Graphical Analysis: Part 2 Prismatic Consideration &amp; Inversion

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Graphical Analysis: Part 2 Prismatic Consideration &amp; 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 2Prismatic Consideration & Inversion

ME 3230

R. R. Lindeke, Ph.D.

ME 3230

Topics:
• Dealing With Crank Sliders
• Position
• Velocity
• Acceleration
• Drivers from outside the Primary Closures

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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)

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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)

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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

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Graphically:
• aC4/D4 – Slider Acceleration = 24.844*10000= 248440 mm/s2
• 3 = atC3/B3/rC/B= 210940/200 = 1054.8 rad/s2

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“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

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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

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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

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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

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

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• 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

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• 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

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

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Graphically:

aF = 5.122*25 = 128.05”/s2

5 = (1.175*25)/2.5 = 11.75 rad/s2 CW

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A Final Issue:
• 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

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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

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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

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A Short Video:

Click to Play Video:

ME 3230