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Prof. Albert Espinoza Mechanical Engineering Department Polytechnic University of Puerto Rico Spring 2015

ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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Page 1: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Prof. Albert Espinoza

Mechanical Engineering Department

Polytechnic University of Puerto Rico

Spring 2015

Page 2: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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)

Page 3: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 4: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 5: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 6: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 7: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 8: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 9: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Mechanisms vs. Structures

Mechanism: M>0

Page 10: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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).

Page 11: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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Page 12: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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.

Page 13: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Linkage Transformation Examples

Behave as crank-rocker with infinite link 4

Slider-crank from Crank-Rocker (Rule #1)

Page 14: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Linkage Transformation Examples

Provides exact simple harmonic motion

Scotch Yoke from Slider-Crank (Rule #4)

Substitution of link by half-joint (2 DOF)

Page 15: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Inversion

Created by grounding a different link in the kinematic chain

Mechanism modifications

Page 16: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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Page 17: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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)

Page 18: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 19: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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

Page 20: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Grashof Condition

Non-Grashof, triple rockers

Class II: 𝑆 + 𝐿 > 𝑃 + 𝑄

• All inversions are triple rockers

(no link can fully rotate)

Page 21: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

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”

Page 22: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Barker’s Classification

Further detailed classification for four-bar mechanism motion

Barker’s Classification

Page 23: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Grashof Condition and Barker’s Classification

Examples

Determine the Grashof Condition and Barker’s Classification

(a) (b)

(c)

Page 24: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Grashof Condition and Barker’s Classification

Examples

Determine the Grashof Condition and Barker’s Classification

(d) Oil Field Pump

Page 25: ME 2220_SP15_LEC02 - Kinematic Fundamentals and Mobility

Grashof Condition and Barker’s Classification

Examples

Determine the Grashof Condition and Barker’s Classification (Motion)

(e) Aircraft Overhead Bin Mechanism