机械原理中四杆机构的运动分析

1、Jounhil nl (Juinpshu histilutr of l rchii(lc, V CB =pV xbc =0.01x36.4=0.364m/s（方向:bf c, co2=V CB /I BC =0.364/0.15=2.43rad/s（方向:逆时 针。2加速度分析a C =aB+an CB +a t CB 大小:? 211 AB co221 BC ?方向: ACBtACtB丄BCa B =3211 AB =102x0.05=5m/s2, anCB =co221 BC =2.432x0.15=0.9m/s2,取力口 速度比例尺Z =0.1（m/s2/imn,作加速度图，如图lc所示

2、。贝lja C =pa xp c =0.1x36=3.6in/s2（方向:P TC，o（-曲柄摇块机构的运动分析图2a所示为一曲柄摇块机构，已知曲柄1长度为1 AB =50nmi,机架4长度为1 AC =150nun, 6=45,曲柄以=10iad/s顺时针匀速转动，求滑块3的角速度co3和角加 速度詔。分析由题意知,此曲柄摇块机构的已知点为B点。学生往往会选择同一构件 （杆2上B、C两点去列矢量方程：VC2=VB2+V C2B2 大小:？ coll AB ?方向：?AB丄BC由于上述矢量方程存在三个未知量，故无法求解。所以正确的求解方法应该是: 选择组成移动副的构件2和构件3上的重合点列矢量

3、方程。由于已知点为B点， 所以重合点选为B点。学生可能直观地认为构件3和构件2没在B点重合，不能列 运动方程。但根据机构运动简图的画法，我们知道构件的尺寸是可以无限扩大 的，为此可以把构件3的尺寸扩大到B点,这样就可以在B 2、B 3点间列出 运动方程。由于组成移动副的两个构件都是活动构件，在加速度分析时有哥氏加速 度存在。解取机构比例尺卩1 =lmm/mm作机构图，如图2a所示,1 BC =pl xBC =120nuiio1速度分析VB3=VB2+VB3B2 大小:？coll AB?方向:丄BC丄AB/BCV B 2=coll AB =10x005=05m/s,取速度比例尺=0.01（m/s

4、/imn.作速度图，如图2b所示。则V B 3=yV xpb 3=0.01 x23.4=0.234in/s（方向:pb 3 , V B3B2=yV xb 2b3=0.01 x44.2=0.442m/s（方向:b 2b3,co2=co3=V B3/1 BC =0.234/0.12=1.95rad/s（方 向：逆 时针。2加速度分析ChinaA 11 L；ChinaAll rightsChina All righls reserved.ChinaAll rightsChinaAll rightsChinaAll rights图1曲柄滑块机构的运动分析DEF图2曲柄摇块机构的运动分析DEF116a

5、11B 3 十a t B 3=aB2 十a k B3B2 十 a r B3B2 大小:231 BC ? co211 AB2co2V B3B2?方向:BtC丄BCBtA丄BC/BCa B =3211 AB =102x0.05=5m/s2, a n B3=co231 BC =1.952x0.12=0.456m/s2, a k B3B2=2co2V B3B2=2xl.95x0.442=1.72nVs2,取加速度比例尺 pa =0.1(m/s2/mm,作加 速度图，如图2c所示。则2=s3=a t B3/1 BC =pa xn b 3, /I BC =0.1 x61.4/0.12=51.2rad/s2

6、(方向 1 逆时针。五、小结平面机构的运动分析是机械原理课程的重点教学内容，也是一个教学难点，学习时必须牢记低副要素的运动特点。对于平 面四杆机构来说，若组成移动副的两个构件都是活动构件，则选择这两个构件的重 合点列矢量方程求解且一定存在哥氏加速度;除此之外就选择同一构件(连杆上 两个点列矢量方程求解,旦一定不存在哥氏加速度。参考文献：1沈晓阳、王平理论力学和机械原理课程运动合成的异同分析J高等函授学报,2010(10 35-37.Teaching Experience of Mechanical Principle in kinematic analysisZHAO Lian-sheng(S

7、choolof Mechanical Engmeenng. Changshu Institute of Technology. Cliangshu 215500, ChinaAbstract: The paper seeks to solve the concision of students when they leani kinematic analysis in the course of “ Mechanical Principle . Motion characteristics of the elements of lower-pairs, the choice of refere

8、nce point, the ex istence terms of Conolis acceleiation, and the examples of the appiopnate kinematic analysis are presented to mi prove the students skills of kinematic analysis and effect of classroom teaching. Key words :mechanical principle; kinematic analysis; four poles organizations(上接第114页Th

9、e nnpiovement of Course Machinery and Electronics System SunulationHMA Wen-bm(Schoolof Mechanical Engineering. Changshu Listmite of Teclinologv. Changshu 215500, cluiiaAbstract: the course of Machinery and Electionics Svstem Simulation” is therJunpoitant professional courses, fiomwhich students can

10、learn system simulation. But duiing the teaching process, the questions deviate fiom engineei ing and lower motivation of learning Accoiding to the problems, a new improvement measure foi it has been intro duced, which include new teaching methods and the measure of impiovement of teaching process .It lias been proved effective Key words :Macliuieiy and Electronics System Smiulation; the improvement of teaching; teaching method参考文献:1 高雷雷“机电传动控制”教学改革与实践J.林区教学，2010(7 :9-