introduction n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Introduction PowerPoint Presentation
Download Presentation
Introduction

Loading in 2 Seconds...

play fullscreen
1 / 9

Introduction - PowerPoint PPT Presentation


  • 469 Views
  • Uploaded on

Introduction. O A. A A. A. B. A. A BA. A B. ω 2. α 2. O 4. O 2. B. Acceleration Polygon for a Four-bar Mechanism

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

PowerPoint Slideshow about 'Introduction' - nikkos


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
introduction
Introduction

Acceleration Polygon for a Four-bar

OA

AA

A

B

A

ABA

AB

ω2

α2

O4

O2

B

Acceleration Polygon for a Four-bar Mechanism

This presentation shows how to construct the acceleration polygon for a four-bar mechanism. It is assumed that the dimensions for the links are known and the analysis is being performed at a given (known) configuration. Since the four-bar has one degree-of-freedom, the angular velocity and acceleration of one of the links must be given as well.

As an example, for the four-bar shown, we first construct the velocity polygon. Then we will learn:

How to construct the acceleration polygon shown on the right

How to extract acceleration information from the polygon

slide2

Example

Acceleration Polygon for a Four-bar

B

A

ω2

α2

O4

O2

Example

This example shows us how to construct the acceleration polygon for a typical four-bar, such as the one shown on this slide.

It is assumed that: all the lengths are known and the four-bar is being analyzed at the configuration shown; the angular velocity and acceleration of the crank are given as well.

slide3

Vector loop, differentiation

Acceleration Polygon for a Four-bar

B

RBA

Vector loop

We follow the same process as we did for the velocity polygon.

We first define four position vectors to obtain a vector loop equation:

RAO2 + RBA = RO4O2 + RBO4

The first time derivative provides the velocity loop equation:

VtA + VtBA = VtB

The second time derivative provides the acceleration loop equation:

AA + ABA = AB

We split each acceleration vector into a normal and a tangential component:

A

RBO4

RAO2

O4

RO4O2

O2

AtA + AnA + AtBA + AnBA = AtB + AnB

We need the velocities to calculate some of the accelerations. Therefore we perform a velocity analysis first.

slide4

Velocity analysis

Acceleration Polygon for a Four-bar

B

VtA

RBA

Velocity polygon

We calculate VtA:

VtA = ω2∙ RAO2

The direction is found by rotating RAO2 90° in the direction of ω2

The direction of VtBA is perpendicular to RBA

The direction of VtB is perpendicular to RBO4

Now we construct the velocity polygon

Next we determine the angular velocities.

We will use these results to calculate the normal components of acceleration vectors.

A

RBO4

RAO2

ω2

O4

RO4O2

O2

A

VtA

OV

VtBA

VtB

B

VtA + VtBA = VtB

slide5

Angular velocities

Acceleration Polygon for a Four-bar

B

VtA

RBA

ω3

Angular velocities

The absolute value of the angular velocities are computed as:

ω3 = VtBA∕ RBA

ω4 = VtB∕ RBO4

RBA has to be rotated 90° cw to head in the direction of VtBA. Therefore ω3 is cw

RBO4 has to be rotated 90° ccw to head in the direction of VtB. Therefore ω4 is ccw

We will use these results to calculate accelerations.

A

RBO4

RAO2

ω2

ω4

O4

RO4O2

O2

A

VtA

OV

VtBA

VtB

B

slide6

Normal components

Acceleration Polygon for a Four-bar

AtA + AnA + AtBA + AnBA = AtB + AnB

B

RBA

ω3

Normal components

We first calculate the magnitude of all the normal components:

AnA = ω22∙ RAO2

AnBA = ω32∙ RBA

AnB = ω42∙ RBO4

The direction of each normal component of acceleration is opposite to the corresponding position vector

A

RBO4

RAO2

ω2

ω4

O4

RO4O2

O2

B

AnBA

A

AnB

AnA

O4

O2

slide7

Tangential components

Acceleration Polygon for a Four-bar

AtA + AnA + AtBA + AnBA = AtB + AnB

B

RBA

Tangential components

We first calculate the magnitude of AtA:

AtA = α2∙ RAO2

The direction is found by rotating RAO2 90° in the direction of α2

We also know that AtBA is on an axis perpendicular to RBA

Similarly we know that AtB is on an axis perpendicular to RBO4

A

RBO4

RAO2

α2

O4

RO4O2

O2

B

AtA

AnBA

A

AnB

AnA

O4

O2

slide8

Acceleration polygon

Acceleration Polygon for a Four-bar

AtA + AnA + AtBA + AnBA = AtB + AnB

Acceleration polygon

Now we are ready to draw the acceleration polygon.

First we select the origin and add AtA and AnA to obtain AA

AnBA is a added at A

We also know the axis of AtBA which would be added to AnBA

AnB is added to the origin

We also know the axis of AtB which would be added to AnB

The two lines intersect at B. We complete the acceleration polygon by drawing the missing accelerations

Next we determine the angular accelerations.

B

AtA

AnBA

A

AnB

AnA

O4

O2

AtA

AnA

OA

AA

AnB

AnBA

A

AB

ABA

AtBA

AtB

B

slide9

Angular accelerations

Acceleration Polygon for a Four-bar

B

α3

RBA

Angular accelerations

The absolute values of the angular accelerations are computed as:

α3 = AtBA∕ RBA

α4 = AtB∕ RBO4

RBA has to be rotated 90° cw to head in the direction of AtBA. Therefore α3 is cw

RBO4 has to be rotated 90° ccw to head in the direction of AtB. Therefore α4 is ccw

This completes the acceleration analysis of this four-bar.

A

RBO4

RAO2

α4

α2

O4

RO4O2

O2

AtA

AnA

OA

AA

AnB

AnBA

AB

ABA

AtBA

AtB