哈工大机械原理大作业——连杆——24号

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哈尔滨工业大学机械原理大作业_连杆26题

哈尔滨工业大学机械原理大作业_连杆26题

1.运动分析题目如图所示机构,已知机构各构件的尺寸为AB=108mm,EF=320mm,BC=CE=CD=200mm,FG=162mm,AD=258mm,AG=514mm,DG=384mm,β=80º,构件1的角速度为ω1=10rad/s,试求构件2上点E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。

2.机构的结构分析,组成机构的基本杆组划分3.各基本杆组的运动分析数学模型(1)RR基本杆组:delt=0xB = xA + AB * Cos(f + delt)yB = yA + AB* Sin(f + delt)vxB = vxA - w * AB * Sin(f + delt)vyB = vyA + w * AB* Cos(f + delt)axB = axA - w ^ 2 * AB * Cos(f + delt):ayB = ayA - w ^ 2 * AB* Sin(f + delt)(2)RRR基本杆组Ci = lbc * Cos(fbc)Si = lbc * Sin(fbc)Cj = lcd * Cos(fcd)Sj = lcd * Sin(fcd)G1 = Ci * Sj - Cj * Siwbc = (Cj * (vxD - vxB) + Sj * (vyD - vyB)) / G1wcd = (Ci * (vxD - vxB) + Si * (vyD - vyB)) / G1vxC = vxB - wbc * lbc * Sin(fbc)vyC = vyB + wbc * lbc * Cos(fbc)G2 = axD - axB + wbc ^ 2 * Ci - wcd ^ 2 * CjG3 = ayD - ayB + wbc ^ 2 * Si - wcd ^ 2 * Sjebc = (G2 * Cj + G3 * Sj) / G1ecd = (G2 * Ci + G3 * Si) / G1axC = axB - ebc * lbc * Sin(fbc) - wbc ^ 2 * lbc * Cos(fbc)ayC = ayB + ebc * lbc * Cos(fbc) - wbc ^ 2 * lbc * Sin(fbc)EB = 2 * BC * Cos(febc)xE = xB + EB* Cos(fbc + febc)yE = yB + EB * Sin(fbc + febc)vxE = vxB – wbc * EB* Sin(fbc + febc)vyE = vyB + wbc * EB* Cos(fbc + febc)axE = axB - wbc ^ 2 * EB * Cos(fbc + delt) - ebc * EB * Sin(fbc + febc)ayE = ayB - wbc ^ 2 * leb * Sin(fbc + delt) + ebc * leb * Sin(fbc + febc) Ci = lef * Cos(fef)Si = lef * Sin(fef)Cj = lfg * Cos(ffg)Sj = lfg * Sin(ffg)G1 = Ci * Sj - Cj * Siwef = (Cj * (vxG - vxE) + Sj * (vyG - vyE)) / G1wfg = (Ci * (vxG - vxE) + Si * (vyG - vyE)) / G1vxF = vxE - wef * lef * Sin(fef)vyF = vyE + wef * lef * Cos(fef)G2 = axG - axE + wef ^ 2 * Ci - wfg ^ 2 * CjG3 = ayG - ayE + wef ^ 2 * Si - wfg ^ 2 * Sjeef = (G2 * Cj + G3 * Sj) / G1efg = (G2 * Ci + G3 * Si) / G1axF = axE - eef * lef * Sin(fef) - wef ^ 2 * lef * Cos(fef)ayF = ayE + eef * lef * Cos(fef) - wef ^ 2 * lef * Sin(fef)4.计算编程Dim xA As DoubleDim yA As DoubleDim vxA As DoubleDim vyA As DoubleDim axA As Double 'A '点加速度x轴分量Dim ayA As Double 'A '点加速度y轴分量Dim xB As Double 'B'点'x轴坐标Dim yB As Double 'B点y轴坐标Dim vxB As Double 'B点速度x轴分量Dim vyB As Double 'B点速度y轴分量Dim axB As Double 'B点加速度x轴分量Dim ayB As Double 'B点加速度y轴分量Dim xC As Double 'C点x轴坐标Dim yC As Double C'点y轴坐标Dim vxC As Double 'C点速度x轴分量Dim vyC As Double 'C点速度y轴分量Dim axC As Double 'C点加速度x轴分量Dim ayC As Double 'C点加速度y轴分量Dim xD As Double 'D点x轴坐标Dim yD As Double 'D点y轴坐标Dim vxD As Double 'D点速度x轴分量Dim vyD As Double 'D点速度y轴分量Dim axD As Double 'D点加速度x轴分量Dim ayD As Double 'D点加速度y轴分量Dim xE As Double 'E点x轴坐标Dim yE As Double 'E点y轴坐标Dim vxE As Double 'E点速度x轴分量Dim vyE As Double 'E点速度y轴分量Dim axE As Double 'E点加速度x轴分量Dim ayE As Double 'E点加速度y轴分量Dim xF As Double 'F点x轴坐标Dim yF As Double 'F点y轴坐标Dim vxF As Double 'F点速度x轴分量Dim vyF As Double 'F点速度y轴分量Dim axF As Double 'F点加速度x轴分量Dim ayF As Double 'F点加速度y轴分量Dim xG As Double 'G点x轴坐标Dim yG As Double 'G点y轴坐标Dim vxG As Double 'G点速度x轴分量Dim vyG As Double 'G点速度y轴分量Dim axG As Double 'G点加速度x轴分量Dim ayG As Double 'G点加速度y轴分量Dim delt As Double ' AB杆初始转角Dim lab As Double 'AB杆长Dim lbc As Double 'BC杆长Dim lcd As Double ' CD杆长Dim lce As Double 'CE杆长Dim lef As Double 'EF杆长Dim lfg As Double 'FG杆长Dim leb As Double 'ED杆长Dim f As Double 'AB杆转角Dim fbc As Double 'BC杆转角Dim fcd As Double 'CD杆转角Dim fce As Double 'CE杆转角Dim fef As Double 'EF杆转角Dim ffg As Double 'FG杆转角Dim fge As Double 'ge杆转角Dim w As Double 'AB杆角速度Dim wbc As Double ' BC角速度Dim wcd As Double 'CD角速度Dim wce As Double 'CE角速度Dim wef As Double 'EF角速度Dim wfg As Double 'FG角速度Dim e As Double 'AB杆角加速度Dim ebc As Double ' BC杆角加速度Dim ecd As Double 'CD杆角加速度Dim ece As Double 'CE杆角加速度Dim eef As Double 'EF杆角加速度Dim efg As Double 'FG杆角加速度Dim LBD As Double 'BD距离Dim leg As Double 'EG距离Dim JCBD As Double '角CBDDim jfeg As Double '角FEGDim fBD As Double 'BD转角Dim feg As Double 'EG转角Dim Ci As DoubleDim Cj As DoubleDim Si As DoubleDim Sj As DoubleDim G1 As DoubleDim G2 As DoubleDim G3 As DoubleDim val As DoubleDim pi As DoubleDim pa As DoubleDim febc As Double '角EBCDim i As DoubleDim fj1 As DoublePrivate Sub Command1_Click() '求点E的轨迹Picture1.Scale (-300, 400)-(10, -15)Picture1.Line (-300, 0)-(10, 0) 'XPicture1.Line (0, 400)-(0, -15) 'YFor i = -300 To 0 Step 50 'X轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 400)-(i, 0)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = 0 To 350 Step 50 'Y轴坐标Picture1.DrawStyle = 2Picture1.Line (0, i)-(-400, i)Picture1.CurrentX = -20: Picture1.CurrentY = i + 7 Picture1.Print iNext iFor fj1 = 0 To 360 Step 0.01f = fj1 * paCall RR1Call RRR1Call RR2Picture1.PSet (xE, yE)Next fj1End SubPrivate Sub Command2_Click() '求构件5的角位移Picture2.Scale (-20, 5)-(380, -0.5)Picture2.Line (-20, 0)-(380, 0) 'XPicture2.Line (0, 3)-(0, -0.5) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture2.DrawStyle = 2Picture2.Line (i, 3)-(i, 0)Picture2.CurrentX = i - 10: Picture2.CurrentY = 0 Picture2.Print iNext iFor i = -0.5 To 3 Step 0.5 'Y轴坐标Picture2.Line (0, i)-(380, i)Picture2.CurrentX = -25: Picture2.CurrentY = i Picture2.Print iNext iFor fj1 = 0 To 360 Step 0.01f = fj1 * paCall RR1Call RRR1Call RR2Call RRR2Picture2.PSet (fj1, ffg)Next fj1End SubPrivate Sub Command3_Click() '求构件5的角速度Picture3.Scale (-20, 10)-(380, -10)Picture3.Line (-20, 0)-(380, 0) 'XPicture3.Line (0, 10)-(0, -10) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture3.DrawStyle = 2Picture3.Line (i, 10)-(i, -10)Picture3.CurrentX = i - 10: Picture3.CurrentY = 0 Picture3.Print iNext iFor i = -8 To 8 Step 2 'Y轴坐标Picture3.Line (0, i)-(380, i)Picture3.CurrentX = -20: Picture3.CurrentY = i Picture3.Print iNext iFor fj1 = 0 To 360 Step 0.01f = fj1 * paCall RR1Call RRR1Call RR2Call RRR2Picture3.PSet (fj1, wfg)Next fj1End SubPrivate Sub Command4_Click() '求构件5的角加速度Picture4.Scale (-20, 300)-(380, -200)Picture4.Line (-20, 0)-(380, 0) 'XPicture4.Line (0, 300)-(0, -200) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture4.DrawStyle = 2Picture4.Line (i, 300)-(i, -200)Picture4.CurrentX = i - 10: Picture4.CurrentY = 0 Picture4.Print iNext iFor i = -200 To 300 Step 50 'Y轴坐标Picture4.Line (0, i)-(380, i)Picture4.CurrentX = -25: Picture4.CurrentY = i + 5 Picture4.Print iNext iFor fj1 = 0 To 360 Step 0.01f = fj1 * paCall RR1Call RRR1Call RR2Call RRR2Picture4.PSet (fj1, efgNext fj1End SubPrivate Sub Form_Load() '赋初值lab = 108lce = 200lbc = 200lcd = 200lef = 320lfg = 162w = 10e = 0delt = 0xA = 0yA = 0vyA = 0axA = 0ayA = 0xD = -178.311284yD = 186.464704vxD = 0vyD = 0axD = 0ayD = 0xG = -514yG = 0vxG = 0vyG = 0axG = 0ayG = 0pi = 3.1415926pa = pi / 180febc = pa * 50End SubPrivate Sub RR1() 'RR基本杆组xB = xA + lab * Cos(f + delt)yB = yA + lab * Sin(f + delt)vxB = vxA - w * lab * Sin(f + delt)vyB = vyA + w * lab * Cos(f + delt)axB = axA - w ^ 2 * lab * Cos(f + delt) - e * lab * Sin(f + delt)ayB = ayA - w ^ 2 * lab * Sin(f + delt) + e * lab * Sin(f + delt)End SubPrivate Sub RR2() 'RR基本杆组leb = 2 * lbc * Cos(febc)xE = xB + leb * Cos(fbc + febc)yE = yB + leb * Sin(fbc + febc)vxE = vxB - wbc * leb * Sin(fbc + febc)vyE = vyB + wbc * leb * Cos(fbc + febc)axE = axB - wbc ^ 2 * leb * Cos(fbc + delt) - ebc * leb * Sin(fbc + febc) ayE = ayB - wbc ^ 2 * leb * Sin(fbc + delt) + ebc * leb * Sin(fbc + febc) End SubPrivate Sub RRR1() 'RRR基本杆组LBD = Sqr((xD - xB) ^ 2 + (yD - yB) ^ 2)If LBD > lbc + lcd And LBD < Abs(lbc - lcd) ThenIf MsgBox("RRR杆组杆长不符合要求", vbOKOnly, "提示") = 1 Then EndEnd IfElseEnd IfIf LBD < lbc + lcd And LBD > Abs(lbc - lcd) Then val = (lbc ^ 2 + LBD ^ 2 - lcd ^ 2) / (2 * lbc * LBD) JCBD = Atn(-val / Sqr(-val * val + 1)) + 2 * Atn(1) ElseEnd IfIf LBD = lbc + lcd ThenJCBD = 0ElseEnd IfIf LBD = Abs(lbc - lcd) ThenIf lbc > lcd ThenJCBD = 0ElseEnd IfIf lbc < lcd ThenJCBD = piElseEnd IfElseEnd IfIf xD > xB And yD >= yB Then '第一象限fBD = Atn((yD - yB) / (xD - xB))ElseEnd IfIf xD = xB And yD > yB ThenfBD = pi / 2ElseEnd IfIf xD < xB And yD >= yB Then '第二象限fBD = pi + Atn((yD - yB) / (xD - xB))ElseEnd IfIf xD < xB And yD < yB Then '第三象限fBD = pi + Atn((yD - yB) / (xD - xB))ElseEnd IfIf xD = xB And yD < yB ThenfBD = 3 * pi / 2ElseEnd IfIf xD > xB And yD <= yB Then '第四象限fBD = 2 * pi + Atn((yD - yB) / (xD - xB))ElseEnd Iffbc = fBD - JCBDxC = xB + lbc * Cos(fbc)yC = yB + lbc * Sin(fbc)If xC > xD And yC >= yD Then '第一象限fcd = Atn((yC - yD) / (xC - xD))ElseEnd IfIf xC = xD And yC >= yD Thenfcd = pi / 2ElseEnd IfIf xC < xD And yC >= yD Then '第二象限fcd = pi + Atn((yC - yD) / (xC - xD))ElseEnd IfIf xC < xD And yC < yD Then '第三象限fcd = pi + Atn((yC - yD) / (xC - xD))ElseEnd IfIf xC = xD And yC < yD Thenfcd = 3 * pi / 2ElseEnd IfIf xC > xD And yC <= yD Then '第四象限fcd = 2 * pi + Atn((yC - yD) / (xC - xD))ElseEnd IfCi = lbc * Cos(fbc)Si = lbc * Sin(fbc)Cj = lcd * Cos(fcd)Sj = lcd * Sin(fcd)G1 = Ci * Sj - Cj * Siwbc = (Cj * (vxD - vxB) + Sj * (vyD - vyB)) / G1 wcd = (Ci * (vxD - vxB) + Si * (vyD - vyB)) / G1 vxC = vxB - wbc * lbc * Sin(fbc)vyC = vyB + wbc * lbc * Cos(fbc)G2 = axD - axB + wbc ^ 2 * Ci - wcd ^ 2 * CjG3 = ayD - ayB + wbc ^ 2 * Si - wcd ^ 2 * Sj ebc = (G2 * Cj + G3 * Sj) / G1ecd = (G2 * Ci + G3 * Si) / G1axC = axB - ebc * lbc * Sin(fbc) - wbc ^ 2 * lbc * Cos(fbc)ayC = ayB + ebc * lbc * Cos(fbc) - wbc ^ 2 * lbc * Sin(fbc)End SubPrivate Sub RRR2() 'RRR基本杆组leg = Sqr((xG - xE) ^ 2 + (yG - yE) ^ 2)If leg > lef + lfg And leg < Abs(lef - lfg) ThenIf MsgBox("RRR杆组杆长不符合要求", vbOKOnly, "提示") = 1 Then EndElseEnd IfElseEnd IfIf leg < lef + lfg And leg > Abs(lef - lfg) Thenval = (lef ^ 2 + leg ^ 2 - lfg ^ 2) / (2 * lef * leg)jfeg = Atn(-val / Sqr(-val * val + 1)) + 2 * Atn(1)ElseEnd IfIf leg = lef + lfg Thenjfeg = 0ElseEnd IfIf leg = Abs(lef - lfg) ThenIf lef > lfg Thenjfeg = 0ElseEnd IfIf lef < lfg Thenjfeg = piElseEnd IfElseEnd IfIf xG > xE And yG >= yE Then '第一象限feg = Atn((yG - yE) / (xG - xE))ElseEnd IfIf xG = xE And yG > yE Thenfeg = pi / 2ElseEnd IfIf xG < xE And yG >= yE Then '第二象限feg = pi + Atn((yG - yE) / (xG - xE))ElseEnd IfIf xG < xE And yG < yE Then '第三象限feg = pi + Atn((yG - yE) / (xG - xE)) ElseEnd IfIf xG = xE And yG < yE Thenfeg = 3 * pi / 2ElseEnd IfIf xG > xE And yG <= yE Then '第四象限feg = 2 * pi + Atn((yG - yE) / (xG - xE)) ElseEnd Iffef = feg - jfegxF = xE + lef * Cos(fef)yF = yE + lef * Sin(fef)If xF > xG And yF >= yG Then '第一象限ffg = Atn((yF - yG) / (xF - xG))ElseEnd IfIf xF = xG And yF >= yG Thenffg = pi / 2ElseEnd IfIf xF < xG And yF >= yG Then '第二象限ffg = pi + Atn((yF - yG) / (xF - xG)) ElseEnd IfIf xF < xG And yF < yG Then '第三象限ffg = pi + Atn((yF - yG) / (xF - xG)) ElseEnd IfIf xF = xG And yF < yG Thenffg = 3 * pi / 2ElseEnd IfIf xF > xG And yF <= yG Then '第四象限ffg = 2 * pi + Atn((yF - yG) / (xF - xG)) ElseEnd IfCi = lef * Cos(fef)Si = lef * Sin(fef)Cj = lfg * Cos(ffg)Sj = lfg * Sin(ffg)G1 = Ci * Sj - Cj * Siwef = (Cj * (vxG - vxE) + Sj * (vyG - vyE)) / G1wfg = (Ci * (vxG - vxE) + Si * (vyG - vyE)) / G1vxF = vxE - wef * lef * Sin(fef)vyF = vyE + wef * lef * Cos(fef)G2 = axG - axE + wef ^ 2 * Ci - wfg ^ 2 * CjG3 = ayG - ayE + wef ^ 2 * Si - wfg ^ 2 * Sjeef = (G2 * Cj + G3 * Sj) / G1efg = (G2 * Ci + G3 * Si) / G1axF = axE - eef * lef * Sin(fef) - wef ^ 2 * lef * Cos(fef)ayF = ayE + eef * lef * Cos(fef) - wef ^ 2 * lef * Sin(fef)End Sub5.结果及分析图1 E点的运动轨迹(1)由图1所示,E点的运动轨迹呈稍倾斜“8字形”。

哈工大机械原理大作业-连杆

哈工大机械原理大作业-连杆

哈工大机械原理大作业-连杆本页仅作为文档页封面,使用时可以删除This document is for reference only-rar21year.MarchHarbin 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)基本杆组的划分①AB即杆件1为原动件②DECB即杆件2、3为RRR型II级杆组,其中CE为同一构件上点。

③ 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 πϕ2~01=原动件杆1的角速度 s rad /236.51=ω原动件1的角加速度01=α运动副A 的位置坐标0,0==A A y x运动副A 的速度0,0==yA xA v v运动副A 的加速度0,0==yA xA a a原动件杆I 的长度mm l 1501=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500400==D D y xD 的速度 0==yD xD v vD 的加速度00==yD xD a a杆长mm l l CD j 500==,mm l l BC i 600== 由关系j j D i i B C l x l x x ϕϕcos cos +=+= j j D i i B C l y l y y ϕϕsin sin +=+= 其中)(20sin cos B D i i i x x l A C B A -==-+ ϕϕ)(2B D i y y l B -=222j BD i l l l C -+=222)()(B D B D BD y y X x l -+-= 可解得C A C B A B i +-++=222arctan 2ϕDC DC j x x y y --=arctan ϕ由上面两个式子可以得到两杆的角速度1)()(G y y s x x c w B D j B D j i i -+-==•ϕ132G s G c Gij i i +==ϕα其中i j j i s c s c G -=1,i i i l c ϕcos =,i i i l s ϕsin =,j j j l c ϕcos =,j j j l s ϕsin =可得E 的位置iB E iB E y y x x ϕϕsin 480cos 480+=+=E 的速度 i i yB E yE i i xB E xE v yv v xv ωϕωϕcos 480sin 480+==-==E 的加速度i i i i xB E xE a xa αϕωϕsin 480cos 4802--== i i i i yB E yE a ya ϕαϕωcos 480sin 4802+-== (3)、构件4、5杆组(II 级杆组RRP )在建立的坐标系中取一参考点K600,0==K K y x 则速度0,0==yK xK v v加速度0,0==yK xK a a杆长mm l EF 600=,设F 位移为s 由s x l x x K i EF E F +=+='cos ϕK i EF E F y l y y =+='sin ϕ由上面两个式子可以得到600600arcsin arcsin'E EF E k i y l y y -=-=ϕ 所以:F 点位移 )600600arcsin(cos 600E E F y x x s -+== 速度F F xv = 加速度F F xa =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 Double Dim pi As DoubleDim pa As Doublepi = 3.1415926pa = pi / 180Dim i As LongDim f1(3600) As Double Dim RR1 As RRDim RR2 As RRDim RRR1 As RRRDim RRP1 As RRPSet RR1 = New RRSet RR2 = New RRSet RRR1 = New RRRSet RRP1 = New RRPFor i = 0 To 3600 Step 1 f1(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.xBRRR1.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.fi RR2.w = RRR1.wi RR2.e = RRR1.ei RR2.L = 480RR2.xA = RR1.xB RR2.yA = RR1.yB RR2.vxA = RR1.vxB RR2.vyA = RR1.vyB RR2.axA = RR1.axB RR2.ayA = RR1.ayB RR2.calRRP1.Li = 600RRP1.Lj = 0RRP1.fj = piRRP1.wj = 0RRP1.ej = 0RRP1.xB = RR2.xB RRP1.yB = RR2.yB RRP1.vxB = RR2.vxB RRP1.vyB = RR2.vyB RRP1.axB = RR2.axB RRP1.ayB = RR2.ayB RRP1.xK = 0RRP1.yK = 600RRP1.vxK = 0RRP1.vyK = 0RRP1.axK = 0RRP1.ayK = 0RRP1.M = 1RRP1.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) 'YFor i = 0 To 360 Step 10 'X轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 700)-(i, 580)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = 580 To 700 Step 10 'Y轴坐标Picture1.DrawStyle = 2Picture1.Line (0, i)-(360, i)Picture1.CurrentX = -10: Picture1.CurrentY = iPicture1.Print iNext iFor i = 0 To 3600 Step 1Picture1.PSet (i / 10, s5(i))Next iEnd SubRR:Public L As DoublePublic f As DoublePublic delt As DoublePublic w As DoublePublic e As DoublePublic xA As DoublePublic yA As DoublePublic vxA As DoublePublic vyA As DoublePublic axA As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic Sub cal()xB = xA + L * 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 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 DoublePublic ayC As DoublePublic xD As DoublePublic yD As DoublePublic vxD As DoublePublic vyD 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 As DoubleDim LBD As DoubleDim JCBD As DoubleDim val As Doublepi = 3.1415926LBD = Sqr((xB - xD) ^ 2 + (yD - yB) ^ 2)If LBD < Li + Lj And 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 DoublePublic 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 vyK As DoublePublic axK As 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 DoublePublic 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)) + 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 = (-Q1 * 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)) - 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 - 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。

哈尔滨工业大学机械原理大作业-连杆机构运动分析

哈尔滨工业大学机械原理大作业-连杆机构运动分析
end;
%打印图像
figure(1);
plot(fi1,sF);
title('位移变化曲线');
figure(2);
plot(fi1,vF);
title('速度变化曲线');
figure(3);
plot(fi1,aF);
title('加速度变化曲线');
六、计算结果
图8:推杆位移变化曲线
图9:推杆速度变化曲线
xD=400;yD=500;vDx=0;vDy=0;aDx=0;aDy=0;
xK=0;yK=600;vKx=0;vKy=0;aKx=0;aKy=0;
l1=150;l2=600;l3=500;lBE=480;l4=600;
fi5=pi;
n1=50;
w1=2*pi*n1/60;
fi1=linspace(0,2*pi,1000);
aEy(i)=aBy(i)-w2(i)^2*lBE*sin(fi2(i))+a2(i)*lBE*cos(fi2(i));
%求F点的运动参数
A1(i)=(yE(i)-yK)*cos(fi5)-(xE(i)-xK)*sin(fi5);
fi4(i)=fi5-asin(A1(i)/l4);
xF(i)=xE(i)+l4*cos(fi4(i));
(2)速度和加速度分析
B点的速度
B点的加速度
2、由“RRRⅡ级杆组”,已知B点和D点的运动参数,可求得构件2、构件3的运动参数
图5
D点的坐标方程
D点的速度
D点的加速度
(1)位置方程
构件2、构件3的长度
先求出 和
其中
则可求得

哈工大机械设计大作业平面连杆机构设计说明书

哈工大机械设计大作业平面连杆机构设计说明书

机械设计基础大作业计算说明书题目:平面连杆机构设计学院:材料学院班号:学号:姓名:日期: 2014年9月30日哈尔滨工业大学机械设计基础大作业任务书题目:平面连杆机构设计设计原始数据及要求:()目录1设计题目2设计原始数据3设计计算说明书3.1计算极位夹角3.2设计制图3.3验算最小传动角4参考文献1 设计题目平面连杆机构的图解法设计2 设计原始数据设计一曲柄摇杆机构。

已知摇杆长度,摆角,摇杆的行程速比系数,要求摇杆靠近曲柄回转中心一侧的极限位置与机架间的夹角为,试用图解法设其余三杆的长度,并检验(测量或计算)机构的最小传动角。

()3 设计计算说明书3.1 计算极位夹角极位夹角代入数值3.2 设计制图3.2.1 在图纸上取一点作为点,从点垂直向上引出一条长为的线段,终点为;3.2.2 从点在左侧引出一条与夹角为的射线;3.2.3 以点为圆心,以为半径画圆,与射线交于点;3.2.4 分别从、两点向下引两条射线,射线与夹角为,两射线交于点,点即为曲柄的回转中心;3.2.5 以点为圆心以为半径画圆;3.2.6 过点向左侧引出一条射线,射线与夹角,与圆交于点;3.2.7 连接,并量取其长度,以为半径画圆,直线,与圆的交点分别为,;3.2.8 在图中量取,,3.3 验算最小传动角3.3.1 在处根据余弦定理3.3.2 在处根据余弦定理所以最小传动角4 参考文献[1]宋宝玉,王瑜,张锋主编.机械设计基础.哈尔滨:哈尔滨工业大学出版,2010.[2]王瑜主编.机械设计基础大作业指导书.哈尔滨:哈尔滨工业大学出版社,2014.。

哈工大机械原理大作业连杆

哈工大机械原理大作业连杆

哈工大机械原理大作业-连杆连杆是机械原理中常见的机构之一,也是机械工程中非常重要的部件。

它由两个旋转接头和一个连接两个旋转接头的杆件组成。

连杆广泛应用于各种机械设备中,如汽车发动机、泵、机床等。

本文将介绍连杆的工作原理、应用以及设计要点。

连杆的工作原理是将旋转运动转化为直线运动或将直线运动转化为旋转运动。

它通过两个旋转接头的运动将杆件上的一个点的运动转化为另一个点的运动。

连杆的运动有两种基本形式:一是曲柄连杆机构,二是摇杆连杆机构。

曲柄连杆机构中,一个旋转接头为曲柄,另一个旋转接头为连杆;摇杆连杆机构中,一个旋转接头为摇杆,另一个旋转接头为连杆。

连杆广泛应用于各种机械设备中。

在汽车发动机中,连杆将曲轴的旋转运动转化为活塞的直线运动,从而驱动汽缸的工作;在泵中,连杆将电机的旋转运动转化为柱塞的直线运动,从而产生压力;在机床中,连杆将电机的旋转运动转化为工作台的直线运动,从而实现加工。

设计连杆时需要考虑一些要点。

首先是连杆的材料选择和尺寸设计。

连杆需要承受较大的力和扭矩,因此需要选择具有较高强度和刚度的材料。

同时,根据应用需求和力学原理,设计连杆的尺寸,以确保其能够承受正常工作条件下的负荷。

其次是连杆的润滑和密封。

连杆在工作过程中需要润滑剂来减少摩擦和磨损,同时需要密封装置来防止润滑剂泄漏。

因此,设计连杆时需要考虑润滑剂的供给和密封装置的设计。

最后是连杆的制造和装配。

连杆的制造需要保证其精度和质量,以确保其运转平稳和可靠。

在装配过程中,需要按照设计要求进行装配,同时进行必要的调试和检测,以确保连杆的工作性能符合要求。

总之,连杆是机械工程中非常重要的部件,广泛应用于各种机械设备中。

设计和制造连杆需要考虑材料选择、尺寸设计、润滑和密封以及制造和装配等方面的要点。

通过合理的设计和制造,可以确保连杆的工作性能和可靠性,从而提高机械设备的工作效率和寿命。

哈工大机械原理连杆和凸轮大作业24题

哈工大机械原理连杆和凸轮大作业24题

班级学号机械原理大作业说明书题目 1、连杆机构运动分析2、凸轮机构设计学生姓名1连杆机构运动分析1.设计题目:一、先建立如下坐标系:二、划分杆组如下,进行结构分析:该机构由I级杆组RR(如图1)、II级杆组RPR(如图2、3)和II级杆组RRP(如图4)组成。

(1)(2)(3)(4)三、运动分析数学模型:(1)同一构件上点的运动分析:如右图所示的原动件1,已知杆1的角速度=10/rad s ω,杆长1l =170mm,A y =0,A x =110mm 。

可求得下图中B 点的位置B x 、B y ,速度xB v 、yB v ,加速度xB a 、yB a 。

θcos 1l xB =,θsin 1l yB =θωυsin 1l xB -=,θωυcos 1l yB =,222B2==-cos =-BxB i d x a l x dt ωϕω2222==-sin =-B yB i B d y a l y dtωϕω。

(2)RPRII 级杆组的运动分析:a. 如右图所示是由2个回转副和1个移动副组成的II 级组。

已知两个外运动副C 、B 的位置(B x 、B y 、c x =110mm 、C y =0)、速度(xB υ,yB υ,xC υ=0,yC υ=0)和加速度(0,0,,==yC xC yB xB a a a a )。

可确定下图中D 点的位置、速度和加速度。

确定构件3的角位移1ϕ、角速度1ω、角加速度1α。

1sin 31..ϕϕl x dt dx C B -= 1s i n 131c o s 13.....2ϕϕϕϕl l x dt x d C B --= 1cos 31..ϕϕl y dt dy C B += 1c o s 131s i n 13.....2ϕϕϕϕl l y dty d C B +-= 根据关系:1111d 122..11.αϕϕωϕϕ====dtd dt , 故可得出: D x =)1cos(4βϕ++l x CD y =)1sin(4βϕ++l y Cb. 如右图所示是由2个回转副和1个移动副组成的II级组。

哈工大-机械原理大作业-连杆机构运动分析

哈工大-机械原理大作业-连杆机构运动分析

机械原理大作业(一)作业名称:连杆机构运动分析设计题目: 20院系:英才学院班级: XXXXXXX设计者:邵广斌学号: XXXXXXXXXX指导教师:林琳设计时间: 2013年05月19日哈尔滨工业大学机械设计1.运动分析题目如图所示机构,已知机构各构件的尺寸为150AB mm =,97β=︒,400BC mm =,300CD mm =,320AD mm =,100BE mm =,230EF mm =,400FG mm =,构件1的角速度为110/rad s ω=,试求构件2上点F 的轨迹及构件5上点G 的位移、速度和加速度,并对计算结果进行分析。

2. 机构分析该机构由原动件AB (Ⅰ级杆组)、BCD (RRR Ⅱ级杆组)和FG (RRP Ⅱ级杆组)组成。

3. 建立坐标系如图3,建立以定点A 为原点的平面直角坐标系A-xy 。

图1 运动机构结构图4. 运动分析数学模型4.1 原动件AB原动件AB 的转角: 10~2ψπ= 原动件AB 的角速度:110/rad s ω=原动件AB 的角加速度: 10α= 运动副A 的位置坐标: 0A x = 0A y =运动副A 的速度: 0xA v = 0yA v = 运动副A 的加速度: 0xA a = 0yA a =原动件AB 长度:150AB l mm =运动副B 的位置坐标: 1B A AB x x l cos ψ=+1B A AB y x l sin ψ=+运动副B 的速度: 11 xB xA AB v v l sin ωψ=-11 yB yA AB v v l cos ωψ=+运动副B 的加速度: 2 1111 xBxA AB AB a a l cos l sin ωψαψ=--21111yB yA AB AB a a l sin l cos ωψαψ=-+4.2 RRR Ⅱ级杆组BCD运动副D 的位置坐标: 320D x mm = 0D y = 运动副D 的速度: 0xD v = 0yD v = 运动副D 的加速度: 0xD a = 0yD a = 杆BC 长度: 400BC l mm = 杆CD 长度:300CD l mm =BC 相对于x 轴转角:200ψ=其中02BC D B A l x x =-() 0 2 BC D B B l y y =-()2220B B C C l C l D l D =+- 222())(BDD B D B l x x y y =-+- CD 相对于x 轴转角: 3C DC Dy y arctanx x ψ-=-求导可得BC 角速度2ω、角加速度2α以及CD 角速度3ω、角加速度3α。

哈工大机械原理大作业一连杆运动分析(02)

哈工大机械原理大作业一连杆运动分析(02)

哈⼯⼤机械原理⼤作业⼀连杆运动分析(02)⼀.设计题⽬⼆. 结构分析与基本杆组划分1.机构的结构分析机构各构件都在同⼀平⾯内运动,活动构件数n=3 P L=4 P H=0则机构的⾃由度为: F = 3n -2P L –P H = 3×3-2×4 = 12.基本杆组划分(1)去除虚约束和局部⾃由度本机构中⽆虚约束或局部⾃由度,此步骤跳过。

(2)拆杆组。

从远离原动件(即杆1)进⾏拆分,就可以得到由杆2,3组成的RRRⅡ级杆组和Ⅰ级机构杆1。

如下图:(3)确定机构的级别由(2)知,机构为Ⅱ级机构。

三. 运动分析数学模型以A为原点建⽴坐标系,如图:原动件AB的转⾓:φ1=0--2π运动副A的位置坐标:x A=0 y A=0 运动副D的位置坐标:x D=d y D=0 则运动副B的位置坐标:x B = acosφ1 y B = asinφ1其中:t=0:0.001:2*pi;a=60;b=90;c=120;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd)曲柄a=50,55,60,65红蓝绿黄b=90,c=120,d=100 t=0:0.001:2*pi; a=50;b=90;c=120;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'r'); hold on;grid on;t=0:0.001:2*pi;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'b'); hold on;t=0:0.001:2*pi;a=60;b=90;c=120;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'g'); hold on;t=0:0.001:2*pi;a=65;b=90;c=120;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'y');a=50,55,60,65红蓝绿黄2.摇杆c=105,115,125,135红蓝绿黄a=50,b=90,d=100 t=0:0.001:2*pi; a=50;b=90;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'r'); hold on;grid on;t=0:0.001:2*pi;a=50;b=90;c=115;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'b');t=0:0.001:2*pi;a=50;b=90;c=125;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'g'); hold on;t=0:0.001:2*pi;a=50;b=90;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'y');c=105,115,125,135红蓝绿黄3.连杆b=80,90,100,110红蓝绿黄a=50,c=120,d=100 t=0:0.001:2*pi; a=50;b=80;c=120;d=100;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'r'); hold on;grid on;t=0:0.001:2*pi;a=50;b=90;c=120;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'b'); hold on;t=0:0.001:2*pi;a=50;b=100;c=120;d=100;xa=0;ya=0;xd=d;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'g'); hold on;t=0:0.001:2*pi;a=50;b=110;c=120;d=100;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'y');b=80,90,100,110红蓝绿黄4.机架d=85,95,105,115 红蓝绿黄a=50,b=90,c=120 t=0:0.001:2*pi; a=50;b=90;c=120;d=85;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'r'); hold on;grid on;t=0:0.001:2*pi;a=50;b=90;c=120;d=95;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'b'); hold on;t=0:0.001:2*pi;d=105;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'g'); hold on;t=0:0.001:2*pi;a=50;b=90;c=120;d=115;xa=0;ya=0;xd=d;yd=0;xb=a.*cos(t);yb=a.*sin(t);m=xd-xb;n=yd-yb;lbd=(m.^2+n.^2).^(1/2);a0=2*b.*(xd-xb);b0=2*b.*(yd-yb);c0=b.^2+lbd.^2-c.^2;dd=2*atan((b0+(a0.^2+b0.^2-c0.^2).^(1/2))./(a0+c0)); plot(dd,'y');d=85,95,105,115 红蓝绿黄。

机械原理大作业一连杆机构运动分析24

机械原理大作业一连杆机构运动分析24

机械原理大作业一课程名称:机械原理设计题目:连杆机构运动分析题号:24院系:机电工程学院班级:*******设计者:********学号:**********指导教师:赵永强唐德威设计时间:2015年6月哈尔滨工业大学连杆机构运动分析题目24、如图1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,x G=25mm,y G=80mm,构件1的角速度为1=10rad/s,试求构件2上点E、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。

图11、建立坐标系建立以点A为原点的固定平面直角坐标系A-xy,如图2所示。

图22、对机构进行结构分析该机构由原动件1(I级机构)、RRR II级杆组(杆2、杆3)和RPR II级杆组(滑块4、杆5)组成。

I级机构如图3所示,II级杆组RRR如图4所示,II级杆组RPR如图5所示。

图3 图4 图53、各基本杆组运动分析的数学模型(1)同一构件上点的运动分析如图6所示的构件AB,,已知杆AB 的角速度=10/rad s ω,AB 杆长i l =280mm,可求得B 点的位置B x 、B y ,速度xB v 、yB v ,加速度xB a 、yB a 。

=cos =280cos B i x l ϕϕ; =sin =280sin B i y l ϕϕ;==-sin =-BxB i B dx v l y dt ωϕω; ==cos =;B yB i B dyv l x dt ωϕω222B 2==-cos =-BxB i d x a l x dt ωϕω; 图62222==-sin =-ByB i B d y a l y dtωϕω(2)RRRII 级杆组的运动分析如图7所示是由三个回转副和两个构件组成的II 级杆组。

已知两杆的杆长2l 、3l 和两个外运动副B 、D 的位置(B x 、B y 、D x 、D y )、速度( xB yB xD yD v v v v 、、、 )和加速度(xB yB xD yD a a a a 、、、)。

哈工大机械原理大作业一连杆-15题

哈工大机械原理大作业一连杆-15题

Harbin Institute of Technology机械原理大作业一课程名称:机械原理设计题目:连杆传动设计院系:船舶与海洋工程学院班级:1513102班分析者:刘康哲学号:*********指导教师:***设计时间:2017.06.051.题目:如图所示机构,已知机构各构件的的尺寸为l AB=100mm,l BD=400mm,l AE=400mm,l BC=200mm,曲柄AB的角速度为w1=10rad/s,试求构件2上点D的轨迹,构件5的角位移、角速度和角加速度,并对结果进行分析。

(题中构件尺寸满足l BD-l AB<l AE<l BD+l AB)。

2.机构的结构分析及基本杆组划分机构各构件都在同一平面内运动,可拆分成一个原动件(构件1),一个RRPⅡ级基本杆组(构件2、3)和一个RPR Ⅱ级基本杆组(构件4、5)。

活动构件数n=5,P L,P H,因此,机构的自由度为F=3*n-2* P L- P H =3*5-2*7=1原动件:RRPⅡ级基本杆组:RPR Ⅱ级基本杆组:3.坐标系的建立以A点为坐标原点,AE方向为x轴正方向,建立平面直角坐标系。

4.各基本杆组的运动分析数学模型(1)原动件:位置分析: X B = X A + L AB * cos(wt) Y B = Y A + L AB * sin(wt) 其中,X A = Y A = 0 ,w = 10rad/s , L AB = 100 mm 速度分析:将上式分别对时间求导,可得速度方程,V XB = V XA – w*L AB *sin(wt)V YB = V YA + w*L AB *cos(wt)其中,V XA = V YA = 0 加速度分析:将上式分别对时间求导,可得速度方程,a XA = a XA – α* L AB * sin(wt) – w*w*L AB *cos(wt)a YA = a YA + α* L AB * cos(wt) – w*w*L AB *sin(wt)其中,a XA = a YA = 0 α= dw/dt(2)RRP Ⅱ级基本杆组:位置分析: X C = LAB*cos(wt)+wt Lab Lbc 222sin *Y C = 0 可求得,β= arcsin (Y B / L BC ) 速度分析:将上式分别对时间求导,可得速度方程,V XC =-w* LAB*cos(wt) – (wt Lab Lbc 222sin * )-1/2 * sin(wt)*wV YC = 0加速度分析:将上式分别对时间求导,可得加速度方程, (3)RPR Ⅱ级基本杆组位置分析: X D = X C + L CD * cos βY D = - L CD * sin β可求得, θ = arctan (-YD / (LAE -XD) ) 速度分析:将上式分别对时间求导,可得速度方程,5.计算编程在MATLAB 中编写如下程序:function [ XD,YD,Q,Q1,Q2] = calculate(t) %UNTITLED4 Summary of this function goes here % Detailed explanation goes here Lab=100; Lbc=200; Lcd=200; Lae=400; w=10;XB=Lab*cos(w.*t); YB=Lab*sin(w.*t); P=asin(YB./Lbc); XC=sqrt(Lbc.^2-YB.^2)+XB; XD=Lcd.*cos(P)+XC; YD=-Lcd.*sin(P); Q=atan(YD./(Lae-XD)); Q1=diff(Q); Q2=diff(Q1); end5.计算结果t 时间 XD D 点横坐标YD D 点纵坐标Q构件5角度Q1 构件5角速度 Q2构件5角加速度5000.10050.003008D点运动轨迹:构件5的角位移线图:构件5的角速度图:构件5的角加速度线图:6.计算结果分析原动件1可做整周回转,同时构件5也可做整周运动。

哈工大机械原理大作业连杆

哈工大机械原理大作业连杆

哈工大机械原理大作业连杆Modified by JACK on the afternoon of December 26, 2020Harbin 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:机电工程学院班级:设计者:学号:指导教师:设计时间:1.运动分析题目(11)在图所示的六杆机构中,已知:AB l =150mm, AC l =550mm, BD l =80mm, DE l =500mm,曲柄以等角速度1w =10rad/s 沿逆时针方向回转,求构件3的角速度、角加速度和构件5的位移、速度、加速度。

2.机构的结构分析建立以点A 为原点的固定平面直角坐标系A-x, y,如下图:机构结构分析该机构由Ⅰ级杆组RR (原动件1)、Ⅱ级杆组RPR (杆2及滑块3)和Ⅱ级杆组RRP (杆4及滑块5)组成。

3.建立组成机构的各基本杆组的运动分析数学模型原动件1(Ⅰ级杆组RR )由图所示,原动件杆1的转角a=0-360°,角速度1w =10rad/s ,角加速度1a =0,运动副A 的位置坐标A x =A y =0,速度(A ,A),加速度(A ,A ),原动件1的长度AB l =150mm 。

求出运动副B 的位置坐标(B x , B y )、速度(B ,B )和加速度(B ,B )。

杆2、滑块3杆组(RPR Ⅱ级杆组)已出运动副B 的位置(B x , B y )、速度(B ,B )和加速度(B ,B ),已知运动副C 的位置坐标C x =0, C y =550mm,速度,加速度,杆长AC l =550mm 。

求出构件2的转角b,角速度2w 和角加速度2a . 构件二上点D 的运动已知运动副B 的位置(B x , B y )、速度(B ,B )、加速度(B ,B ),已经求出构件2的转角b ,角速度2w 和角加速度2a ,杆BD 的长度BD l =80mm 。

哈工大机械原理凸轮大作业-24江湖青衫版

哈工大机械原理凸轮大作业-24江湖青衫版

实用标准H a r b i n I n s t i t u t e o f T e c h n o l o g y机械原理大作业(二)课程名称:机械原理设计题目:凸轮机构设计-24院系:机电工程学院班级: 1108***班设计者:江湖青衫学号: 11108*****指导教师:×××设计时间: 2013年6月哈尔滨工业大学机电工程学院一、设计题目 如右图所示直动从动件盘形凸轮机构,选择一组凸轮机构的原始参数,据此设计该凸轮机构。

二. 凸轮推杆升程、回程运动方程及推杆位移、速度、加速度线图凸轮推杆正弦加速度升程运动方程:)]2sin(21[s oo h Φ-Φ=πϕπϕ)2sin(2)]2cos(1[v 2121ϕππϕπωωo oh a oo h ΦΦ=Φ-Φ=凸轮推杆余弦加速度回程运动方程:[]})('cos1{2s s o o hΦ+Φ-Φ+=ϕπ[])('cos '2)](['sin'2v 21221s o o o h a s o o o hΦ+Φ-ΦΦ-=Φ+Φ-ΦΦ-=ϕππϕππωω1、推杆位移线图% t表示转角 ,s表示位移t=0:0.01:5*pi/6; %升程阶段s=120*[(6*t)/(5*pi)-1/(2*pi)*sin(12*t/5)]; hold onplot(t,s);t= 5*pi/6:0.01:10*pi/9; %远休止阶段s=120;hold onplot(t,s);t=10*pi/9:0.01:5*pi/3; %回程阶段s=60*[1+cos(9*(t-10*pi/9)/5)];hold onplot(t,s);t=5*pi/3:0.01:2*pi; %近休止阶段s=0;hold onplot(t,s);grid onhold offtitle('推杆位移线图','FontSize',20);2、推杆速度线图% t表示转角 ,令ω1=1t=0:0.01:5*pi/6; %升程阶段v=144*1*[1-cos(12*t/5)]/pihold onplot(t,v);t= 5*pi/6:0.01:10*pi/9; %远休止阶段v=0hold onplot(t,v);t=10*pi/9:0.01:5*pi/3; %回程阶段v=-108*1*sin(9*(t-10*pi/9)/5)hold onplot(t,v);t=5*pi/3:0.01:2*pi; %近休止阶段v=0hold ontitle('推杆速度线图','FontSize',20);3、推杆加速度线图% t表示转角 ,令ω1=1t=0:0.001:5*pi/6; %升程阶段a=345.6*sin(12*t/5)/pi;hold onplot(t,a);t=5*pi/6:0.01:10*pi/9; %远休止阶段a=0;hold onplot(t,a);t=10*pi/9:0.001:5*pi/3; %回程阶段a=-194.4*cos(9*(t-10*pi/9)/5);hold onplot(t,a);t=5*pi/3:0.001:2*pi; %近休止阶段a=0;hold ontitle('推杆加速度线图','FontSize',20);三. 绘制凸轮机构的sdds-ϕ线图% t表示转角ϕ,x(横坐标)表示速度ds/dϕ,y(纵坐标)表示位移s t=0:0.001:5*pi/6; % 升程阶段x= 144/pi-144*cos(12*t/5)/pi;y= 120*[(6*t)/(5*pi)-1/(2*pi)*sin(12*t/5)];hold onplot(x,y,'-r');t= 5*pi/6:0.01:10*pi/9; %远休止阶段x=0;y=120;hold onplot(x,y,'-r');t=10*pi/9:0.001:5*pi/3; % 回程阶段x=-180*1*sin(9*(t-10*pi/9)/5);y=60*[1+cos(9*(t-10*pi/9)/5)];hold onplot(x,y,'-r');t=14*pi/9:0.01:2*pi; %近休止阶段x=0;y=0;hold onplot(x,y,'-r');grid onhold offtitle('凸轮机构的ds/dφ-s线图','FontSize',20);四.按许用压力角确定凸轮基圆半径和偏距1. 求切点转角(1)在图-4中,右侧曲线为升程阶段的类速度-位移图,作直线D t d t与其相切,且位移轴正方向呈夹角[α1]=400,但是为了安全起见,可缩小需用压力角至[α1]=300,则切点处的斜率与直线D t d t的斜率相等,故右侧曲线斜率可以表示为k Dtdt=tan600, 通过编程求其角度。

哈工大机械原理大作业连杆机构运动分析完美满分版哈尔滨工业大学

哈工大机械原理大作业连杆机构运动分析完美满分版哈尔滨工业大学

连杆机构运动分析说明书院(系)机电工程学院专业机械设计制造及其自动化姓名李乾学号1130810904班号1308109指导教师唐德威、赵永强日期2015年6月20日哈尔滨工业大学机电工程学院2015年6月一、题目如图1所示机构,已知机构各构件的尺寸为l AB=200mm,l BD=700mm,l AC=400mm,l AE=800mm,构件1的角速度为ω1=10rad/s,试求构件2上点D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。

(题中构件尺寸满足l BD-l AB<l AE<l BD+l AB)。

图 1 机构运动简图二、建立数学模型分析1.建立坐标系建立以点A为原点的平面直角坐标系A-x,y,如图2所示图 2 建立坐标系2.对机构进行结构分析该机构由Ⅰ级机构AB、两个RPRⅡ级基本杆组BCD、ED组成。

杆组拆分结果如图3、图4、图5所示。

图 3 Ⅰ级杆组AB图 4 RPRⅡ级基本杆组BCD图 5 RPRⅡ级基本组DE3.确定已知参数和求解流程(1)原动件AB(I级杆组)已知原动件1的转角φ=0~360°运动副A的运动参数x A=0y A=0原动件AB的长度l AB = 200mm代入I级杆组子程序,得到运动副B的位置坐标(x B,y B)根据《机械原理》第三版书中第36页的公式推导可知:A,B两点坐标在x轴,y轴上投影,得方程x B = x A+l AB*cosφy B = y A+l AB*sinφ(2)BCD(RPR II级杆组)已知运动副B的位置坐标(x B,y B)运动副C的坐标位置:x C=l AC=400mmy C=0代入RPR II级杆组子程序,求出构件2上D点的位置坐标(x D,y D)根据《机械原理》第三版书中第339页的公式推导可知:当杆件处于图所示位置,即x B>x D并且y B≥y D时,l j杆角位移:φj=arctan B0s+A0C0 A0s−B0C0式中:A0=x B-x DB0=y B-y DC0=l i+l ks=√A02+B02−C02而当x B<x D并且y B≥y D时,φj=arctan B0s+A0C0A0s−B0C0+180o 当x B<x D并且y B<y D时,φj=arctan B0s+A0C0A0s−B0C0+180o 当x B>x D并且y B<y D时,φj=arctan B0s+A0C0A0s−B0C0+360o图 6 RPR II级杆组分析内移动副C的位置:x C=x B-l i sinφjy C=y B-l i cosφj导杆上E点的位置:x E=x C+(l j-s)cosφjy E=y C+(l j-s)sinφj(3)DE(RPR II级杆组)已知运动副D的位置坐标(x D,y D),运动副E的坐标:x E=l AE=800mmy E=0代入RPR II级杆组子程序,求出构件5的转角φ5。

哈工大机械原理大作业——连杆机构运动分析报告162014

哈工大机械原理大作业——连杆机构运动分析报告162014
机械原理大作业——连杆
课程名称: 机械原理

系:
能源科学与工程学院

级:
完成
者:

号:

号:
16
任课教师: 丁刚
完成内容: 在完成题目计算要求的同时, 扩展了内容, 程序为该结构的通用程 序,可解决机构在不同条件下的运动情况,文本最末为几种情况的分析
哈尔滨工业大学
16、如图所示机构,已知机构各构件的尺寸为
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这是同一根杆, °,λ,λ
现在来看 , , ,由题目得, , 和 是未知的,但不影响整 体,不然给 和 一个初值 °, ,当然,这是可以随意更改的。
基于以上的基本原理,
syms theta theta1 theta2 lamuda lamuda1 lamuda2 sigma sigma1 sigma2 beta beta1 beta2 l1 l11 l2 l21 t output i

机械原理大作业连杆机构.

机械原理大作业连杆机构.

Harbin Institute of Technology大作业设计说明书课程名称:机械原理设计题目:连杆机构设计院系:班级:设计者:学号:指导教师:设计时间:2013-6-13哈尔滨工业大学1.连杆题目(16):如图所示机构,已知机构各构件的尺寸为==100AC CE l l mm ,==200BC CD l l mm ,90BCD ∠=︒,构件1的角速度为1=10/w rad s ,试求构件5的角位移、角速度和角加速度,并对计算结果进行分析。

1.1 机构的运动分析AB 为原动件,AB 转动通过转动导杆机构带动杆BCD 转动,BCD 转动通过转动导杆机构带动杆DE 摆动。

1.2 机构的结构分析杆组可以划分为一个RR I级杆组(杆1)、RRPII级杆组(滑块2,杆3)、RPRII 级杆组(滑块4,杆5)1)RRI级杆组1:2)RRPII级杆组2,3:3)RPRII级杆组4,5:1.3 机构各杆组的运动分析数学模型1)RRI 级杆组1:B 点位移方程:abab ·cos ·sin B A ab B A ab x x l y y l ϕϕ=+⎧⎨=+⎩B 点速度方程: xB xA ab ab ab yB yA ab ab abv v l Sin v v l Cos ωϕωϕ=-⎧⎨=+⎩B 点加速度方程:22 cos sin xB xA ab ab AB yB yA ab ab AB a a l a a l ωϕωϕ⎧=-⎪⎨=+⎪⎩2)RRPII 级杆组2,3: 以A 点为参考点-y -y Cos C A ab C A abAO ϕϕ=(x -x )Sin ()0.5ab MO Cos ϕ=2=tan ()++1bc ab arc MO ϕϕ-+B C bc bc l Cos ϕ=x x+B C bc bc l Sin ϕ=y y-=B Aabs Cos ϕx x1xA xC ab ab Q v v w sSin ϕ=--2=+yA yC ab ab Q v v w sCos ϕ-3=+bc bc ab bc bc ab Q l Sin Sin l Cos Cos ϕϕϕϕ123-+=ab abbc Q Sin Q Cos w Q ϕϕ123()+()=-bc bc bc bc s Q l Cos Q l Sin v Q ϕϕ224=--2bc ab bc bc ab s ab ab Q w l Cos w sCos v w Sin ϕϕϕ 225=--2bc ab bc bc ab s ab ab Q w l Sin w sSin v w COs ϕϕϕ45=-+bc ab ab Q Sin Q COs αϕϕ3)RPRII 级杆组4,5:de l =arctan arctan arctan arctan 2 0.5 y D Ede D E D Ede D E D Ede D E D Ede D E de dey y x x y y x x y y x x y y x x ϕϕπϕπϕπϕπϕ-=--=+--=+--=+-=第一象限第二象限第三象限第四象限轴正半轴1.5 y π⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪=⎪⎩轴负半轴()yDde xD dededev Cos v Sin l ϕϕω-=() 2yD de de de de dede xD xD yD dea Cos a Sin v Cos v Sin w l ϕϕϕϕα--+=1.4 机构各杆组的编程1)对RRI级杆组1::xB = lab * Cos(fab)yB = lab * Sin(fab)vxB = -wab * lab * Sin(fab)vyB = wab * lab * Cos(fab)axB= -wab ^ 2 * lab * Cos(fab) - eab * lab * Sin(fab )ayB = -wab ^ 2 * lab * Sin(fab) + eab * lab * Cos(fab)1)对RRPII级杆组2,3:A0 = (xC - xA) * Sin(fab) - (yC - yA) * Cos(fab)M0 = 0.5 * Cos(fab)fi = Atn(M0 / Sqr(-M0 * M0 + 1)) + fabxB = xC + lbc * Cos(fbc)yB = yC + lbc * Sin(fbc)s = (xB - xA) / Cos(fab)Q1 = vxA - vxC - wab * s * Sin(fab)Q2 = vyA - vyC + wab * s * Cos(fab)Q3 = lbc * Sin(fbc) * Sin(fab) + lbc * Cos(fbc) * Cos(fab)wbc = (-Q1 * Sin(fab) + Q2 * Cos(fab)) / Q3vs = -(Q1 * lbc * Cos(fbc) + Q2 * lbc * Sin(fbc)) / Q3Q4 = wbc ^ 2 * lbc * Cos(fbc) - wab ^ 2 * s * Cos(fab) - 2 * vs * wab * Sin(fab) Q5 = wbc ^ 2 * lbc * Sin(fbc) - wab ^ 2 * s * Sin(fab) + 2 * vs * wab * Cos(fab) ebc = (-Q4 * Sin(fab) + Q5 * Cos(fab))2)对RPRII级杆组4,5:lde = Sqr((xD - xE) ^ 2 + (yD - yE) ^ 2)If xD>xE And yD>yE Then '第一象限fde = Atn((yD - yE) / (xD - xE))ElseEnd IfIf xD<xE And yD>= yE Then '第二象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD<xE And yD<yE Then '第三象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD>xE And yD<= yE Then '第四象限fde = Atn((yD - yE) / (xD - xE)) + 2 * piElseEnd IfIf xD = xE And yD>yE Then 'y轴正向fde = 0.5 * piElseEnd IfIf xD = xE And yD<yE Then 'y轴负向fde = 1.5 * piElseEnd Ifwde = (vyD * Cos(fde) - vxD * Sin(fde)) / ldeede = ((ayD * Cos(fde) - axD * Sin(fde) - 2 * (vxd * Cos(fde) + vyd * Sin(fde)) * wde) / lde2.计算编程以A为坐标原点,建立坐标系计算编程,源代码如下:Option Explicit '定义自变量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 atd As DoubleDim adn 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 lbc As Double '杆BC的长度Dim lcd As Double '杆CD的长度Dim fab As Double '杆AB的角位移Dim fbc As Double '杆BC的角位移Dim fde As Double '杆DE的角位移Dim fj1 As Double '循环变量Dim wab As Double '杆AB的角速度Dim wbc As Double '杆BC的角速度Dim wde As Double '杆DE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 ede As Double '杆DE的角加速度Dim lab As Double 'AB的距离Dim lde As Double 'DE的距离Dim s As Double 's的长度Dim vs As Double 's的速度Dim Q1 As Double 'RRP中的Q1Dim Q2 As Double 'RRP中的Q2Dim Q3 As Double 'RRP中的Q3Dim Q4 As Double 'RRP中的Q4Dim Q5 As Double 'RRP中的Q5Dim A0 As Double '杆组的中间变量Dim M0 As DoubleDim 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() '赋值Form1.WindowState = 2lbc = 200lcd = 200wab = 10eab = 0xA = 0yA = 0vxA = 0vyA = 0axA = 0ayA = 0xC = 0yC = -100vxC = 0vyC = 0axC = 0ayC = 0xE = 0yE = -200vxE = 0vyE = 0axE = 0ayE = 0pi = 4 * Atn(1)pa = pi / 180fj1 = 0End SubPrivate Sub Command1_Click()Set Picture1.Picture = NothingPicture1.Scale (-20, 8)-(400, -2)Picture1.Line (-20, 0)-(400, 0) 'XPicture1.Line (0, 8)-(0, -2) 'YFor i = -20 To 400 Step 50 'X轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 8)-(i, -2)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = -2 To 8 Step 1 'Y轴坐标Picture1.DrawStyle = 2Picture1.Line (-20, i)-(400, i)Picture1.CurrentX = -10: Picture1.CurrentY = iPicture1.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture1.PSet (fj1, fde), vbRedNext fj1End SubPrivate Sub Command2_Click()Set Picture2.Picture = NothingPicture2.Scale (-20, 30)-(400, -4)Picture2.Line (-20, 0)-(400, 0) 'XPicture2.Line (0, 30)-(0, -4) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture2.DrawStyle = 2Picture2.Line (i, 30)-(i, -4)Picture2.CurrentX = i - 10: Picture2.CurrentY = 0 Picture2.Print iNext iFor i = -4 To 30 Step 4 'Y轴坐标Picture2.Line (0, i)-(400, i)Picture2.CurrentX = -20: Picture2.CurrentY = iPicture2.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture2.PSet (fj1, wde), vbRedNext fj1End SubPrivate Sub Command3_Click() '杆5的角加速度Set Picture3.Picture = NothingPicture3.Scale (-20, 10000)-(400, -20000)Picture3.Line (-20, 0)-(400, 0) 'XPicture3.Line (0, 10000)-(0, -20000) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture3.DrawStyle = 2Picture3.Line (i, 10000)-(i, -20000)Picture3.CurrentX = i - 10: Picture3.CurrentY = 0Picture3.Print iNext iFor i = -20000 To 10000 Step 2500 'Y轴坐标Picture3.Line (0, i)-(400, i)Picture3.CurrentX = -25: Picture3.CurrentY = i + 5 Picture3.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture3.PSet (fj1, ede), vbRedNext fj1End SubPrivate Sub RRP() 'Ⅱ级杆组RRP(滑块2、杆3)A0 = (xC - xA) * Sin(fab) - (yC - yA) * Cos(fab)M0 = 0.5 * Cos(fab)fbc = Atn(M0 / Sqr(-M0 * M0 + 1)) + fabxB = xC + lbc * Cos(fbc)yB = yC + lbc * Sin(fbc)If fab = pi / 2 Thens = lbc - (yA - yC)ElseIf fab = 3 * pi / 2 Thens = lbc + (yA - yC)Elses = (xB - xA) / Cos(fab)End IfQ1 = vxA - vxC - wab * s * Sin(fab)Q2 = vyA - vyC + wab * s * Cos(fab)Q3 = lbc * Sin(fbc) * Sin(fab) + lbc * Cos(fbc) * Cos(fab)wbc = (-Q1 * Sin(fab) + Q2 * Cos(fab)) / Q3vs = -(Q1 * lbc * Cos(fbc) + Q2 * lbc * Sin(fbc)) / Q3Q4 = wbc ^ 2 * lbc * Cos(fbc) - wab ^ 2 * s * Cos(fab) - 2 * vs * wab * Sin(fab) Q5 = wbc ^ 2 * lbc * Sin(fbc) - wab ^ 2 * s * Sin(fab) + 2 * vs * wab * Cos(fab) ebc = (-Q4 * Sin(fab) + Q5 * Cos(fab))atd = lbc * ebcadn = wbc ^ 2 * lbcxD = xC + lbc * Sin(fbc)yD = yC - lbc * Cos(fbc)vxD = wbc * lbc * Cos(fbc)vyD = wbc * lbc * Sin(fbc)axD = -adn * Sin(fbc) + atd * Cos(fbc)ayD = adn * Cos(fbc) + atd * Sin(fbc)End SubPrivate Sub RPR() 'Ⅱ级杆组RPR(滑块4、杆5)lde = Sqr((xD - xE) ^ 2 + (yD - yE) ^ 2)If xD>xE And yD>yE Then '第一象限fde = Atn((yD - yE) / (xD - xE))ElseEnd IfIf xD<xE And yD>= yE Then '第二象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD<xE And yD<yE Then '第三象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD>xE And yD<= yE Then '第四象限fde = Atn((yD - yE) / (xD - xE)) + 2 * piElseEnd IfIf xD = xE And yD>yE Then 'y轴正向fde = 0.5 * piElseEnd IfIf xD = xE And yD<yE Then 'y轴负向fde = 1.5 * piElseEnd Ifwde = (vyD * Cos(fde) - vxD * Sin(fde)) / ldeede = (ayD * Cos(fde) - axD * Sin(fde) - 2 * (vxD * Cos(fde) + vyD * Sin(fde)) * wde) / ldeEnd Sub3.计算结果(绘制构件运动图线):构件5角位移构件5角速度图像构件5角加速度图像4. 构件运动属性随主动件的转动变化列举(角度变化为Δθ=1°):角构件5角位移(rad) 构件5角速度(rad/s) 构件5角加速度(rad/s2) 度0 5.651270995 14.76627109 -14893.44451 5.676722398 14.40057723 -14666.11492 5.701544405 14.04504413 -14437.389243 5.72575477 13.69968448 -14207.738354 5.749371219 13.36445541 -13977.573885 5.772411362 13.03926746 -13747.254436 5.794892617 12.72399249 -13517.091297 5.81683215 12.41847071 -13287.353848 5.838246824 12.12251691 -13058.274499 5.859153164 11.83592579 -12830.0532810 5.879567318 11.55847668 -12602.8619311 5.899505042 11.28993752 -12376.8475912 5.918981679 11.0300683 -12152.1361113 5.938012146 10.77862402 -11928.8350114 5.956610932 10.53535705 -11707.0360415 5.974792092 10.30001929 -11486.8174616 5.992569251 10.07236376 -11268.2460217 6.009955602 9.85214605 -11051.3786618 6.026963915 9.639125439 -10836.2639619 6.043606546 9.433065763 -10622.9434620 6.059895442 9.23373613 -10411.4526421 6.07584215 9.040911451 -10201.8219122 6.091457833 8.854372832 -9994.07727823 6.106753275 8.673907846 -9788.24104524 6.121738897 8.499310711 -9584.3322825 6.136424764 8.330382385 -9382.36725426 6.150820603 8.166930588 -9182.35977527 6.164935808 8.008769777 -8984.32146228 6.178779457 7.855721072 -8788.26195529 6.192360321 7.707612149 -8594.18907930 6.205686876 7.564277103 -8402.10897231 6.218767314 7.425556294 -8212.02617732 6.231609553 7.291296171 -8023.94371133 6.244221248 7.161349095 -7837.86311234 6.256609803 7.035573143 -7653.78447335 6.268782379 6.913831914 -7471.70646336 6.2807459 6.795994337 -7291.62634137 9.32E-03 6.68193447 -7113.53996638 2.09E-02 6.571531308 -6937.44180339 0.032262796 6.464668594 -6763.32492540 4.35E-02 6.361234628 -6591.18102541 5.45E-02 6.26112209 -6421.00041842 6.53E-02 6.164227865 -6252.77205243 0.075988655 6.070452866 -6086.48352644 8.65E-02 5.979701878 -5922.121145 9.69E-02 5.891883397 -5759.66971946 0.107072173 5.806909481 -5599.11303747 0.117135002 5.724695602 -5440.43344848 0.127056689 5.645160514 -5283.6121149 0.136841842 5.568226113 -5128.62898950 0.146494934 5.493817317 -4975.46289451 0.15602031 5.421861945 -4824.09151752 0.16542219 5.352290598 -4674.4914853 0.174704676 5.285036556 -4526.63838254 0.183871758 5.220035668 -4380.50684555 0.192927313 5.157226257 -4236.07056556 0.201875114 5.096549022 -4093.30236857 0.210718833 5.037946952 -3952.17425558 0.219462042 4.981365236 -3812.65746359 0.228108222 4.926751184 -3674.72251360 2.37E-01 4.874054148 -3538.3392761 2.45E-01 4.823225447 -3403.4769962 2.53E-01 4.774218296 -3270.10438263 2.62E-01 4.726987743 -3138.18965364 2.70E-01 4.681490596 -3007.70056765 2.78E-01 4.637685368 -2878.60449466 0.286188901 4.595532216 -2750.86846467 0.294174009 4.554992884 -2624.45921368 0.302089743 4.51603065 -2499.34323469 0.309938826 4.478610274 -2375.48682770 0.31772392 4.442697951 -2252.85614471 0.325447626 4.40826126 -2131.41723472 0.333112495 4.375269121 -2011.13608873 0.34072102 4.343691751 -1891.97868474 0.348275646 4.313500625 -1773.91102575 0.355778768 4.284668431 -1656.89918776 0.363232735 4.257169038 -1540.9093577 0.370639852 4.230977453 -1425.90784578 0.37800238 4.206069791 -1311.86118679 0.385322539 4.182423239 -1198.73611280 0.392602512 4.160016021 -1086.49961881 0.399844443 4.13882737 -975.118993782 0.407050442 4.118837498 -864.561859383 0.414222585 4.100027562 -754.796196684 0.421362914 4.08237964 -645.790384785 0.428473444 4.065876704 -537.51323386 0.435556158 4.050502591 -429.934014487 0.442613013 4.036241977 -323.022498488 0.44964594 4.023080356 -216.748983889 0.456656844 4.011004013 -111.08433290 0.463647609 4 -691 0.470620095 3.990056116 98.5319262392 0.477576142 3.981160883 202.538697793 0.484517572 3.973303522 306.046868294 0.491446186 3.966473937 409.082259795 0.49836377 3.960662689 511.669926496 0.505272094 3.955860979 613.834118797 0.512172913 3.952060626 715.598246298 0.519067966 3.949254046 816.984839299 0.525958984 3.947434235 918.0155099 100 0.532847682 3.946594748 1018.710912 101 0.539735765 3.946729679 1119.0907 102 0.546624931 3.947833642 1219.173486 103 0.553516865 3.94990175 1318.976795 104 0.560413247 3.9529296 1418.51702 105 0.567315749 3.956913248 1517.809379 106 0.574226034 3.961849192 1616.867862107 0.581145763 3.967734353 1715.705182 108 0.588076591 3.97456605 1814.332729 109 0.595020166 3.982341987 1912.760512 110 0.601978135 3.991060224 2010.997111 111 0.608952141 4.000719164 2109.049616 112 0.615943826 4.011317523 2206.923577 113 0.622954827 4.022854314 2304.622942 114 0.629986782 4.035328821 2402.150002 115 0.637041327 4.048740578 2499.505327 116 0.644120098 4.063089343 2596.687713 117 0.651224729 4.078375073 2693.694114 118 0.658356857 4.094597902 2790.519586 119 0.665518118 4.111758112 2887.157222 120 0.672710146 4.129856105 2983.598094 121 0.679934581 4.148892379 3079.831188 122 0.687193059 4.168867498 3175.843346 123 0.69448722 4.189782061 3271.619206 124 0.701818704 4.211636674 3367.141145 125 0.709189153 4.234431917 3462.389221 126 0.716600208 4.258168314 3557.34112 127 0.724053512 4.282846299 3651.972103 128 0.73155071 4.308466182 3746.254959 129 0.739093444 4.335028115 3840.159964 130 0.74668336 4.362532057 3933.654834 131 0.754322102 4.390977737 4026.704696 132 0.762011312 4.420364617 4119.272056 133 0.769752632 4.450691856 4211.316778 134 0.777547703 4.481958271 4302.796065 135 0.785398163 4.514162296 4393.664455 136 0.793305647 4.547301949 4483.873819 137 0.801271785 4.581374785 4573.373368 138 0.809298203 4.616377864 4662.10968 139 0.817386522 4.652307705 4750.026723 140 0.825538357 4.689160253 4837.065902 141 0.833755314 4.726930833 4923.166108 142 0.842038989 4.765614119 5008.263792 143 0.850390973 4.805204089 5092.293039 144 0.85881284 4.845693993 5175.185667 145 0.867306155 4.887076317 5256.871335 146 0.875872468 4.929342744 5337.277666 147 0.884513316 4.972484126 5416.330391 148 0.893230216 5.016490453 5493.953504 149 0.902024668 5.06135082 5570.069429 150 0.910898153 5.107053405 5644.599218151 0.91985213 5.153585444 5717.462746 152 0.928888034 5.200933207 5788.578936 153 0.938007278 5.249081987 5857.865994 154 0.947211244 5.298016082 5925.241659 155 0.95650129 5.347718782 5990.623467 156 0.965878741 5.398172371 6053.929029 157 0.975344892 5.449358117 6115.076322 158 0.984901004 5.50125628 6173.983991 159 0.994548301 5.553846118 6230.571658 160 1.004287974 5.607105897 6284.760239 161 1.01412117 5.661012913 6336.472274 162 1.024048999 5.715543509 6385.63225 163 1.034072529 5.770673107 6432.166939 164 1.044192782 5.826376242 6476.005722 165 1.054410737 5.882626595 6517.080922 166 1.064727325 5.939397042 6555.328128 167 1.07514343 5.9966597 6590.686511 168 1.085659886 6.054385983 6623.099128 169 1.096277477 6.112546661 6652.513223 170 1.106996934 6.171111921 6678.880498 171 1.117818939 6.230051439 6702.157385 172 1.128744116 6.289334449 6722.305279 173 1.139773039 6.348929824 6739.290767 174 1.150906226 6.408806148 6753.085818 175 1.162144138 6.468931809 6763.66796 176 1.173487184 6.529275074 6771.020416 177 1.184935716 6.589804184 6775.132227 178 1.19649003 6.650487441 6775.998326 179 1.208150368 6.711293297 6773.619598 180 1.219916916 6.772190444 6768.002899 181 1.231789807 6.83314791 6759.161041 182 1.243769119 6.894135142 6747.112752 183 1.255854878 6.955122103 6731.882601 184 1.268047058 7.016079353 6713.500888 185 1.280345581 7.076978139 6692.00351 186 1.292750322 7.137790478 6667.431793 187 1.305261104 7.198489235 6639.832297 188 1.317877708 7.2590482 6609.256601 189 1.330599866 7.319442166 6575.761054 190 1.343427271 7.379646989 6539.406511 191 1.356359571 7.439639663 6500.258049 192 1.369396376 7.49939837 6458.384664 193 1.382537261 7.558902541 6413.858952 194 1.395781764 7.618132907 6366.756784195 1.409129392 7.677071537 6317.156965 196 1.42257962 7.735701887 6265.14089 197 1.436131897 7.794008824 6210.792195 198 1.449785647 7.851978666 6154.196404 199 1.46354027 7.909599198 6095.440585 200 1.477395147 7.966859697 6034.612997 201 1.491349642 8.023750943 5971.802751 202 1.505403102 8.080265231 5907.099479 203 1.519554865 8.136396376 5840.593006 204 1.533804258 8.192139715 5772.373037 205 1.5481506 8.247492105 5702.528856 206 1.562593208 8.302451918 5631.149038 207 1.577131396 8.357019033 5558.321178 208 1.59176448 8.411194821 5484.131631 209 1.606491779 8.464982136 5408.665274 210 1.621312618 8.518385297 5332.005281 211 1.636226333 8.571410069 5254.232914 212 1.651232268 8.624063647 5175.427339 213 1.666329783 8.676354634 5095.665446 214 1.681518254 8.728293022 5015.021703 215 1.696797074 8.779890169 4933.568012 216 1.712165661 8.831158781 4851.373588 217 1.727623451 8.882112888 4768.504855 218 1.743169909 8.932767829 4685.025357 219 1.758804528 8.983140227 4600.995675 220 1.77452683 9.033247978 4516.473375 221 1.790336369 9.083110231 4431.512952 222 1.806232735 9.132747374 4346.165793 223 1.822215553 9.182181025 4260.480158 224 1.838284488 9.231434017 4174.501155 225 1.854439246 9.280530396 4088.270741 226 1.870679574 9.329495413 4001.827724 227 1.887005267 9.378355524 3915.207766 228 1.903416166 9.427138389 3828.443409 229 1.91991216 9.475872882 3741.56409 230 1.936493191 9.524589094 3654.596168 231 1.953159255 9.573318347 3567.56296 232 1.969910403 9.622093209 3480.484768 233 1.986746743 9.670947517 3393.378917 234 2.003668445 9.719916394 3306.259795 235 2.02067574 9.769036285 3219.138886 236 2.037768924 9.818344983 3132.024812 237 2.054948362 9.867881672 3044.923365 238 2.072214486 9.917686961 2957.837547239 2.089567801 9.967802939 2870.767604 240 2.107008888 10.01827322 2783.71105 241 2.124538404 10.06914301 2696.662703 242 2.142157087 10.12045915 2609.614703 243 2.159865759 10.17227021 2522.556533 244 2.177665326 10.22462654 2435.475033 245 2.195556786 10.27758035 2348.354411 246 2.213541228 10.33118583 2261.176243 247 2.231619839 10.38549918 2173.919471 248 2.249793905 10.44057878 2086.560394 249 2.268064815 10.49648522 1999.072648 250 2.286434068 10.55328149 1911.427181 251 2.304903273 10.61103303 1823.59222 252 2.323474156 10.66980789 1735.533225 253 2.342148565 10.72967686 1647.212842 254 2.360928473 10.79071361 1558.590832 255 2.379815985 10.85299484 1469.624007 256 2.398813342 10.91660043 1380.266139 257 2.417922928 10.9816136 1290.467866 258 2.437147274 11.0481211 1200.176582 259 2.456489068 11.11621339 1109.336319 260 2.475951156 11.18598479 1017.887606 261 2.495536555 11.25753373 925.7673213 262 2.515248456 11.33096292 832.9085287 263 2.535090233 11.40637958 739.2402931 264 2.555065453 11.48389563 644.687483 265 2.57517788 11.56362796 549.1705546 266 2.595431489 11.64569862 452.605317 267 2.615830472 11.73023508 354.9026776 268 2.636379246 11.81737043 255.9683687 269 2.657082469 11.9072437 155.7026516 270 2.677945045 12 54 271 2.698972136 12.09579084 -49.25123962 272 2.720169177 12.19477433 -154.16921 273 2.741541882 12.29711538 -260.8789293 274 2.763096261 12.40298597 -369.5127 275 2.784838628 12.5125653 -480.2105427 276 2.806775619 12.62603996 -593.1206523 277 2.828914201 12.7436041 -708.3998776 278 2.851261686 12.86545951 -826.2142221 279 2.873825745 12.99181568 -946.7393647 280 2.896614422 13.1228898 -1070.161198 281 2.919636147 13.25890668 -1196.676381 282 2.942899748 13.40009858 -1326.492899283 2.966414465 13.54670498 -1459.830631 284 2.990189962 13.69897215 -1596.92191 285 3.014236339 13.85715265 -1738.012069 286 3.03856414 14.02150458 -1883.359973 287 3.063184366 14.19229075 -2033.238495 288 3.088108479 14.36977744 -2187.934953 289 3.113348406 14.55423308 -2347.751462 290 3.138916547 14.74592648 -2513.005186 291 3.164825766 14.94512472 -2684.028462 292 3.191089393 15.15209073 -2861.168755 293 3.217721209 15.36708025 -3044.788411 294 3.244735433 15.59033836 -3235.264152 295 3.272146701 15.82209544 -3432.986263 296 3.299970033 16.06256236 -3638.357411 297 3.328220797 16.31192507 -3851.791015 298 3.356914659 16.57033834 -4073.709092 299 3.386067523 16.83791857 -4304.539493 300 3.415695457 17.11473584 -4544.712426 301 3.445814605 17.40080484 -4794.656164 302 3.476441079 17.69607489 -5054.791817 303 3.50759084 18.00041893 -5325.527077 304 3.53927955 18.31362154 -5607.248798 305 3.571522411 18.63536601 -5900.314316 306 3.60433397 18.96522071 -6205.041424 307 3.637727914 19.30262471 -6521.696914 308 3.671716826 19.64687324 -6850.483668 309 3.706311923 19.99710298 -7191.526309 310 3.741522764 20.35227795 -7544.855479 311 3.777356945 20.71117638 -7910.390911 312 3.813819753 21.07237927 -8287.923564 313 3.850913824 21.43426153 -8677.097172 314 3.888638771 21.79498642 -9077.389742 315 3.926990817 22.15250437 -9488.095643 316 3.965962423 22.50455722 -9908.309074 317 4.005541936 22.84868866 -10336.90988 318 4.045713253 23.18226199 -10772.55272 319 4.086455532 23.50248579 -11213.66079 320 4.127742946 23.80644791 -11658.42518 321 4.169544512 24.09115807 -12104.81101 322 4.211823995 24.35359852 -12550.57119 323 4.254539912 24.59078188 -12993.26869 324 4.297645637 24.79981492 -13430.30738 325 4.341089626 24.97796606 -13858.97143 326 4.384815759 25.12273412 -14276.47267327 4.428763801 25.23191553 -14680.00441 328 4.472869976 25.30366675 -15066.80024 329 4.517067644 25.33655883 -15434.19516 330 4.561288069 25.32962121 -15779.68675 331 4.60546124 25.28237224 -16100.99329 332 4.649516748 25.19483483 -16396.10616 333 4.693384663 25.06753614 -16663.334 334 4.736996405 24.90149146 -16901.33649 335 4.780285573 24.69817319 -17109.14645 336 4.823188699 24.45946661 -17286.17927 337 4.865645929 24.18761506 -17432.23011 338 4.907601592 23.88515729 -17547.45927 339 4.949004657 23.55486017 -17632.36742 340 4.98980907 23.19964997 -17687.7623 341 5.029973974 22.82254493 -17714.71923 342 5.069463816 22.42659175 -17714.53761 343 5.108248339 22.01480796 -17688.69562 344 5.146302494 21.5901315 -17638.805 345 5.183606256 21.15537851 -17566.56784 346 5.220144385 20.71320962 -17473.73648 347 5.255906129 20.2661045 -17362.07765 348 5.290884894 19.81634447 -17233.34134 349 5.325077886 19.36600218 -17089.23479 350 5.358485742 18.91693765 -16931.40148 351 5.391112162 18.47079954 -16761.40501 352 5.422963541 18.0290308 -16580.7174 353 5.454048622 17.59287767 -16390.71142 354 5.484378157 17.16340106 -16192.65635 355 5.513964598 16.74148976 -15987.71668 356 5.54282181 16.32787449 -15776.95316 357 5.570964803 15.92314255 -15561.32571 358 5.5984095 15.52775234 -15341.69776 359 5.62517252 15.14204762 -15118.84162 360 5.651270995 14.76627109 -14893.44455.计算结果分析:1、原动件AB转动一周,构件5随之转动一周。

哈工大机械原理大作业(连杆机构)

哈工大机械原理大作业(连杆机构)
三、建立数学模型
建立坐标系:以C为原点,水平方向为X轴,CA所在直线为Y轴建立直角坐标系(如图4)。
取曲柄1水平且位于A点右侧为初始时刻,设曲柄1角速度为w,由题意知w= =8.5π rad/s………………(1)
设曲柄1转角为θ,则B点坐标:
xB=ιABcosθ=ιABcoswt
yB=H1+ιABsinθ=H1+ιABsinwt………………(2)
form=1:length(t)-1
ddxF(m)=(dxF(m+1)-dxF(m))/0.0001;
end
ddxF(length(t))=ddxF(length(t)-1);
figure
plot(t,ddxF)
title('¼ÓËÙ¶ÈͼÏñ');
xlabel('t /s'),ylabel('v /(m/s^2)');
输出图像:
xE(m)=yE(m)/k(m);
xF(m)=xE(m)-(-H^2+lEF^2-yE(m)^2+2*yE(m)*H)^(1/2)+0.1142;
end
form=1:length(t)-1
dxF(m)=(xF(m+1)-xF(m))/0.0001;
end
dxF(length(t))=dxF(length(t)-1);
∵ιEF+ιCE>H且ιCE<H
∴E点始终在F点的右下方
∴xF<xE,所以x2舍去,只取xF=x1……………(8)
∴点F坐标为(xF,H)
当t=0时,可得F点初始位置坐标,不妨设为(xo,H)。
则F点位移(通过计算,t=0时,得xo=-0.1142)
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Harbin Institute of Technology
机械原理设计说明书
课程名称:机械原理
设计题目:连杆机构运动分析
指导老师:陈明丁刚
班级:
设计者:
学号:
指导教师:
设计时间:
一、运动分析题目
如图1-6所示是曲柄滑块机构,各构件长度分别为a、b,偏距为e,连杆BC上
一点到铰链B的距离为l
m ,试研究各构件长度a、b、l
m
及偏距e的变化对点m
的轨迹的影响规律。

二、机构结构分析及基本杆组划分
1.除去虚约束力和局部自由度,计算机构的自由度并确定原动件
本机构中无虚约束或局部自由度。

机构各杆件都在同一平面运动,活动构件数
n=3,P
L =4,P
H
=0,则机构的自由度为:F=3n-2P
L
-P
H
=33-24-10=1。

原动
件为曲柄AB。

2.拆分杆组
从远离原动件(即杆AB,如图1)进行拆分,可得到由杆BC和C点处滑块组成的RRPⅡ级杆组(如图2),剩下的就是Ⅰ级机构杆AB。

3.确定机构的级别
由上可知,机构为Ⅱ级机构
⨯⨯⨯⨯⨯
三、各基本杆组的运动分析数学模型
(1)原动件AB (Ⅰ级杆组)
原动件AB 的转角为:i ϕ=0~2π;角速度为:s rad /101=ω 角加速度为:01=ε
假定运动副A 的位置坐标为:x A =0,y A =0 A 点与机架相连,
即该点的速度和加速度都为0。

原动件AB 长度为l i ,从而可求得运动副B 点的位置坐标:i
ϕcos l x x i A B +=,
i i A B l y y ϕsin +=
(2)杆BC 和C 点的滑块(RRP Ⅱ级杆组)
RRP Ⅱ级杆组是由两个构件两个转动副及一个外移动副组成的。

已知两杆长为l i 和l j (l j 杆垂直于滑块导路),外回转副B 的参数,滑块导路方向角和计算位移时的参考点K 的位置和导路的运动参数,求内运动副C 的运动参数。

位置方程:内回转副C 的位置方程为:
j j j K i i B C j j j K i i B C l s y l y y l s x l x x ϕϕϕϕϕϕcos sin sin sin cos cos ++=+=-+=+=
消去s
可得:j i
j
i l l A ϕϕ++=0arcsin
其中
j K B j K B y y x x A ϕϕcos )(sin )(0---=
为保证机构的存在,应满足装配条件i j l l A ≤+0,求得
i ϕ后,可求得x C 和y C ,而后求得滑块的位移s :
j j j K C j j j K C l y y l x x s ϕϕϕϕsin /)cos (cos /)sin (--=+-=
滑块D 点的位移方程为:
j
K D j K D s y y s x x ϕϕsin cos +=+=
(3)求M 点位置坐标
要画出点m 的轨迹图,需要求出点m 的位置坐标。

假定以A 点为原点,则点B 的位置坐标为:
1
1sin cos ααa y a x B B == 点C
的位置坐标为:
e
y b a x C C =+=21cos cos αα 而点m 是杆BC 上的一点,且到B 点的距
离为l m ,则B,C,m 三点满足关系:C
B m
B B
C B m m y y y y x x x x b l --=--= 从而求得m 点的位置坐
标为:m
m m m m l b
e a b l y l a x +-=+=12
1sin )1(cos cos ααα
四、建立坐标系
五、计算编程
利用matlab 编程,代码如下
f1=[0:pi/180:2*pi];%原动件AB 杆转角范围0~2*pi w1=1;%AB 杆的角速度 e1=0;%AB 杆的角加速度
xA=0;yA=0;%运动副A 的位置
e=200;%偏距e ,改变e 的大小可以观察偏距e 对m 点轨迹的影响 l1=50;%杆AB 的长,可通过修改l1来观察长度a 对m 点轨迹的影响 l2=100;%杆BC 的长,可通过修改l2来观察长度b 对m 点轨迹的影响
lm=30;%BC 杆上的m 点到B 点的距离,可通过修改lm 来观察长度l m 对m 点轨迹的影响
r1=w1*f1;%AB 杆与x 轴正方向的夹角
r2=asin((e -l1*sin(r1))/l2);%360。

-BC 杆与x 轴正方向所夹锐角 xm=lm*cos(r2)+l1*cos(r1);%m 点的横坐标
ym=(1-lm/l2)*l1*sin(r1)+e*lm/l2;%m 点的纵坐标 plot(xm,ym); hold off grid on
title('点m 的轨迹图','FontSize',20);
六、计算结果
(1)构件a长度对m运动轨迹的影响(b=100,e=20,l
m
=30)
(2)构件B长度对m运动轨迹的影响(a=20,e=20,l
m
=100)
(3)l
m
长度对m点运动轨迹的影响(a=50,b=100,e=20)
(3)e的大小对m点运动轨迹的影响(a=50,b=100,l
m =30
a=20 a=50 a=80 a=100 b=20 b=50 b=80 b=100 l=20 l=50 l=80 l=90
e=20 e=50 e=40 e=50
七、计算结果分析
a的长度对m运动轨迹的影响
当构件长度a增大时,m点轨迹图的横向的左右两端会变长,纵向的上下两端也会变宽,并且左端的凸起会略微向下移动,右端凸起会略微向右移,而上端和下端凸起都会向左移。

图形整体上下左右都变大。

b的长度对m运动轨迹的影响
当构件长度b增大时,m点轨迹图的横向的左右两端长度基本保持不变,纵向的上下两端会变宽,并且左端的凸起会略微向右和向下移动,右端凸起会略微向右和向下移,而上端和下端凸起都向右移。

图形整体上下变宽,向右移。

L m的长度对m运动轨迹的影响
长度l
增大时,m点轨迹图的横向的左右两端会略微增长,纵向的上下两端会变m
窄,并且左端和右端的凸起都会向上和向右移动,而上端的凸起会向右移和向下移,下端的凸起会向右和向上移。

图形整体上下变窄,向右移动。

e的长度对m运动轨迹的影响
e增大时,m点轨迹图的横向的左右两端会略微变长,纵向的上下两端不变,并且左端和右端的凸起都会向左和向上移动,而上端的凸起会向右移和向上移,下端的凸起会向左和向上移。

图形整体向上移动。

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