Vector analysis Mechanics Engineering Dynamics .pptx

Vector Analysis, Mechanical Engineering Dynamics

1. How Does This Mechanism Move?

2. Angular Velocity Vector If is always perpendicular to , Since needs to be perpendicular to both and , is found using the right hand rule.  

3. Angular Acceleration Vector  

4. Determining Cross Products i j k + -  

5. velocity of any point on a link with respect to another point on the same link is always perpendicular to the line joining these points on the configuration (or space) diagram. Velocity Diagram

6. With “A” Chosen as the reference point, the velocity of “B” can be calculated using vector analysis using the following equation: Where “ ω ” is the angular- velocity vector normal to the plane of the motion in the sense determined by the right hand rule. Velocity Diagram i j k + -

7. Workshop Exercise 1: The sharper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the velocity of the slider block C at the instant θ = 60⁰, if link AB is rotating at 4 rad/s.

8.   Cross products First consider link AB: Coordinates   i j k Workshop Exercise 1 Solution:

9.   Consider link BC: -i +j Cross products Coordinates i j k

10.   Consider slider C:     Substituting v C /A , v B /A , v C /B in the relative velocity equation Cross products Coordinates     i j k

11. Equating i & j components : ….equation 1   ….equation 2 Cross products Coordinates              i j k

12. Determining acceleration of rigid bodies using vector analysis

13. The rotation of the gear is controlled by the horizontal motion of end A of the rack AB as shown in the figure. If the piston rod has a constant velocity of V A = 300i mm/s during a short interval of motion, determine: a)The angular velocity and acceleration of the gear, and b) The angular velocity and acceleration of AB at the instant where x = 800 mm Workshop Exercise 2:

14. Calculating :  ] Calculating :  ]   C         O Given: - Equation (1)   Workshop Exercise 2 Solution: Velocity Vector Analysis

15. i = i = - Equation (3) j = j  - Equation (4)   C         O Substituting in the relative velocity equation (1) for   = - Equation (2)  

16. C         O From Equation (3): )=  From Equation (4):  

17. Calculating : ] +   C       O Given: Since A has constant linear velocity   - Equation (5)   Acceleration Vector Analysis Workshop Exercise 2 Solution:

18. Calculating :  ] +   C         O

19. - Equation (6)   Substituting in the relative acceleration in Equation (5) for   C         O

20. i = i - Equation (7) j = j ; - Equation (8) Solving the two simultaneous equations (7 and 8)     Equating i and j components: - Equation (6)     .

21. The rack R of the rack-and –pinion gear is moving upward with a constant velocity of 1200 mm/s. Using vector analysis, find the angular velocity and angular acceleration of bar BC. Workshop Exercise 3:

22. Velocity of point B in terms of velocity of point C: Given:     Velocity of point C in terms of Gear velocity:         Velocity of point B in terms of angular velocity of the arm AB:     r C /D r B /A r B /C Workshop Exercise 3 Solution: Velocity Vector Analysis

23. Equating i and j components: i = i ) = j = j) = Solving two simultaneous equations  &      

24. Calculating angular acceleration of the link BC:     r C /D r B /A r B /C Workshop Exercise 3 Solution: Acceleration Vector Analysis

25. Calculating angular acceleration of the link BC:         r C /D r B /A r B /C

26. Substitute for , from above: = i = i ) = -120 - Eq (1) j = j) = - Eq (2) Multiply Eq (2) by 2 and solve the two simultaneous equations to find and :   &