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P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Introduction Acceleration Polygon for a Four-bar Mechanism This presentation shows how to.

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Presentation on theme: "P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Introduction Acceleration Polygon for a Four-bar Mechanism This presentation shows how to."— Presentation transcript:

1 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Introduction 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: 1. How to construct the acceleration polygon shown on the right 2. How to extract acceleration information from the polygon OAOA A BA ABAB A A B A O4O4 B O2O2 ω2ω2 α2α2

2 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-barExample 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. A O4O4 B O2O2 ω2ω2 α2α2

3 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-barVector loop, differentiation A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 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: R AO 2 + R BA = R O 4 O 2 + R BO 4 The first time derivative provides the velocity loop equation: V t A + V t BA = V t B The second time derivative provides the acceleration loop equation: A A + A BA = A B We split each acceleration vector into a normal and a tangential component: A t A + A n A + A t BA + A n BA = A t B + A n B We need the velocities to calculate some of the accelerations. Therefore we perform a velocity analysis first. ►

4 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Velocity polygon We 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 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. Velocity analysis A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 ω2ω2 VtAVtA VtBVtB V t BA A B OVOV VtAVtA ► ► ► ► V t A + V t BA = V t B

5 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Angular velocities The absolute value of the angular velocities are computed as: ω 3 = V t BA ∕ R BA ω 4 = V t B ∕ R BO 4 R BA has to be rotated 90° cw to head in the direction of V t BA. Therefore ω 3 is cw R BO 4 has to be rotated 90° ccw to head in the direction of V t B. Therefore ω 4 is ccw We will use these results to calculate accelerations. Angular velocities A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 ω2ω2 VtAVtA VtBVtB V t BA A B OVOV VtAVtA ω3ω3 ω4ω4 ► ►

6 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Normal components We first calculate the magnitude of all the normal components: A n A = ω 2 2 ∙ R AO 2 A n BA = ω 3 2 ∙ R BA A n B = ω 4 2 ∙ R BO 4 The direction of each normal component of acceleration is opposite to the corresponding position vector Normal components A t A + A n A + A t BA + A n BA = A t B + A n B AnBAnB A n BA AnAAnA A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 A O4O4 B O2O2 ω2ω2 ► ω3ω3 ω4ω4

7 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Tangential components We first calculate the magnitude of A t A : A t A = α 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of α 2 We also know that A t BA is on an axis perpendicular to R BA Similarly we know that A t B is on an axis perpendicular to R BO 4 Tangential components A t A + A n A + A t BA + A n BA = A t B + A n B AnBAnB A n BA AnAAnA AtAAtA A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 A O4O4 B O2O2 ► ► ► α2α2

8 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Acceleration polygon Now we are ready to draw the acceleration polygon. First we select the origin and add A t A and A n A to obtain A A A n BA is a added at A We also know the axis of A t BA which would be added to A n BA A n B is added to the origin We also know the axis of A t B which would be added to A n B The two lines intersect at B. We complete the acceleration polygon by drawing the missing accelerations Next we determine the angular accelerations. Acceleration polygon AnBAnB A n BA AnAAnA AtAAtA A O4O4 B O2O2 OAOA AnBAnB AnAAnA AtAAtA A BA A t BA ABAB AtBAtB A A t A + A n A + A t BA + A n BA = A t B + A n B A B ► ► ► ► ► ►

9 P. Nikravesh, AME, U of A Acceleration Polygon for a Four-bar Angular accelerations The absolute values of the angular accelerations are computed as: α 3 = A t BA ∕ R BA α 4 = A t B ∕ R BO 4 R BA has to be rotated 90° cw to head in the direction of A t BA. Therefore α 3 is cw R BO 4 has to be rotated 90° ccw to head in the direction of A t B. Therefore α 4 is ccw This completes the acceleration analysis of this four-bar. Angular accelerations OAOA A n BA AnAAnA AtAAtA A BA A t BA ABAB AtBAtB A A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 α2α2 α3α3 α4α4 AnBAnB ► ►


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