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 3_CHAP3.7-singularity

3.7 Singular configuration Definition Singular configuration The configuration of a mechanism,beyond which the motion can not continue or more than one possible motion can occur (1)Bifurcation-A branching of motion to two possible paths (2)Lock-up-in this configuration,the mechanism can not be driven in certain direction

Definition 3.7 Singular configuration Singular configuration The configuration of a mechanism, beyond which the motion can not continue or more than one possible motion can occur (1) Bifurcation — A branching of motion to two possible paths (2) Lock-up — in this configuration, the mechanism can not be driven in certain direction

Example 3.7.1 Slider-crank lock up bifurcation

Example 3.7.1 Slider-crank lock up bifurcation

Example 3.7.2 Four-bar linkage bifurcation

Example 3.7.2 Four-bar linkage bifurcation w w w

Example 3.7.3 Two body slider-crank model(1) The distance constraint equation 0-ot 0-0 Jacobian matrix Determinant of Jacobian D,=2(x-c0s0)

Example 3.7.3 Two body slider-crank model(1) 0 ( cos ) sin ( , ) 2 2 2             t x l t  w    q The distance constraint equation 1 l wt x         0 1 2(x cos ) 2x sin q         x 0 q 2( cos )( sin ) 2 sin cos              w            x x   2(x  cos ) q Jacobian matrix Determinant of Jacobian

Example 3.7.3 Two body slider-crank model(1) (1)1>1 1 y2↑ 2 0-0t =t X2 X x=cos0+cos20+(12-1) x cos0-/cos20+(12-1) D,=2(x-c0s8) =2(x-cos0) =2Vcos20+(12-1)≠0 =-2Vc0s20+(12-1)≠0 Solution ofΦ，A=-Φ，is unique For g=go,the mechanism will stay in one of two distinct configurations

Example 3.7.3 Two body slider-crank model(1) cos cos ( 1) 2 2 x      l  (1) l >1 cos cos ( 1) 2 2 x      l  x1 1 l wt x x1 y1 x2 y2 l 1 x y1 x2 y2 wt 2 cos ( 1) 0 2( cos ) 2 2       l x  q  2 cos ( 1) 0 2( cos ) 2 2        l x  q  For q = q 0 , the mechanism will stay in one of two distinct configurations qq  t Solution of  is unique

Example 3.7.3 Two body slider-crank model(1) (2)1=1 X1 y2 0=o0t X2 X2 X x=2cos0 x=0 D,=2(x-c0s0)=2c0s0 D,=2(x-c0s0)=-2c0s0 In case that 0-回-0@=8。 Two solution branches (bifurcation)are possible

Example 3.7.3 Two body slider-crank model(1) x  2cos (2) l =1 x1 1 l  wt x x1 y1 x2 y2 l 1 x=0 y1 x2 y2 wt q  2(x  cos )  2cos q  2(x  cos )  2cos , 0 2  q   In case that                w 0 , 0 1 0 0 q t Two solution branches (bifurcation) are possible

Example 3.7.3 Two body slider-crank model(1) (3)1<1 XI y2 X1 X2 X x=cos0+Vcos20-(1-12) x=cos0-Vcos20-(1-12) D,=2(x-c0s0)=2Vc0s20-(1-1)D,=2(x-c0s0)=-2Vcos20-(1-1) In case that日=arcsinl,cos0=V-,x=V1-P,|电，=0 82a- No solution (lock up)

Example 3.7.3 Two body slider-crank model(1) (3) l <1 x l 1 x1 y1 x2 y2 2( cos ) 2 cos (1 ) 2 2  x       l q arcsin , cos 1 , 1 , 0 2 2 In case that   l    l x   l q  x l 1 x1 y1 y2 x2 cos cos (1 ) 2 2 x       l cos cos (1 ) 2 2 x       l 2( cos ) 2 cos (1 ) 2 2  x       l q              w 0 , 0 1 0 2 1 2 t l l q  No solution (lock up)

Example 3.7.4 Two body slider-crank model(2) X2+ X1 The constraint equations 1 y2 02 X2 -x2+cos+lsin sin-lcos =0 9= X2 =0 功-0t 42

Example 3.7.4 Two body slider-crank model(2)  0             t l x l t  w     1 1 2 2 1 2 sin cos cos sin (q, ) The constraint equations 1 l 1 x2 x1 y1 y2 x2 2  0        2 2 1   q x

-sin-1 Icos 中 cos 0 Isin ②=-lsin4, 10 0 Φ，9=-D -sing -1 Icos cos 0 Isin 元2 = 0 100 西

1 2 1 2 sin 1 cos cos 0 sin 1 0 0 ll             q 2 q  lsin qq  t  1 2 1 1 2 22 sin 1 cos 0 cos 0 sin 0 1 0 0 ll x       w                            

Redundant constraint Consider a four-bar parallelogram mechanism y2'↑ B X 身 x2-c0s42-C0s4, 2 y2-sin2-sin DOF=5-4=1 1= Φ水= x2-cos43-1 =0 中2 y2-sin L中 Φ=(x2+cosp2-2)2+(y2+sin2)2-1=0 DOF=5-5=0

Redundant constraint B x3 y ’ 3 ’ x1 ’ y1 ’ x2 ’ y2 ’ x y 1 3 A  0                      2 3 2 3 2 2 1 2 2 1 3 2 2 2 1 sin cos 1 sin sin cos cos          y x y x y x k q   Consider a four-bar parallelogram mechanism ( cos 2) ( sin ) 1 0 2 2 2 2 5  x2  2   y     k DOF = 5 – 4 = 1 DOF = 5 – 5 = 0