Mechanics of Machines Dr. Mohammad Kilani

1. Mechanics of MachinesDr. Mohammad Kilani Class 4 Velocity Analysis

2. DERIVATIVE OF A ROTATING VECTOR

3. Derivative of a Rotating Unit Vector • A unit vector in the θdirection, uθ, is a vector of unity magnitude and an angle θwith the x- axis. It is written as: • If uθ rotates, it angle θ changes with time, the time derivative of uθ is found by applying the standard differentiation rules on the expression of uθ above uθ sinθj θ cosθi

4. Derivative of a Rotating Unit Vector • Noting that • the time derivative of the vector uθ is uθ sinθj θ cosθi

5. Derivative of a Rotating Unit Vector • The derivative of a unit vector whose angle with the x-axis is θ, (θ changes in time), is a vector whose angle with the x-axis is θ+π/2 and whose magnitude is dθ/dt. • If we define a vector ωas a vector in the k direction of magnitude ω = dθ/dt, then duθ/dt uθ θ + π/2 θ

6. Derivative of a Rotating Vector • A vector rθ = r uθ, is a general vector of magnitude r pointing in the θ direction. The time derivative of rθ is found by applying the normal differentiation rules on the expression for rθ duθ/dt uθ θ + π/2 θ

7. Derivative of a Rotating Vector • Given a vector rθ = ruθwhich rotates relative to the reference coordinates, the time derivative of rθ has two components; a component in the direction of uθ and a component normal to uθ in the direction of uθ+π/2 • The magnitude of the component of drθ /dt in the direction of uθ is equal to dr/dt; that is the time derivative of the length of rθ. • The magnitude of the component of drθ /dt in the direction of direction of uθ+π/2is equal to ωr (dr/dt) uθ ω x rθ θ + π/2 rθ θ

8. VELOCITY ANALYSIS OF FOUR BAR MECHANISMS

9. Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain • The loop closure equation of a 4-bar kinematic chain is written as • When all the links in the chain are of constant lengths, the equation above reduces to

10. Derivative of the Loop Closure Equation for a Four Bar Mechanism • For a four bar mechanism with link 1 fixed we have dθ1/dt =ω1 = 0. • The vector equation above contains two scalar equations and can be solved for two unknowns. Knowledge of dθ2/dtallows the calculation of dθ3/dtand dθ4/dt .

11. Derivative of the Loop Closure Equation for a Four Bar Mechanism • Eliminate dθ3/dtby carrying out a dot product with uθ3 on both sides of the equation • Alternatively, eliminate dθ4/dtby carrying out a dot product with uθ4 on both sides of the equation

12. Angular Velocity Ratio and Mechanical Advantage • The angular velocity ratio mV is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism with link 2 as the input and link 4 as the output this is expressed as • The efficiency of a four bar linkage is defined as the output power over the input power, • Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have

13. Angular Velocity Ratio and Mechanical Advantage • The mechanical advantage is defined as the ratio between the output force to the input force

14. VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS

15. Velocity Analysis of a Slider-Crank Mechanism • Design parameters: r2, r3, r4, θ1. • Position analysis parameters: • r1, θ2, θ3 • Velocity analysis parameter. • Find dr1/dt, dθ2/dt , dθ3/dt • To eliminate ω3 dot product both sides by uθ3 r3 • To eliminate dr1/dtdot product both sides by u(θ1+π/2) r4 rp r2 r1

16. Velocity Analysis of an Inverted Slider-Crank Mechanism • Given r1 , r2, θ1, θ2 , ω2 • Position analysis: Find r3, θ3 • Velocity analysis: Find dr3/dt, dθ3/dt • To eliminate ω3 and find dr1/dtdot product both sides by uθ3 • To eliminate dr1/dt and find ω3 dot product both sides by u(θ3+π/2) θ3 r2 r3 r1 θ2 θ1

17. Velocity Analysis of an Inverted Slider-Crank Mechanism • Given r1 , r2, r4 , θ1, θ2 , ω2 • Position analysis: Find r3, θ3 , θ4 • Velocity analysis: Find dr3/dt, dθ3/dt , dθ4/dt r3 r4 r2 r1

18. Velocity Analysis of an Inverted Slider-Crank Mechanism r3 r4 r2 r1

19. Example r5 s b a r4 r2 r1 r5 b a s r4 r2 r1

20. Example r5 s b a r4 r2 r1 r5 b a s r4 r2 r1

21. HW#36-17, 6-18 (b+c), 6-21 (b), 6-44.

22. METHOD OF INSTANT CENTERS

23. Relative Velocities Between Two Points on a Rigid Body • Given any two points A and B that lie on a rigid body, let the line AB be a line passing through A and B, then the components of the velocity of A and the velocity of B on the line AB must be equal. • The velocity of point B relative to A must be normal to the line AB A A B B

24. Instant Center Relative to the Ground A • If line AA’ is drawn normal to the direction of vA, then the velocity of any point on the rigid body that falls on line AC must be perpendicular to the line AA’ • In a similar manner, if line BB’ is drawn perpendicular to the direction of vB, then the velocity of any point on the rigid body that falls on this line must be perpendicular to the line BB’ • If point I is the intersection of lines AA’ and BB’, then the velocity of point I must be perpendicular to both AA’ and BB’. This can only happen when the velocity of point I is zero. I B’ B A’

25. Instant Center Relative to the Ground C A • Point I determined as before is called the instant center of zero velocity of the rigid body with respect to the ground. The direction of the velocity of point A is normal to the line IA. The direction of the velocity of line B is normal to the line IB. • The direction of the velocity of any other point C on the same body must be normal to the line IC. • The magnitude of the velocity of any point on the link is proportional to its distance from the point I. I B’ B A’

26. Velocity of a Point on a Link byInstant Center Method • Consider two points A and B on a rigid link. Let vA and vBbe the velocities of points A and B at a given instant. If vA is known in magnitude and direction and vBin direction only, then the magnitude of vBmay be determined by the instantaneous centre method. • Draw AI and BI perpendiculars to the directions vA and vBrespectively. Let these lines intersect at I, which is known as instantaneous centre or virtual centre of the link. The complete rigid link rotates about the centre I at the given instant. The relations shown may be used to determine the magnitude of the velocity of point B.

27. Instant Center Between Two Rigid Bodies A • In the foregoing discussion, the velocities of points A and B were assumed to be reference to a coordinate system attached to the ground. The resulting center is called an instant center relative to the ground. • The velocities of points A and B could as well be taken relative to coordinate system attached to another body. The resulting point in this case will have a zero velocity with respect to that body. In other words, point I will have the same velocity for both bodies. I B’ B A’

28. Instant Center Between Two Rigid Bodies A • A instant center between two bodies is a: • A point on both bodies • A point at which the two bodies have no relative velocity. • A point about which one body may be considered to rotate around the other body at a given instant. I B’ B A’

29. Instant Center Between Two Rigid Bodies A • When two links are connected to one another by a revolute joint, the center of the connecting joint is an instant center for the two links. • When two links are not connected, an instant center between the two links will also exist and can be determined if the velocities of both links is known. A’

30. Instant Center Between Two Rigid Bodies • The number of instant centers in a constrained kinematic chain is equal to the number of possible combinations of two links. • The number of pairs of links or the number of instantaneous centers is the number of combinations of L links taken two at a time. Mathematically, number of instant centers is,

31. Types of Instantaneous Centers • The instant centers for a mechanism are of the following three types : • Primary (Permanent) instant centers. They can be fixed or moving • Secondary Instant centers (Not permanent)

32. Instant Centers of a Four Bar Mechanism D • Consider a four bar mechanism ABCD as shown. The number of instant centers (N) in a four bar mechanism is given by • The instant centers I12 and I14 are fixed instant centers as they remain in the same place for all configurations of the mechanism. The instant centers I23 and I34 are permanent instant centers as they move when the mechanism moves, but the joints are of permanent nature. The instantaneous centresI13 and I24 are neither fixed nor permanent as they vary with the configuration of the mechanism. 3 2 C A 4 B 1 (Ground)

33. Location of Instant Centers • When the two links are connected by a pin joint (or pivot joint), the instant center lies on the center of the pin as. Such an instant center is of permanent nature. If one of the links is fixed, the instant center will be of fixed type. • When the two links have a pure rolling contact (without slipping), the instantaneous centre lies on their point of contact, as this point will have the same velocity on both links. • When the two links have a sliding contact, the instant center lies at the center of curvature of the path of contact. This points lies at the common normal at the point of contact.

34. Kennedy (or Three Centers in Line) Theorem • Kennedy’s theorem states that any three bodies in plane motion will have exactly three instant centers, and the three centers will lie on the same straight line. • Note that this rule does not require that the three bodies be connected in any way. We can use this rule, in conjunction with the linear graph, to find the remaining lCs which are not obvious from inspection.

35. Example: Instant Centers of a Four Bar Mechanism • Draw a circle with all links numbered around the circumference • Locate as many ICs as possible by inspection. All pin joints will be permanent ICs . Connect the links numbered on the circle to create a linear graph and record those found. • Identify a link combination for which the IC has not been found, and draw a dotted line connecting those two link numbers. Identify two triangles on the graph which each contains the dotted line and whose two other sides are solid lines representing the ICs the already found. Use Kennedy’s theorem to locate the needed IC.

36. Example: Instant Centers of a Four Bar Mechanism

37. Example: Instant Centers of a Slider-Crank Mechanism

38. Example: Instant Centers of a Slider-Crank Mechanism

39. Example: Instant Centers of a Cam and Follower: Common Normal Method

40. Example: Instant Centers of a Cam and Follower: Effective Link Method

41. Velocity Analysis with Instant Centers • Once the instant centers (ICs) of a linkage with respect to the ground link have been found, they can be used for a very rapid graphic analysis for that link. • Note that some of the ICs may be very far removed from the links. For example, if links 2 and 4 are nearly parallel, their extended lines will intersect at a point far away and not be practically available for velocity analysis.

42. Velocity Analysis with Instant Centers • From the definition of the instant center, both links sharing the instant center will have identical velocity at that point. • Instant center I13 involves the coupler (link 3) which is in general plane motion, and the ground link which is stationary. All points on the ground link have zero velocity in the global coordinate system, which is embedded in link 1. Therefore, I13 must have zero velocity at this instant, and it can be considered to be an instantaneous "fixed pivot" about which link 3 is in pure rotation with respect to link 1. • A moment later, I13 will move to a new location and link 3 will be "pivoting" about a new instant center.

43. Velocity Analysis with Instant Centers • If ω2is known for the mechanism shown, the magnitude of the velocity of point A can be computed as vA = ω2O2AIts direction and sense can be determined by inspection. • Note that point A is also instant center and it has the same velocity as part of link 2 and as part of link 3. Since link 3 is effectively pivoting about I13 at this instant, the angular velocity ω3 can be found by ω3 = vA /AI13 . Once ω3 is known, the magnitude of vB can also be found from vB = ω3AI13 • Once vB is known, ω4 can also be found from ω4 = vB /BO4 = vB /BO4 . Finally, vC (or the velocity of any other point on the coupler) can be found from vC = ω3CI13