1 / 28

# Graphical Analysis: Part 2 Prismatic Consideration & Inversion - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Graphical Analysis: Part 2 Prismatic Consideration & Inversion' - savanna

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Graphical Analysis: Part 2Prismatic Consideration & Inversion

ME 3230

R. R. Lindeke, Ph.D.

ME 3230

• Dealing With Crank Sliders

• Position

• Velocity

• Acceleration

• Beyond the 4-Bar Linkage

• Drivers from outside the Primary Closures

ME 3230

• 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

• 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

• 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

ME 3230

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

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

ME 3230

ME 3230

ME 3230

• 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

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

• 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

• 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

ME 3230

• aB = 2.828*25 = 70.7”/s2340

ME 3230

ME 3230

• 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

• 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

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

ME 3230

aF = 5.122*25 = 128.05”/s2

5 = (1.175*25)/2.5 = 11.75 rad/s2 CW

ME 3230

• 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

• 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 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: Kinematics Modeler

Click to Play Video:

ME 3230