机械原理课后习题2-3.ppt

* * 2.3 在图示的摆动导杆机构中，已知lAB =30mm , lAC =100mm , lBD =50mm , lDE =40mm ,φ1=45o,曲柄1以等角速度ω1=10 rad/s沿逆时针方向回转。求D点和E点的速度和加速度及构件3的角速度和角加速度（用相对运动图解法）。 解： VB2==10*0.03=0.3m/s ?v=(0.3m/s)/(30mm)=0.01(m/s)/mm p b2 b3 e VB3=?pb3=0.01*24.5=0.245m/s Ve=0.01*16.6=0.166m/s ?Vd=0.01*14.57=0.1457m/s ?3=245/123=2rad/s =10*10*0.03=3m/s2 ＝ =2*2*0.123=0.492m/s2 ＝ =2*2*0.17=0.68m/s2 ?a=(3m/s2)/30=0.1(m/s2)/mm P’ b2’ k’ n3 b3’ e =10*10*0.03=3m/s2 ＝ =2*2*0.123=0.492m/s2 ＝ =2*2*0.17=0.68m/s2 ?a=(3m/s2)/30=0.1(m/s2)/mm P’ b2’ k’ n3 b3’ e =10*10*0.03=3m/s2 ＝ =2*2*0.123=0.492m/s2 ＝ =2*2*0.17=0.68m/s2 ?a=(3m/s2)/30=0.1(m/s2)/mm P’ b2’ k’ n3 b3’ e =0.1*11.54=1.154M/S2 =0.1*10.43=1.043m/s2 ?=1.043/0.123=8.47m/s2 ae=0.1*7.82=0.782m/s2 ? ad=0.1*6.85=0.685m/s2 *