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Prof. Albert Espinoza
Mechanical Engineering Department
Polytechnic University of Puerto Rico
Spring 2015
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2
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Circuits and Branches in Linkages
Circuit: All possible orientations of the links that can be
realized without disconnecting any of the joints
Branch: A continuous series of positions of the mechanism on a
circuit between two stationary configurations.
These stationary configurations divide the circuit into a series
of branches
Branch 1
Branch 2
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Crank-Slider Position Analysis
This configuration is a variation of the four-bar linkage
In a general configuration, theres an offset between the slider
and the crank pivot (R4)
By using four vectors, we can constrain R1 and R4 to always be
oriented horizontally and vertically, respectively (1=0, 4=90)
The direction of R3 is determined (arbitrarily) for
convenience
Given:
Link Lengths: a, b, c, 1, 4 Input: 2 Find:
3, d
NOTE: c is commonly known as the offset of the crank-slider
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2 3 4 1 0R R R R
32 4 1 0jj j j
ae be ce de
2 2 3 3
4 4 1 1
cos sin cos sin
cos sin cos sin 0
a j b j
c j d j
Real part (x component):
2 3 4 1cos cos cos cos 0a b c d
1 4but: 0, 90 , so:
2 3cos cos 0 a b d
Imaginary part (y component):
2 3 4 1sin sin sin sin 0ja jb jc jd
1 4but: 0, 90 , and the 's divide out so: j
2 3sin sin 0 a b c
1
23
1 2 31
sinarcsin
cos cos
a c
b
d a b
Vector Loop: Crank-Slider Position Solution
2
23
2 2 32
sinarcsin 180
cos cos
a c
b
d a b
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Vector Loop: Crank-Slider Position Solution
As is the case for the four-bar linkage, there are two possible
solutions for 3 and d for a given input angle, 2
These solutions define the open and crossed circuits
Note that in the crossed case, d is negative
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Example 1: Crank-Slider
Given:
2 3 21.4 in, 4 in, offset 1 in, 45L L
Find:
3 and for the open and crossed positions.d
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8
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Slider-Crank Position Analysis
Same configuration as before, except that in this case, the
mechanism is driven at the slider
For this reason, in this particular case:
Given:
Link Lengths: a, b, c, 1, 4 Input: d
Find:
2, 3
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Vector Loop: Slider-Crank Position Solution
2 3 4 1 0 R R R R
32 4 1 0 jj j jae be ce de
cos sinje j
2 2 3 3 4 4 1 1cos sin cos sin cos sin cos sin 0 a j b j c j d
j
3 2
3 2
cos cos
sin sin
b a d
b a c
2 22 2 23 3 2 2sin cos cos sin b a d a c
2 22
2 2cos sin b a d a c
2 2 2 2
2 22 sin 2 cos 0 a b c d ac ad
2 2 2 2
1 2 3 2 2 K a b c d K ac K ad
1 2 2 3 2sin cos 0 K K K
1 3
2
1 3
2
A K K
B K
C K K
2
22 2
2 tan2
sin ;
1 tan2
2 2
22 2
1 tan2
cos
1 tan2
2 2 2tan tan 02 2
A B C
1,2
2
2
42arctan
2
B B AC
A
NOTE: Once 2 is known, then 3 can be calculated using either
equation in Box 2 (Use second eq. for convenience) This equation
will yield two possible solutions for 3, (same as crank-slider
case) but only one solution will be compatible with the given
information (must check to see which one)
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Vector Loop: Slider-Crank Position Solution
As is the case for the four-bar linkage, there are two possible
solutions for a given d
These solutions represent two different branches:
For a given d, there are two combinations of 2 and 3
Changes between these two branches can occur once the
slider-crank reaches the extreme positions (Top Dead Center and
Bottom Dead Center)
In real systems, stored energy in the crank (a flywheel) is used
to carry the mechanism through these extreme points without
switching branches
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Example 2: Slider-Crank
Given:
2 31.4 in, 4 in, offset 1 in, 2.5 inL L d
Find:
2 3all possible solutions for and .
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13
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Vector Loop: Inverted Slider Crank Solution
2 3 4 1 0R R R R
32 4 1 0jj j j
ae be ce de
2 2 3 3
4 4 1 1
cos sin cos sin
cos sin cos sin 0
a j b j
c j d j
2 3 4cos cos cos 0a b c d
2 3 4sin sin sin 0a b c
2 4
3
sin sin
sin
a cb
2 42 3 4
3
sin sincos cos cos 0
sin
a ca c d
3 4
2 2
2 2
sin sin cos cos
sin cos cos sin
sin
P a a d
Q a a d
R c
; 2 ; S R Q T P U Q R
1,2
2
4
42arctan
2
T T SU
S
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Example 3: Inverted Slider-Crank
Given:
1 2 4 26 in, 2 in, 4 in, 90 , 30L L L
Find:
3 4all possible solutions for , , and b.
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