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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 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



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



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




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



Acceleration of pt b l.jpg
Acceleration of Pt. B:

  • aB = 2.828*25 = 70.7”/s2340

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: E

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

ME 3230



Graphically23 l.jpg
Graphically: E

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: E

  • 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 E

  • 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



Using this modeler l.jpg
Using this Modeler Kinematics 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: Kinematics Modeler

Click to Play Video:

ME 3230