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Graphical Analysis: Part 2 Prismatic Consideration & Inversion

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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

Graphical Analysis: Part 2Prismatic Consideration & Inversion

ME 3230

R. R. Lindeke, Ph.D.

ME 3230

topics
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
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
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
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

graphically
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
“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
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
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
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
Acceleration of Pt. B:
  • aB = 2.828*25 = 70.7”/s2340

ME 3230

leads to
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
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
Leads to:

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

ME 3230

graphically23
Graphically:

aF = 5.122*25 = 128.05”/s2

5 = (1.175*25)/2.5 = 11.75 rad/s2 CW

ME 3230

a final issue
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
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

using this modeler
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
A Short Video:

Click to Play Video:

ME 3230

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