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P. Nikravesh, AME, U of A Instant Centers for a Four-bar Introduction Velocity Analysis with Instant Centers for a Four-bar Mechanism This presentation.

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Presentation on theme: "P. Nikravesh, AME, U of A Instant Centers for a Four-bar Introduction Velocity Analysis with Instant Centers for a Four-bar Mechanism This presentation."— Presentation transcript:

1 P. Nikravesh, AME, U of A Instant Centers for a Four-bar Introduction Velocity Analysis with Instant Centers for a Four-bar Mechanism This presentation shows how to perform velocity analysis for a four-bar mechanism with the method of instant centers. 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 (or one other velocity information) must be given as well. For a given four-bar mechanism the velocity analysis consists of two steps: 1. Finding the instant centers 2. Finding velocities Note that the instant center method is a graphical method.

2 P. Nikravesh, AME, U of A Instant Centers for a Four-bar Four-bar mechanism Assume that for this four-bar mechanism all the link lengths are known and the angular velocity of the crank is given as ω 2 ccw. In the configuration shown we can perform a velocity analysis with the instant center method. Four-bar mechanism O2O2 A O4O4 ω2ω2 B

3 P. Nikravesh, AME, U of A Instant Centers for a Four-bar Number of instant centers The first task is to determine how many instant centers exist for a four-bar. The number of links in a four-bar is n = 4 Between n links, there are n (n − 1) ∕ 2 instant centers. That means in a fourbar there are 4 (4 − 1) ∕ 2 = 6 instant centers. A small circle will help us keep track of locating each center. On the circumference of the circle we put as many marks as the number of links. Each time we find a center between two links, we draw a line between the corresponding marks on the circle. Number of instant centers O2O2 A O4O4 ω2ω2 B ►

4 P. Nikravesh, AME, U of A Instant Centers for a Four-bar O 2 = I 1,2 O 4 = I 4,1 A = I 2,3 B = I 3,4 Finding the instant centers Four of the centers are already known: They are the four pin joints. We don’t know I 1,3 but we know that it lies on the same line as I 4,1 and I 3,4. I 1,3 also lies on the same line as I 2,1 and I 3,2. The point of intersection is I 1,3. I 2,4 is also unknown but it lies on the same line as I 3,4 and I 2,3. I 2,4 also lies on the same line as I 4,1 and I 1,2. The point of intersection is I 2,4. Now we have found all 6 centers Finding the instant centers ω2ω2 I 1,3 I 2,4 ► ► ► ► ► ► ►

5 P. Nikravesh, AME, U of A Instant Centers for a Four-bar R AI 1,3 Finding velocities A (or I 2,3 ) is a point on link 2, therefore: V A = ω 2 ∙ R AI 1,2 Its direction is obtained by rotating R AI 1,2 90° in the direction of ω 2. A (or I 2,3 ) is also a point on link 3, which rotates around I 1,3, this means: V A = ω 3 ∙ R AI 1,3 Since we already know V A, we can solve for ω 3 : ω 3 = V A ∕ R AI 1,3 Finding ω 3, knowing ω 2 O 2 = I 1,2 O 4 = I 4,1 ω2ω2 ω3ω3 VAVA I 1,3 R AI 1,2 I 2,4 A = I 2,3 B = I 3,4 ► ►

6 P. Nikravesh, AME, U of A Instant Centers for a Four-bar R BI 4,1 R BI 1,3 Finding velocities Since we know ω 3, we can find V B. B is a point on link 3: V B = ω 3 ∙ R BI 1,3 Its direction is obtained by rotating R BI 1,3 90° in the direction of ω 3. B is also a point on link 4 which yields: V B = ω 4 ∙ R BI 1,4 We already know V B so we can solve for ω 4 : ω 4 = V B ∕ R BI 1,4 Finding ω 4, knowing ω 3 ω3ω3 VAVA ω4ω4 VBVB O 2 = I 1,2 O 4 = I 4,1 ω2ω2 I 1,3 I 2,4 A = I 2,3 B = I 3,4 ► ►

7 P. Nikravesh, AME, U of A Instant Centers for a Four-bar R I 2,4 I 1,2 Another approach We could have determined ω 4 without knowing ω 3 : I 2,4 is a point on link 2, therefore: V I 2,4 = ω 2 ∙ R I 2,4 I 1,2 Its direction is obtained by rotating R I 2,4 I 1,2 90° in the direction of ω 2. I 2,4 is also a point on link 4, which rotates around I 4,1. This means: V I 2,4 = ω 4 ∙ R I 2,4 I 4,1 We already know V I 2,4 so we can solve for ω 4 : ω 4 = V I 2,4 ∕ R I 2,4 I 4,1 Finding ω 4, knowing ω 2 I 1,3 V I 2,4 R I 2,4 I 4,1 ω4ω4 O 2 = I 1,2 O 4 = I 4,1 ω2ω2 I 2,4 A = I 2,3 B = I 3,4 ► ►

8 P. Nikravesh, AME, U of A Instant Centers for a Four-bar Velocity of a coupler point The instant center method makes it easy to find the velocities of additional points. For example suppose link 3 is a triangular plate and we want to determine the velocity of P. Since P is a point on link 3, we have: V P = ω 3 ∙ R PI 1,3 The direction is found by rotating R PI 1,3 90° in the direction of ω 3. Velocity of coupler point P R PI 4,1 VPVP O 2 = I 1,2 A = I 2,3 B = I 3,4 O 4 = I 4,1 ω2ω2 I 1,3 I 2,4 ►


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