上海交通大学：《复杂系统动力学计算机辅助分析》课程教学资源_Chapter 8_Modeling for Example1

Dynamic analysis of crank-slider y X2 X3 g 1=2m,1=3.5m Initial condition 4(0)=[-103.1416-0.25001.500] 4(0)=[0 -30300-3017.1429 000]' Kinematic Constraints No. Kind o R 0, so R 0, C 1 ax 1 -1/2 0 L o] ay 1/2 0 Number of constraint equations:2 -cos sim4 Φ0=八-2

Dynamic analysis of crank-slider 1 l  2m , 2 l  3.5m Initial condition (0)  1 0 3.1416 0.25 0 0 1.5 0 0 T q    (0) 0 30 30 0 30 17.1429 0 0 0 T q    Kinematic Constraints No. Kind i P i s Q i s Ri  i j P j s Q j s Rj  j C 1 ax 1 1 / 2 0 l      0 2 ay 1 1 / 2 0 l      0 Number of constraint equations: 2 (1) 1 1 1 cos 2 ax l   x   , (1) 1 1 1 sin 2 ay l   y  

No. Kind i siQ R 59 R 2 -4/2 0 0 Number of constraint equations:2 Φ(1,2) X2 cos -sin c0s1 y2」 sind, coso, sin No. Kind s 。 R 0, 9 飞 0 C 3 0 Number of constraint equations:2 中(2,3) cos sin， cos No. Kind SiP e R 0 so 0 C ay 0 o 6 a 3 0 Number of constraint equations:2 Φw3》=y3, Φ(3)=中3 Constraint equations X-2c0s4=0 ysin=0 -_2c0s2-x1- x22 cos=0

No. Kind i P i s Q i s Ri  i j P j s Q j s Rj  j C 3 r 1 1 / 2 0 l      2 2 / 2 0 l      Number of constraint equations: 2 (1,2 ) 2 2 2 2 1 1 1 1 2 2 2 1 1 1 cos sin / 2 cos sin / 2 sin cos 0 sin cos 0 r x l x l y y                                                    No. Kind i P i s Q i s Ri  i j P j s Q j s Rj  j C 4 r 2 2 / 2 0 l      3 0 0       Number of constraint equations: 2 ( 2,3) 3 2 2 2 2 3 2 2 2 cos sin / 2 sin cos 0 r x x l y y                                  No. Kind i P i s Q i s Ri  i j P j s Q j s Rj  j C 5 ay 3 0 0       0 6 a 3 0 Number of constraint equations: 2 ( 3) 3 ay   y , (3) 3 a    Constraint equations cos 0 2 1 1 1    l x sin 0 2 1 1 1    l y cos 0 2 cos 2 1 1 2 1 2 2       l x l x

为-号sm所-y-2 2 in0 x3-x2-2 os2=0 为冷如明=0 y3=0 43=0 The Jacobian matrix 「101，sin4,/200 0 000 0 1-1cos41/2 00 0 00 0 -101sing,/2101,sinp,/2 00 0 0-1-11cos4/201 Φ，= -12cosφ2/2000 0 0 0 -10 12 sin/2 1 00 0 0 0 0-1-12cos42/201 0 0 0 0 00 0 01 0 0 0 00 0 001 -1cos4,6212 -4sin4,4212 -1cos4,d212-12c0s,,212 -1sin4,42/2-2sin9,,2/2 Y= -l2c0s42,212 -2sin4,2/2 0 0 Body number 1 2 3 Mass(Kg) 200 35 25 Moment of Inertia 450 35 0.02 (Kg.m2)

sin 0 2 sin 2 1 1 2 1 2 2       l y l y cos 0 2 2 2 3  2    l x x sin 0 2 2 2 3  2    l y y 0 y3  0 3  The Jacobian matrix              0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 cos / 2 0 1 0 0 0 0 1 0 sin / 2 1 0 0 0 1 cos / 2 0 1 cos / 2 0 0 0 1 0 sin / 2 1 0 sin / 2 0 0 0 0 1 cos / 2 0 0 0 0 0 0 1 0 sin / 2 0 0 0 0 0 0 2 2 2 2 1 1 2 2 1 1 2 2 1 1 1 1         l l l l l l l l  q                0 0 sin / 2 cos / 2 sin / 2 sin / 2 cos / 2 cos / 2 sin / 2 cos / 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2 2 2 2 2 1 1 1 2 1 1 1 2 1 1 1                         l l l l l l l l  Body number 1 2 3 Mass (Kg) 200 35 25 Moment of Inertia (Kg.m2) 450 35 0.02

mg+F 0 L m28 M= 0= 0 0 m38 m3 0 J3) 0 0 元≤0 0 x3>0且x30且1.5≤x3≤5 6-x3 1100001-sin2π(x3-5.25] x3>0且5<x3≤5.5 L=41450.0

       3 3 3 2 2 2 1 1 1 J m m J m m J m m M ,         0 0 0 0 0 3 2 1 m g m g L m g F A Q   3 3 3 3 3 3 3 3 3 0 0 0 0 1.5 282857 62857 0 1.5 5 6 110000 1 sin 2 5.25 0 5 5.5 x x x F x x x  x x x                              且 且 且 L  41450.0