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Introduction

Introduction. Instant Centers of Velocities for a Six-bar Mechanism Part 2: Velocity Analysis.

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Introduction

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  1. Introduction Instant Centers of Velocities Instant Centers of Velocities for a Six-bar Mechanism Part 2: Velocity Analysis In Part 1 of this presentation, we found all the fifteen instant centers for the six-bar mechanism as shown. In this presentation we use some of these centers to perform several velocity analyses. We show the process in separate simply stated problems. I5,6 (6) I3,4 (5) (3) I1,5 I4,5 I2,5 I2,3 (4) I3,5 I6,1 (2) I2,4 I4,6 I2,6 I3,6 I1,2 I4,1 (1) I1,3

  2. Problem 1 Instant Centers of Velocities O6 O2 O4 Problem 1 Known: Assume that the angular velocity of link 2 is given. Unknown: Determine the angular velocity of link 6. Question: What is the best choice of IC’s, among the fifteen, to determine the unknown velocity? ► (6) (5) (3) (4) (2) ω2

  3. Problem 1: the best choice Instant Centers of Velocities I5,6 I3,4 (5) (3) I1,5 I4,5 I2,5 I2,3 (4) I3,5 I2,4 I4,6 I3,6 I4,1 I1,3 The best choice We know the angular velocity of link 2 and we want to determine the angular velocity of link 6. We choose the centers between links 2, 6 and 1 (the ground). These centers are I1,2, I2,6 , and I1,6 : We can simply ignore all the other centers. As the matter of fact, we can ignore all the other bodies as well! Now we can perform the analysis. (6) ► I6,1 ► (2) I2,6 I1,2 ► (1)

  4. Problem 1: velocity analysis Instant Centers of Velocities Velocity analysis Check: The three centers we selected must be on a straight line! Step 1:I2,6 is a point on link 2 which rotates about I1,2. Knowing the angular velocity of link 2 we determine the velocity of I2,6 : VI2,6 = ω2 ∙ RI2,6 I1,2 Step 2: I2,6 is also a point on link 6 that rotates about I6,1. Knowing the velocity of I2,6, we determine the angular velocity of link 6: ω6 = VI2,6∕ RI2,6 I6,1 ► ω6 VI2,6 (6) ω2 ► I6,1 (2) RI2,6 I6,1 I2,6 RI2,6 I1,2 I1,2 (1) ►

  5. Problem 2 Instant Centers of Velocities O6 O2 O4 Problem 2 Known: Assume that the angular velocity of link 2 is given. Unknown: Determine the angular velocity of link 5. Question: What is the best choice of IC’s, among the fifteen, to determine the unknown velocity? ► (6) (5) (3) (4) (2) ω2

  6. Problem 2: the best choice Instant Centers of Velocities I5,6 (6) I3,4 (3) I4,5 I2,3 (4) I3,5 I6,1 I2,4 I4,6 I2,6 I3,6 I4,1 I1,3 The best choice We know the angular velocity of link 2 and we want to determine the angular velocity of link 5. We choose the centers between links 2, 5 and 1 (the ground). These centers are I1,2, I2,5 , and I1,5 : We can simply ignore all the other centers and bodies. Now we can perform the analysis. (5) ► I1,5 I2,5 ► (2) I1,2 (1)

  7. Problem 2: velocity analysis Instant Centers of Velocities Velocity analysis Check: The three centers we selected must be on a straight line! Step 1:I2,5 is a point on link 2 which rotates about I1,2. Knowing the angular velocity of link 2 we determine the velocity of I2,5 : VI2,5 = ω2 ∙ RI2,5 I1,2 Step 2: I2,5 is also a point on link 5 that rotates about I1,5. Knowing the velocity of I2,5, we determine the angular velocity of link 5: ω5 = VI2,5∕ RI2,5 I1,5 ► ω5 VI2,5 (5) I1,5 ω2 I2,5 ► RI2,5 I1,5 (2) RI2,5 I1,2 I1,2 (1) ►

  8. Problem 3 Instant Centers of Velocities O6 O2 O4 Problem 3 Known: Assume that the velocity of point P on link 3 is given. Unknown: Determine the angular velocity of link 6. Question: What is the best choice of IC’s, among the fifteen, to determine the unknown velocity? ► VP (6) (5) (3) (4) (2)

  9. Problem 2: the best choice Instant Centers of Velocities I5,6 I3,4 (5) I1,5 I4,5 I2,5 I2,3 (4) I3,5 (2) I2,4 I4,6 I2,6 I1,2 I4,1 The best choice We know the velocity of a point on link 3, and we want to determine the angular velocity of link 6. We choose the centers between links 3, 6 and 1 (the ground). These centers are I1,3, I3,6, and I1,6 : We can simply ignore all the other centers and bodies. Now we can perform the analysis. (6) (3) ► I6,1 ► I3,6 (1) I1,3

  10. Problem 3: velocity analysis Instant Centers of Velocities Velocity analysis Check: The three centers we selected must be on a straight line! Step 1: Link 3 rotates about about I1,3. Knowing the velocity of point P, we first determine the angular velocity of link 3: ω3 = VP∕ RPI1,3 I3,6 is a point on link 3 which rotates about I1,3. Knowing the angular velocity of link 3 we determine the velocity of I3,6: VI3,6 = ω3 ∙ RI3,6 I1,3 Step 2: I3,6 is also a point on link 6 that rotates about I1,6. Knowing the velocity of I3,6, we determine the angular velocity of link 6: ω6 = VI3,6∕ RI3,6 I1,6 ► VP ω6 (6) ► (3) ω3 I6,1 ► RI3,6 I6,1 I3,6 RPI1,3 (1) VI3,6 RI3,6 I1,3 ► I1,3

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