1(t)XD2(t) = XB2(t) - LBD * A12(t) * (Cos(1(t) + LBD * (A1(t) 2 * (Sin(1(t)2(t) = Arcsin(-(XD(t) / LDE)A2(t) = -XD1(t) / (LDE * Sqr(1 - (-(XD(t) / LDE) 2)A22(t) = -XD2(t) / (LDE * Sqr(1 - (-XD(t) / LDE) 2) - (XD1(t) 2 * XD(t) / (LDE 3 * (1 - (XD(t) / LDE) 2) 1.5)Else: XD(t) = XB(t) + LBD * (Sin(1(t)XD1(t) = XB1(t) + LBD * A1(t) * (Cos(1(t)XD2(t) = XB2(t) + L
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