Exam (elaborations) MECHANISM

107學年度第一學期成大機械工程學系機動學(一)第二次期中考 11/27/2018
解題助教：柯昀昇 研究室：91805 分機號碼：62231-17

1. [40%] For the mechanism shown below, where 𝜌2 = 20𝑐𝑚, 𝜌3 = 10𝑐𝑚.
(a) [5%] Determine the number of independent loops.
(b) [10%] Derive the rolling contact equation (θ2i = 0°, θ3i = 0°)
(c) [25%] Solve for the positions of the mechanism when θ2 = 45° using Newton-Raphson’s method.
Perform one iteration with θ3 = -70° and r1 = 36 cm as the initial estimates.
*Use four significant digits for all your computation.




Solution
(a)
The number of independent loops is determined by the following equation.
L  J  N  1  4  4  1  1 [5]

(b)
The equation for the rolling contact is

2 (2  k )   3 (3  k ) [10]

(d)
The loop equation of the mechanism shown in Fig.2 is

r1  r2  rk  0 (1)

the real component of Eq. (1) is

r2 cos 2  rk cos k  0 (2)

, and the imaginary component of Eq. (1) is

r2 sin 2  rk sin k  r1  0 (3)

There’s rolling contact between link 3 and link 4, the rolling contact equation is

2 (2  k )   3 (3  k ) (4)

and 𝜌2 = 20, 𝜌3 = 10, 𝜃2𝑖 = 𝜃3𝑖 = 0° ，so

0.4 2 2
ki  180  cos 1 ( )  112.1556 [3] (5)
30
Substituting 𝜌2 = 20, 𝜌3 = 10, 𝜃2𝑖 = 𝜃3𝑖 = 0, and 𝜃𝑘𝑖 = 112.1556° into Eq. (4) can be arranged as

below
1 1
k  ( 22  33 )   ki  (20 2  103 )  112.1556 [3] (9)
 2  3 30
Applying the given parameters to Eq. (2), Eq. (3)

F1 (r , r1 )  8 2 cos 2  30cos k (10)

F2 (r , r1 )  8 2 sin 2  30sin k  r1 (11)

Substituting Eq. (9) into Eq. (10) and Eq. (11), they become

1
F1 (3, r1)  8 2 cos 2  30cos( (202  103)  112.1556 ) (12)
30
1
F2 (3, r1)  8 2 sin 2  30sin( (202  103)  112.1556 )  r1 (13)
30
So the residuals 𝜀1 and 𝜀2 are

F1 (3, r1)  F1 (70 ,36)  1  0.0116 [1]


F2 (3, r1)  F2 (70 ,36)   2  0.9105 [1]

Now assuming 𝛥𝜃3 = 𝜃3 − 𝜃3′ and 𝛥𝑟1 = 𝑟1 − 𝑟1′ , and applying Newton-Raphson’s method to obtain the
following simultaneous equations.
 F1 (3, r1) F1 (3, r1) 
   r1   3      0.0116 
 3
    1    [3]
 F2 (3, r1) F2 (3, r1)   r1   2   0.9105
   r1 
 3

where
F1 (3, r1) 1 F1 (3, r1)
 10sin( (20 2  103 )  112.1556 )  8.7612 [1]  0 [1]
3 30 r1
F2 (3, r1) 1 F2 (3, r1)
 10cos( (20 2  103 )  112.1556 )  4.8209 [1]  1 [1]
3 30 r1