me 2220_sp15_lec02 - kinematic fundamentals and mobility
TRANSCRIPT
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Prof. Albert Espinoza
Mechanical Engineering Department
Polytechnic University of Puerto Rico
Spring 2015
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Mobility (Degree of Freedom)
Mobility determines the “character” of the links assemblage
Degree of freedom (DOF) of a system can be defined as:
– The number of inputs which need to be provided in order to create a predictable output;
– The number of independent coordinates required to define its position
Kutzbach’s (modified Gruebler’s) Equation
𝑀 = 3 𝐿 − 1 − 2𝐽1 − 𝐽2
L= number of links
J1=number of full joints (e.g., pins, sliders)
J2=number of half-joints (e.g., roll-slide, cams, gears)
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
4L
1 4J
2 0J
Num. Links = 4
Num. full joint, (1 DOF) = 3 pins , 1 slide
Num. half joint, (2 DOF) = 0
3 4 1 2 4 0
1
M
M
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
8L
1 10J
2 0J
Num. Links = 8
Num. full joint, (1 DOF) = 7 (single pins)+ 2 (multiple joint) +1 (slide)
Num. half joint, (2 DOF) = 0
3 8 1 2 10 0
1
M
M
Multi-Nodes
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
6L
1 7J
2 1J
Num. Links = 6
Num. full joint, (1 DOF) = 5 (single pins)+ 2 (multiple joint)
Num. half joint, (2 DOF) = 1
3 6 1 2 7 1
0
M
M
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
Num. Links = 9
Num. full joint, (1 DOF) = 7(single pins)+ 2 (Multiple Joint) +2(slider)
Num. half joint, (2 DOF) = 0
L= 9
J1 = 11
J2 = 0
M = 3(9-1) - 2(11)
M = 2
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
Num. Links = 9
Num. full joint, (1 DOF) = 9 pins , 2 slide
Num. half joint, (2 DOF) = 1
L = 9
J1= 11
J2 = 1
M = 3(9-1) - 2(11) - 1
M = 1
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Mobility (Degree of Freedom)
Example
1 23 1 2M L J J
Num. Links = 3
Num. full joint, (1 DOF) = 2 pins
Num. half joint, (2 DOF) = 1
L = 3
J1= 2
J2 = 1
M = 3(3-1) - 2(2) - 1
M = 1
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Mechanisms vs. Structures
Mechanism: M>0
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Paradoxes
Specific cases due to unique geometric considerations, a complete motion
analysis is needed instead
Kutzbach’s (or Gruebler’s) may lead to wrong answers (does not consider the
geometry of the linkage).
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Linkage Transformation
Transforms basic links into other mechanisms
Rules
1. Pin joints can be replaced by sliders with no change in DOF (provided we keep at
least 2 pins).
2. A full joint can be replaced by a half-joint an increasing the DOF by 1.
3. Removal of a link reduces DOF by 1.
4. Combination of (2) and (3) keeps same DOF.
5. Any higher-order link (e.g., ternary) can be partially “shrunk” by coalescing
nodes in a multiple joint (with no change in DOF).
6. Complete shrinkage of higher-order link is equivalent to removing it. This
creates a multiple joint and reduces the DOF.
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Linkage Transformation Examples
Behave as crank-rocker with infinite link 4
Slider-crank from Crank-Rocker (Rule #1)
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Linkage Transformation Examples
Provides exact simple harmonic motion
Scotch Yoke from Slider-Crank (Rule #4)
Substitution of link by half-joint (2 DOF)
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Inversion
Created by grounding a different link in the kinematic chain
Mechanism modifications
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The Four-Bar Linkage
• The most fundamental linkage is the 4-bar
It consists of 3 moving links and 1 fixed link
The Links are:
• Ground/Base link (fixed)
• Input Link (“driver” mover and connected to ground)
• Output Link (“driven” mover and connected to ground)
• Coupler or Floating Link (moving and connects driver to driven link)
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Grashof Condition
Predicts rotation behavior of four-bar linkages
Grashof Linkage if
𝑆 + 𝐿 ≤ 𝑃 + 𝑄
S = length of shortest link
L = length of longest link
P = length of one remaining link
Q = length of other remaining link
Driver
Frame
Coupler
Follower
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Grashof Condition
Crank and/or rocking motion depending on grounded link
Class I: 𝑆 + 𝐿 < 𝑃 + 𝑄
• Grounding either link adjacent
to S yields a crank-rocker (1
and 2)
• Grounding S yields a double-
crank (3)
• Grounding the link opposite S
yields a Grashof-double
rocker
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Grashof Condition
Non-Grashof, triple rockers
Class II: 𝑆 + 𝐿 > 𝑃 + 𝑄
• All inversions are triple rockers
(no link can fully rotate)
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Grashof Condition
Special-case Grashof, will have “change points”
Class III: 𝑆 + 𝐿 = 𝑃 + 𝑄
• All are either double-
cranks or crank-
rockers, but will have
“change points”
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Barker’s Classification
Further detailed classification for four-bar mechanism motion
Barker’s Classification
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Grashof Condition and Barker’s Classification
Examples
Determine the Grashof Condition and Barker’s Classification
(a) (b)
(c)
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Grashof Condition and Barker’s Classification
Examples
Determine the Grashof Condition and Barker’s Classification
(d) Oil Field Pump
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Grashof Condition and Barker’s Classification
Examples
Determine the Grashof Condition and Barker’s Classification (Motion)
(e) Aircraft Overhead Bin Mechanism