Download presentation

Presentation is loading. Please wait.

Published byMaximo Darden Modified over 2 years ago

1
P. Nikravesh, AME, U of A Instant Centers of Velocities Introduction 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. (2) (3) (4) (5) (6) (1) I 1,2 I 2,3 I 3,4 I 4,1 I 2,4 I 1,3 I 4,5 I 5,6 I 6,1 I 4,6 I 1,5 I 2,6 I 2,5 I 3,5 I 3,6

2
P. Nikravesh, AME, U of A Instant Centers of Velocities O2O2 O4O4 O6O6 ω2ω2 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? Problem 1 ► (2) (3) (4) (5) (6)

3
P. Nikravesh, AME, U of A Instant Centers of Velocities I 1,3 I 4,6 I 2,3 I 3,4 I 4,1 I 2,4 I 4,5 I 5,6 I 1,5 I 2,5 I 3,5 I 3,6 Problem 1: the best choice 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 I 1,2, I 2,6, and I 1,6 : 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. ► ► (2) (3) (4) (5) (6) (1) I 1,2 I 2,6 ► I 6,1

4
P. Nikravesh, AME, U of A Instant Centers of VelocitiesProblem 1: velocity analysis Velocity analysis Check: The three centers we selected must be on a straight line! Step 1: I 2,6 is a point on link 2 which rotates about I 1,2. Knowing the angular velocity of link 2 we determine the velocity of I 2,6 : V I 2,6 = ω 2 ∙ R I 2,6 I 1,2 Step 2: I 2,6 is also a point on link 6 that rotates about I 6,1. Knowing the velocity of I 2,6, we determine the angular velocity of link 6: ω 6 = V I 2,6 ∕ R I 2,6 I 6,1 ► ► (2) (6) (1) I 1,2 I 2,6 ► I 6,1 ω2ω2 V I 2,6 R I 2,6 I 1,2 R I 2,6 I 6,1 ω6ω6

5
P. Nikravesh, AME, U of A Instant Centers of Velocities O2O2 O4O4 O6O6 ω2ω2 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? Problem 2 ► (2) (3) (4) (5) (6)

6
P. Nikravesh, AME, U of A Instant Centers of VelocitiesProblem 2: the best choice I 6,1 (3) (4) (6) I 2,3 I 3,4 I 4,1 I 2,4 I 1,3 I 4,5 I 5,6 I 4,6 I 2,6 I 3,5 I 3,6 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 I 1,2, I 2,5, and I 1,5 : We can simply ignore all the other centers and bodies. Now we can perform the analysis. (2) (5) (1) I 1,2 I 1,5 I 2,5 ► ►

7
P. Nikravesh, AME, U of A Instant Centers of VelocitiesProblem 2: velocity analysis Velocity analysis Check: The three centers we selected must be on a straight line! Step 1: I 2,5 is a point on link 2 which rotates about I 1,2. Knowing the angular velocity of link 2 we determine the velocity of I 2,5 : V I 2,5 = ω 2 ∙ R I 2,5 I 1,2 Step 2: I 2,5 is also a point on link 5 that rotates about I 1,5. Knowing the velocity of I 2,5, we determine the angular velocity of link 5: ω 5 = V I 2,5 ∕ R I 2,5 I 1,5 ► ► ► ω2ω2 V I 2,5 R I 2,5 I 1,2 R I 2,5 I 1,5 ω5ω5 (2) (5) (1) I 1,2 I 1,5 I 2,5

8
P. Nikravesh, AME, U of A Instant Centers of Velocities O2O2 O4O4 O6O6 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? Problem 3 ► (2) (3) (4) (5) (6) VPVP

9
P. Nikravesh, AME, U of A Instant Centers of VelocitiesProblem 2: the best choice 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 I 1,3, I 3,6, and I 1,6 : We can simply ignore all the other centers and bodies. Now we can perform the analysis. (4) I 2,3 I 3,4 I 4,1 I 2,4 I 4,5 I 5,6 I 4,6 I 2,6 I 3,5 (2) (5) I 1,2 I 1,5 I 2,5 ► ► I 6,1 (3) (6) I 1,3 I 3,6 (1)

10
P. Nikravesh, AME, U of A Instant Centers of VelocitiesProblem 3: velocity analysis Velocity analysis Check: The three centers we selected must be on a straight line! Step 1: Link 3 rotates about about I 1,3. Knowing the velocity of point P, we first determine the angular velocity of link 3: ω 3 = V P ∕ R P I 1,3 I 3,6 is a point on link 3 which rotates about I 1,3. Knowing the angular velocity of link 3 we determine the velocity of I 3,6 : V I 3,6 = ω 3 ∙ R I 3,6 I 1,3 Step 2: I 3,6 is also a point on link 6 that rotates about I 1,6. Knowing the velocity of I 3,6, we determine the angular velocity of link 6: ω 6 = V I 3,6 ∕ R I 3,6 I 1,6 ► ► ► ω3ω3 V I 3,6 R I 3,6 I 1,3 R I 3,6 I 6,1 ω6ω6 I 6,1 (3) (6) I 1,3 I 3,6 (1) VPVP R P I 1,3 ►

Similar presentations

Presentation is loading. Please wait....

OK

Acceleration analysis (Chapter 4)

Acceleration analysis (Chapter 4)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on airbag working principle of ups Ppt on child labour in nepal Ppt on tourist places in india Ppt on traffic light controller using verilog Ppt on case study of apple Ppt on artificial intelligence in machines Ppt on water pollution causes Ppt on formal education Ppt on steps Ppt on agriculture in indian economy