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Graphical Analysis: Part 2 Prismatic Consideration & Inversion. ME 3230 R. R. Lindeke, Ph.D. Topics:. Dealing With Crank Sliders Position Velocity Acceleration Beyond the 4-Bar Linkage Drivers from outside the Primary Closures. Lets Look at Problem 2.6.

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Graphical Analysis: Part 2 Prismatic Consideration & Inversion

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Graphical analysis part 2 prismatic consideration inversion l.jpg

Graphical Analysis: Part 2Prismatic Consideration & Inversion

ME 3230

R. R. Lindeke, Ph.D.

ME 3230


Topics l.jpg

Topics:

  • Dealing With Crank Sliders

    • Position

    • Velocity

    • Acceleration

  • Beyond the 4-Bar Linkage

    • Drivers from outside the Primary Closures

ME 3230


Lets look at problem 2 6 l.jpg

Lets Look at Problem 2.6

  • This is a Simple Crank Slider and lets add a point D4 (we will define vC4/D4 as the Slider Velocity)

  • Link AB = 60mm, Link BC = 200 mm

  • 2 = 100 rad/s (about 955 RPM)

ME 3230


What we know l.jpg

What we “Know”

  • Velocity Pole (o) includes points A2, A3, D3, D4

  • vB2/A2 = 2 x rB2/A2 = 100 X 60 = 6000 mm/s (direction is normal to A2B2) = vB2 = vB3

  • vC4/D4 is Horizontal

  • vC3 + vC3/B3 = vA2 + vB2/A2

  • vC3/B3 = 3 x rC3/B3 (direction is normal to B3C3)

ME 3230


The scaled model velocity polygon l.jpg

The Scaled Model & Velocity Polygon:

ME 3230


Summarizing velocity analysis l.jpg

Summarizing Velocity Analysis:

  • Vc/d = slider velocity = 45 * 100 = 4500 mm/s

  • Vc/b = 75*100 = 7500 mm/s = 3 * rc/b

  • Therefore, 3 = 7500/200 = 37.5 rad/s

  • Its direction is from rc/b to vc/b or CW

ME 3230


Acceleration analysis what we know l.jpg

Acceleration Analysis, What we ‘know’:

ME 3230


Continuing l.jpg

Continuing:

ME 3230


Graphically l.jpg

Graphically:

  • aC4/D4 – Slider Acceleration = 24.844*10000= 248440 mm/s2

  • 3 = atC3/B3/rC/B= 210940/200 = 1054.8 rad/s2

ME 3230


One that involved imaging l.jpg

One that Involved “Imaging”

ME 3230


Drawn to scale l.jpg

Drawn to scale:

ME 3230


Knowns l.jpg

“Knowns”:

  • vB2 = vB2/A2 + vA2 = 2 x rB/A + 0= 20 rad/s * 0.5” = 10”/s (115-90)= 10  25”/s

  • vC3 = vB2 + vC3/B3  wherevC3/B3 = 3 x rC/B (direction normal to rC/B)

  • And vC4 = vD4 + vC4/D4 = 0 + 4 x rC/D where direction is normal to rC/D

  • vF5 = vE + vF5/E5  wherevE5/F5 = 5 x rF/E (Direction Normal to rF/E)

  • vF/’G’ = Slider Velocity is Horizontal

ME 3230


We build velocity image to get pt e l.jpg

We “Build” Velocity Image to get Pt E

Used Similar Triangles rotated 90 CW and Scaling factor: 1.318/2 for each side (diameters of construction circles)

vE/B = 1.186* 5 = 5.93”/s

vE/C = 1.582 * 5 = 7.91”/s

vC/B = 1.318 * 5 = 6.59”/s

ME 3230


With values l.jpg

With Values:

  • vc = 1.639*5 = 8.195”/s344.03

  • vE = 2.938*5 = 14.69”/s7.76

  • 3 = vC3/B3/rC/B = 6.59/2 = 3.295 rad/s CCW

  • 4 = vC4/D4/rC/D = 8.915/.95 = 8.626 rad/s CW

ME 3230


Finding the f velocities l.jpg

Finding the “F” Velocities:

  • vF5/E5 = .401*5 = 2.005”/s98.107 

  • 5 = vF5/E5/rF/E = 2.005/2.5 = 0.802 rad/s CCW

  • vF6/’G6’ = Slider velocity = 2.967*5 = 14.835”/s

ME 3230


On to acceleration knowns l.jpg

On to Acceleration (Knowns)

ME 3230


Acceleration of pt b l.jpg

Acceleration of Pt. B:

  • aB = 2.828*25 = 70.7”/s2340

ME 3230


Attacking accel of pt c l.jpg

Attacking Accel. Of Pt C:

ME 3230


Leads to l.jpg

Leads to:

  • aC’ = 121.475308.45 in/s2

  • atC4/D4 =3.952*25 = 98.8 in/s2

  • 4 = 98.8/.95 = 104 rad/s2

  • atC3/B3 = 2.726*25 = 68.15 in/s2

  • 3 = 68.15/2 = 34.075 rad/s2

ME 3230


Next we will us an acceleration image approach for pt e l.jpg

Next We will us an “Acceleration Image” approach for Pt. E

  • The Acceleration Image Triangle “side ratio” is aC/B/rC/B = 2.861/2 = 1.4305

  • So: aE/B = 1.4305*1.8 =2.5749

  • & aE/C = 1.4305 * 2.4 = 3.4332

  • This lets us set Point e’ for E’s acceleration and branching to Pt. F’s acceleration

ME 3230


Leads to21 l.jpg

Leads to:

Thus aE = 5.354*25 = 133.85”/s2 @ 347.354

ME 3230


On to accel of pt f and the slider l.jpg

On to Accel. of Pt. F (and the slider)

ME 3230


Graphically23 l.jpg

Graphically:

aF = 5.122*25 = 128.05”/s2

5 = (1.175*25)/2.5 = 11.75 rad/s2 CW

ME 3230


A final issue l.jpg

A Final Issue:

  • Driven Link Not in the Primary Linkage Closure (Primary 4-Bar Linkage)

  • Typically this means “solution by Inversion” is required

    • We will drive the mechanism from the primary linkage

    • We will use a Unit Rate for the inverted driver

ME 3230


The rest of the technique l.jpg

The Rest of the Technique

  • We Need to learn the position of a Point of Interest where the primary linkage meets the (real) Driver Linkage (since we will not necessarily have the actual geometry of interest in the inverted system)

  • This “Coupler Point” then must be traced

  • The Pose most closely approximating the desired “real driver” position when driven from the inversion is selected.

  • Using this geometry, we solve the Velocity/Acceleration problem working from the inverted driver link

  • We Find the “Proposed” Velocity of the Actual Driver Link

  • We generate an Inverted Velocity of the Driver Link to real Driver Velocity “Scaling Ratio”

  • All calculated Velocities and Acceleration found are multiplied by this scaling ratio to get actual values for the links and points

ME 3230


Coupler curve of pt d generated using catia s dmu kinematics modeler l.jpg

Coupler Curve of Pt D generated using Catia’s DMU Kinematics Modeler

ME 3230


Using this modeler l.jpg

Using this Modeler

  • We can get Coupler Curves of Points of Interest

  • We can get plots of actual Velocities, Accelerations, etc. From the Linkage in Motion (graphically)

  • We can generate Videos of the Linkage in Motion

ME 3230


A short video l.jpg

A Short Video:

Click to Play Video:

ME 3230


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