機(jī)械原理大作業(yè)1

1、原題目：（30）題：如圖1-1所示機(jī)構(gòu)，已知機(jī)構(gòu)各構(gòu)件的尺寸為AB=100mm，BC=2.73AB，CD=1.36AB，CG=2.32AB，BG=4.9AB，AF=2.36AB，AD=2.87AB，DF=2AB，GE=1.45AB，EF=2.82AB，GM=1.36AB，MK=1.91AB，KD=0.54AB，KF=2.18AB，HF=3.1AB，DH=3.63AB，構(gòu)件1的角速度為。試求構(gòu)件2上G點(diǎn)的軌跡及構(gòu)件4、構(gòu)件6和構(gòu)件8的角位移、角速度和角加速度，并對(duì)計(jì)算結(jié)果進(jìn)行分析。解答如下：1.建立直角坐標(biāo)系以H點(diǎn)為直角坐標(biāo)系的原點(diǎn)建立直角坐標(biāo)系X-Y,如圖1-1所示。 圖1-12.機(jī)構(gòu)結(jié)構(gòu)分析

2、該機(jī)構(gòu)由I級(jí)桿組RR（原動(dòng)件AB）、II級(jí)桿組RRR（桿2、9）、II級(jí)桿組RRR（桿5、6）、II級(jí)桿組RRR（桿7、8）、II級(jí)桿組RPR（桿4、滑塊3）組成。分別如圖所示。3.各基本桿組運(yùn)動(dòng)分析1. I級(jí)桿組RR（原動(dòng)件AB） 如圖所示，已知原動(dòng)件桿1的轉(zhuǎn)角，角速度，角加速度，運(yùn)動(dòng)副A的位置坐標(biāo)為xA = 542.3mm，yA = 41.5mm，速度為，加速度為，原動(dòng)件桿1的長(zhǎng)度lAB = 100mm。1）位置分析 2）速度和加速度分析 將已知參數(shù)帶入可求出。2. II級(jí)桿組RRR（桿2、9）如圖所示，已求出運(yùn)動(dòng)副B的位置、速度、加速度，運(yùn)動(dòng)副D的位置坐標(biāo)為，速度為，加速度為，桿長(zhǎng)。（1

3、）桿2的角位置、角速度、角加速度 B、D連線與x軸的夾角為： 桿BC的位置角為： 桿BC的角速度為： 式中 桿BC的角加速度為 ： 式中 將已知量帶入可求得。3. II級(jí)桿組RRR（桿5、6）如圖所示，已求出G點(diǎn)的位置、速度、加速度，K點(diǎn)的位置坐標(biāo)為，速度為，加速度為，桿長(zhǎng)。1） 位置分析 運(yùn)動(dòng)副M的位置方程為： 桿MK的轉(zhuǎn)角位為： 且式中 ） 則桿GM的轉(zhuǎn)角位為： 。2） 速度分析 桿MK的角速度為： 桿GM的角速度為： 式中 運(yùn)動(dòng)副M的速度為： 3） 加速度分析 桿MK的角加速度為： 桿GM的角加速度為： 式中 運(yùn)動(dòng)副M的加速度為： 4. II級(jí)桿組RRR（桿7、8） 如圖所示，已求出G點(diǎn)

4、的位置、速度、加速度，F(xiàn)點(diǎn)的位置坐標(biāo)為，速度為，加速度為，桿長(zhǎng)。 1）位置分析 運(yùn)動(dòng)副E的位置方程為： 桿EF的轉(zhuǎn)角位為： 式中 ） 桿GE的轉(zhuǎn)角位為： 。 2）速度分析 桿EF的角速度為： 桿EG的角速度為： 式中 運(yùn)動(dòng)副E的速度為： 3）加速度分析 桿EF的角加速度為： 桿GE的角加速度為： 式中 運(yùn)動(dòng)副E的加速度為： 5. II級(jí)桿組RPR（桿4、滑塊3）1）位置分析 桿4的轉(zhuǎn)角為： 式中 2）速度分析 桿4的角速度為： 式中 3）加速度分析桿4的角加速度為： 式中 4.程序設(shè)計(jì) 采用VB編程，代碼如下： Private f4(3600) As Double Private w4(360

5、0) As Double Private e4(3600) As Double Private f6(3600) As Double Private w6(3600) As Double Private e6(3600) As Double Private f8(3600) As Double Private w8(3600) As Double Private e8(3600) As Double Private xG(3600) As Double Private yG(3600) As Double Private xB(3600) As Double Private yB(3600)

6、As Double Private xC(3600) As Double Private yC(3600) As Double Private Sub Form_Load() Dim i As Single Dim pi As Double Dim pa As Double Dim RR1 As RR Dim RR2 As RR Dim RRR1 As RRR Dim RRR2 As RRR Dim RRR3 As RRR Dim RPR1 As RPR Dim d1 As Double Dim d2 As Double Dim ss As Double Dim vxG(3600) As Do

7、uble Dim vyG(3600) As Double Dim axG(3600) As Double Dim ayG(3600) As Double Dim vG As Double Dim aG As Double Set RR1 = New RR Set RR2 = New RR Set RRR1 = New RRR Set RRR2 = New RRR Set RRR3 = New RRR Set RPR1 = New RPR pi = 3.1415926 pa = pi / 180For i = 0 To 3600 Step 1 RR1.xA = 542.3: RR1.yA = 4

8、1.57 RR1.vxA = 0: RR1.vyA = 0 'A點(diǎn) RR1.axA = 0: RR1.ayA = 0 RR1.f = i * pa / 10: RR1.delt = pi RR1.w = 10: RR1.e = 0 RR1.L = 100 RR1.cal xB(i) = RR1.xB: yB(i) = RR1.yB RRR1.xD = RR1.xB: RRR1.yD = RR1.yB RRR1.vxD = RR1.vxB: RRR1.vyD = RR1.vyB RRR1.axD = RR1.axB: RRR1.ayD = RR1.ayB RRR1.xB = 303: R

9、RR1.yB = 200 'D位置 RRR1.vxB = 0: RRR1.vyB = 0 RRR1.axB = 0: RRR1.ayB = 0 RRR1.Lj = 273: RRR1.Li = 136 RRR1.M = -1 RRR1.cal xC(i) = RRR1.xC: yC(i) = RRR1.yC RR2.xA = RRR1.xD: RR2.yA = RRR1.yD RR2.vxA = RRR1.vxD: RR2.vyA = RRR1.vyD RR2.axA = RRR1.axD: RR2.ayA = RRR1.ayD RR2.f = RRR1.fj - 0.07157 *

10、pi: RR2.delt = pi RR2.w = RRR1.wj: RR2.e = RRR1.ej RR2.L = 490 RR2.cal xG(i) = RR2.xB: yG(i) = RR2.yB RRR2.xB = RR2.xB: RRR2.yB = RR2.yB RRR2.vxB = RR2.vxB: RRR2.vyB = RR2.vyB RRR2.axB = RR2.axB: RRR2.ayB = RR2.ayB RRR2.xD = 249.8: RRR2.yD = 209.5 'K位置 RRR2.vxD = 0: RRR2.vyD = 0 RRR2.axD = 0: RR

11、R2.ayD = 0 RRR2.Li = 191: RRR2.Lj = 136 RRR2.M = 1 RRR2.cal f6(i) = RRR2.fi / pa: w6(i) = RRR2.wi: e6(i) = RRR2.ei RRR3.xD = RR2.xB: RRR3.yD = RR2.yB RRR3.vxD = RR2.vxB: RRR3.vyD = RR2.vyB RRR3.axD = RR2.axB: RRR3.ayD = RR2.ayB RRR3.xB = 310: RRR3.yB = 0 'F位置 RRR3.vxB = 0: RRR3.vyB = 0 RRR3.axB

12、= 0: RRR3.ayB = 0 RRR3.Li = 282: RRR3.Lj = 145 RRR3.M = -1 RRR3.cal f8(i) = RRR3.fi / pa: w8(i) = RRR3.wi: e8(i) = RRR3.ei vxG(i) = RR2.vxB: vyG(i) = RR2.vyB axG(i) = RR2.axB: ayG(i) = RR2.ayB LHG(i) = Sqr(xG(i) - 0) 2 + (yG(i) - 0) 2) If xG(i) > 0 And yG(i) >= 0 Then '第一象限 f4(i) = Atn(yG(

13、i) - 0) / (xG(i) - 0) vG = vxG(i) * Sin(f4(i) + vyG(i) * Cos(f4(i): w4(i) = vG / LHG(i) aG = axG(i) * Sin(f4(i) + ayG(i) * Cos(f4(i): e4(i) = aG / LHG(i) End If If xG(i) = 0 Then f4(i) = pi / 2 w4(i) = vxG(i) / LHG(i) e4(i) = axG(i) / LHG(i) End If If xG(i) < 0 And yG(i) > 0 Then '第二象限 ss = Atn(-(yG(i) - 0) / (xG(i) - 0) f4(i) = pi + Atn(yG(i) - 0) / (xG(i) - 0) vG = vxG(