Download presentation

Presentation is loading. Please wait.

Published byRudolf Green Modified over 2 years ago

1
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar Introduction Velocity Polygon for a Four-bar Mechanism This presentation shows how to construct the velocity polygon for a given 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 of one of the links must be given as well. As an example, for the four-bar shown on the left we will learn: 1. How to construct the polygon shown on the right 2. How to extract velocity information from the polygon A O4O4 B O2O2 ω2ω2 V t A V t B V t BA A B OVOV

2
P. Nikravesh, AME, U of A Velocity Polygon for a Four-barExample 1 The first example shows us how to construct the velocity 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; i.e., all the angles are known. Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction). Introduction A O4O4 B O2O2 ω2ω2

3
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: R AO 2 + R BA = R O 4 O 2 + R BO 4 The time derivative of this equation yields the velocity loop equation: V AO 2 + V BA = V O 4 O 2 + V BO 4 Since R O 4 O 2 is fixed to the ground, V O 4 O 2 = 0. Since any velocity vector with respect to a fixed point only needs one subscript, we can further simplify the velocity loop equation: A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 Vector loop O2O2 ω2ω2 V A + V BA = V B For clarification purposes we assign superscript “t” to these vectors indicating they are tangential : V t A + V t BA = V t B ►

4
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar We can calculate V t A : V t A = ω 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of ω 2 : The direction of V t BA is perpendicular to R BA : The direction of V t B is perpendicular to R BO 4 : V A and lines of action A O4O4 B R BO 4 RO4O2RO4O2 R BA V t A O2O2 V t A + V t BA = V t B R AO 2 ω2ω2 ► ► ►

5
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar To construct the velocity polygon we select an origin and draw V t A : V t B starts at the origin. We know the line of action: V t BA starts at A. We also know the line of action: Now we can complete the velocity polygon: Note that this polygon represents the velocity loop equation shown above! Velocity polygon A O4O4 B R BO 4 RO4O2RO4O2 V t A V t B V t BA V t A O2O2 V t A + V t BA = V t B A B OVOV R AO 2 ω2ω2 R BA ► ► ► ►

6
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar We can determine ω 3 : ω 3 = V t BA / R BA R BA has to be rotated 90° clockwise to point in the same direction as V t BA. Therefore ω 3 is clockwise: To determine ω 4 : ω 4 = V t B / R BO 4 R BO 4 has to be rotated 90° counter-clockwise to point in the same direction as V t B. Therefore ω 4 is ccw: ω4ω4 ω3ω3 Angular velocities A O4O4 B R BO 4 RO4O2RO4O2 R BA O2O2 V t A V t B V t BA V t A R AO 2 ω2ω2 A B OVOV ► ►

7
P. Nikravesh, AME, U of A Velocity Polygon for a Four-barExample 2 The second example shows how to construct the velocity polygon for another typical four-bar. In addition, we learn how to determine the velocity of a coupler point from the polygon. Similar to the first example, it is assumed that all the lengths are known and the four-bar is being analyzed at the configuration shown; i.e., all the angles are known. Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction). P A O4O4 B O2O2 ω2ω2

8
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar P Velocity analysis We define four position vectors to obtain a vector loop equation: R AO 2 + R BA = R O 4 O 2 + R BO 4 Note that we first construct the vector loop containing the primary links and points. Point P is not a primary point--it will be consider later for analysis. Since the vectors have constant lengths the time derivatives are tangential velocities: V t AO 2 + V t BA = V t O 4 O 2 + V t BO 4 Discard zero vectors and subscripts referring to non-moving points: V t A + V t BA = V t B A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 ω2ω2 Vector loop ►

9
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar A O4O4 R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 V A and lines of action We can calculate V t A : V t A = ω 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of ω 2 : The direction of V t BA is perpendicular to R BA : The direction of V t B is perpendicular to R BO 4 : P B R BA ► ► ► V t A + V t BA = V t B

10
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Velocity polygon A O4O4 R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 To construct the velocity polygon we select the origin and draw V t A : V t B starts at the origin. We know the line of action: V t BA starts at A. We also know the line of action: Now we can complete the velocity polygon: Note that this polygon represents the velocity loop equation shown above! P B R BA ► ► ► ► V t A + V t BA = V t B

11
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Velocity of coupler point P A O4O4 B R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 P In order to determine the Velocity of point P we rotate the line AP 90° and move it to point A in the velocity polygon: Next we rotate the line BP 90° and move it to point B in the velocity polygon: The point of intersection is point P in the velocity polygon. Now we can draw V P : P VPVP R BA ► ► ►

12
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Angular velocities A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 V t A O2O2 ω2ω2 P P V t P We can determine ω 3 : ω 3 = V t BA / R BA R BA has to be rotated 90° clockwise to point in the same direction as V t BA. Therefore ω 3 is clockwise: To determine ω 4 : ω 4 = V t B / R BO 4 R BO 4 has to be rotated 90° clockwise to point in the same direction as V t B. Therefore ω 4 is clockwise: ω3ω3 ω4ω4 ► ►

Similar presentations

OK

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Tenth Edition Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P.

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Tenth Edition Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on girl child education in india Ppt on programmable logic array design Ppt on solar energy utilization Ppt on power generation using footsteps prayer Ppt on as 14 amalgamation tape Ppt on surface water runoff Ppt on computer hardware components Ppt on materialized view in oracle Ppt on single phase motors Ppt on global warming and climate change