Mechanics of Machines Dr. Mohammad Kilani

1. Mechanics of MachinesDr. Mohammad Kilani Class 3 Position Analysis

2. TYPES OF MECHANISM ANALYSESE

3. Position Analysis • Position Analysis • Given θ12find θ23, θ34and θ14 • Find maximum and minimum values of θ14 . • Find the location of each point on the mechanism for a given value of θ12and the curve traced by the a point during motion. θ34 θ23 θ41 θ12

4. Velocity Analysis • Position Analysis • Given θ12and ω12find ω23, ω34and ω14 • Find the speed of a point in the mechanism θ 34, ω34 θ 23, ω23 θ 41, ω41 θ 12, ω12

5. Acceleration Analysis • Position Analysis • Given θ12, ω12 and α12find α23, α34and α14 • Find the acceleration of a point in the mechanism θ 34, ω34 , α34 θ 23, ω23 , α23 θ 41, ω41 , α41 θ12, ω12, α12

6. Question • Why is acceleration analysis important? θ 34, ω34 , α34 θ 23, ω23 , α23 θ 41, ω41 , α41 θ12, ω12, α12

7. GRAPHICAL POSITION ANALYSIS

8. Graphical Position Analysis • Use graphical methods, (pen, ruler, compass and protractors to solve position analysis problems. • CAD drafting methods may also be used as a convenient alternative. θ34 θ23 θ41 θ12

9. VECTOR POSITION ANALYSIS

10. Vector Position Analysis • Use vector loop closure equation to solve position analysis problems

11. Loop Closure Equation • Use vector loop closure equation to solve position analysis problems

12. Planar Four Bar Loop Closure Equation • How many scalar equations can be written from the vector loop-closure equation? • How many unknowns can be solved for

13. Planar Four Bar Loop Closure Equation • How many scalar equations can be written from the vector loop-closure equation?: Two scalar equations could be written from a vector equation • How many unknowns can be solved for: Two unknowns: In a standard problem, θ1 and θ2are given. θ3 and θ4are to be solved for. • We will focus on planar mechanisms in this course

14. Planar Four Bar Loop Closure Equation • The closure condition expresses the condition that a loop of a linkage closes on itself. • For the four-bar linkage shown the closure equation is

15. Planar Four Bar Loop Closure Equation • The closure condition expresses the condition that a loop of a linkage closes on itself. • For the four-bar linkage shown the closure equation is • The following two scalar equations are produced

16. Analytical Solution when θ2 is known • The above two equations could be solved for two unknowns. One approach is to eliminate one of the unknown angles by isolating the trigonometric function involving the angle on the left-hand side of the equation. • Squaring both sides of both equations and using the identity we obtain

17. Analytical Solution when θ2 is known • To obtain an explicit expression for θ4 in terms of θ2 and the constant angle θ1 , we combine the coefficients of cos θ4 and sin θ4 in the equation above as follows: • where

18. Analytical Solution when θ2 is known • The above equation can be solved for θ4 by using the following half angle identities • After substitution and simplification, we get • where

19. Analytical Solution when θ2 is known • Solving for t we obtain

20. Analytical Solution when θ2 is known Given the values of θ1 and θ2 for a four bar mechanism of known r1, r2, r3 and r4 the output angle θ4 is calculated as: The ± sign identifies the two possible assembly modes of the linkage

21. Analytical Solution when θ2 is known Given the values of θ1 and θ2 for a four bar mechanism of known r1, r2, r3 and r4 the output angle θ4 is calculated as: The ± sign identifies the two possible assembly modes of the linkage

22. Analytical Solution when θ2 is known • Note that –π/2 ≤ tan-1 (t) ≤ –π/2. Therefore, θ4 will have the range –π ≤ θ4 ≤ –π. • Unless the linkage is a Grashof type II linkage in one of the extreme positions of its motion range, there are two valid solutions for θ4. These correspond to two assembly modes or branches for the linkage.

23. Analytical Solution when θ2 is known • Because of the square root in the expression for t, it can be complex if (A2 + B2) < C2. In this case, the mechanism is a Grashof type II linkage in one of the extreme positions of its motion range and it cannot be assembled in specified values of θ1 and θ2 . • The assembly configurations would then appear as shown below

24. Analytical Solution when θ2 is known • After θ4 is known, an expression for θ3 can be obtained by solving the loop closure equation to obtain: • Note that it is essential that the sign of the numerator and denominator be maintained to determine the quadrant in which the angle θ3 lies. This can be done by using the ATAN2 function. The form of this function is: • ATAN2 (sin θ3, cosθ3)

25. Analytical Solution when θ2 is known • Once all of the angular quantities are known, it is relatively straightforward to compute the coordinates of any point on the mechanism. • In particular, the coordinates of Q,P, and R are given by

26. Analytical Solution when θ3 is known • If the coupler angle θ3 is given, and θ2 and θ3 are to be determined, graphical solution will require an iterative trial and error solution. • The analytical procedure, in contrast, follows exactly the same procedure as when θ2 is given. It starts by writing the loop closure equations with θ2 replacing θ3 as the variable to be eliminated. • The equations set are of exactly the same form except that the indices 2 and 3 are interchanged. Therefore, we can use directly the position solution derived for the case of known θ2 while interchanging the indices 2 and 3.

27. Analytical Solution when θ3 is known • When the coupler angle θ3 is given, there is an assembly-mode ambiguity similar to that occurine when θ2 is given. • It is necessary to know the appropriate mode of the linkage before the analysis is begun;. The mode is determined by the + or – sign used for the square root term when calculating t. • Once the assembly mode is determined, it remains the same for any position of the input link unless the linkage is a class III linkage, and passes through a singular (indeterminate) position .

28. Example: Position Analysis of a Four Bar Linkage • A four bar linkage with r1 = 1, r2 = 2, r3 = 3.5, r4 = 4, and θ1 = 0, find θ3 and θ4 for each of the solution branches when the driving crank is in the positions θ2 = 0, π/2, π,and --π/2.

29. Example: Position Analysis of a Four Bar Linkage • A four bar linkage with r1 = 1, r2 = 2, r3 = 3.5, r4 = 4, and θ1 = 0, find θ3 and θ4 for each of the solution branches when the driving crank is in the positions θ2 = 0, π/2, π,and --π/2.

30. POSITION ANALYSIS FOR A RIGID BODY WHEN TWO POINTS ARE KNOWN

31. Position Analysis for a Rigid Body When Two Points are Knows • Given the kinematic properties of one point on a rigid body and the angular position, angular velocity, and angular acceleration of the body, we can compute the position, velocity, and acceleration of any defined point on the rigid body. • For the rigid body shown. Assume that A and B are two points attached to an arbitrary link, say link 5, and a third point is defined relative to the line between points A and B by the angle β and the distance rC/A, which is represented as r6. Then the position of point C can be computed directly if rA and θ5 are known.

32. Position Analysis for a Rigid Body When Two Points are Knows • The position of point C is given as: • If θ5 If is known, the equation above can be used to calculate the location of point C directly. • We often know the position vectors of two points A and B on the rigid body. The value of θ5 can be calculated from the from the x and y components of the position vectors for A and B using

33. POSITION ANALYSIS FOR A SLIDER-CRANK MECHANISM

34. HW#2 (Prob. 4-6 and 4-7, with data in row (a) Table P4-1).(4-10, 4-12, 4-18(f))

35. Position Analysis for a Slider-Crank Mechanism • Next to the four-bar linkage, the slider-crank is probably the most commonly used mechanism. • It appears in all internal combustion engines and in numerous industrial and household devices.

36. Position Analysis for a Slider-Crank Mechanism • To develop the closure equations, locate vectors r2 and r3 as was done in the regular four-bar linkage. • One of the other two vectors is taken in the direction of the slider velocity and the other is taken perpendicular to the velocity direction. The loop closure equation is Offset r3 r4 rp r2 r1

37. Position Analysis for a Slider-Crank Mechanism • Writing the loop closure equation in terms of the vector angles, we obtain • Two scalar equations are produced. The equations can be solved for two unknowns. r3 r4 rp r2 r1

38. Position Analysis for a Slider-Crank Mechanism • Unlike the four-bar linkage loop closure equations where all link lengths are known, the piston displacement r1 is an unknown in the slider-crank equation. The constraint resulting from a known r1 is replaced by the constraint θ4 = θ1 + π/2. • The following problem statements are possible • Crank angle θ2 given , find θ3 and r1 • Piston displacement r1 given, find θ2 and θ3 • Coupler angle θ3 given, find θ2 and r1. r3 r4 rp r2 r1

39. Analytical Solution when θ2 is known • The analytical solution procedure follows the same major steps as in the four-bar linkage case. To eliminate θ3 , first isolate it in the loop closure equations as follows: • Squaring both sides of both equations and using the identity we obtain r3 r4 rp r2 r1

40. Analytical Solution when θ2 is known • The expression gives r1 in a quadratic expression involving θ2 and the other known variables. To obtain a solution, collect together the coefficients of the different powers of r1 as follows r3 r4 rp r2 r1

41. Analytical Solution when θ2 is known • The expression for A could be simplified by noting that r3 r4 rp r2 r1

42. Analytical Solution when θ2 is known • Solving for r1 gives • The ± sign indicates two possible assembly modes for the same θ2.

43. Analytical Solution when θ2 is known • Because of the square root in the expression for r1, it becomes complex when A2 < 4B. • If this happens, the mechanism cannot be assembled in the position specified.

44. Analytical Solution when θ2 is known • Once a value for r1 is determined, the closure equations can be solved for θ3 to give • As in the case of the four-bar linkage, it is essential that the signs of the numerator and denominator in the above expression be maintained to determine the quadrant in which the angle θ3 lies. r3 r4 rp r2 r1

45. Analytical Solution when θ2 is known • Once all of the angular quantities are known, it is relatively straightforward to compute the coordinates of any point on the vector loops used in the closure equations. • In particular, the coordinates of Q and P are given by r3 r4 rp r2 r1

46. Analytical Solution when r1 is known • The analytical solution procedure follows the same major steps as in the previous case. • After eliminating θ3 from the loop closure equation, we simplify the resulting equation as follows • where r3 r4 rp r2 r1

47. Analytical Solution when r1 is known • The trigonometric half-angle identities can be used to solve the equation above for θ2 . Using these identities and simplifying gives • Solving for t gives r3 r4 rp r2 r1

48. Analytical Solution when r1 is known • The ± sign indicates two possible assembly mode. Typically, there are two solutions for θ2 and they are both valid. • Because tan-1 has a valid range of π/2 ≤ tan-1 (t) ≤ –π/2, θ2 will have the range –π ≤ θ2 ≤ –π.

49. Analytical Solution when r1 is known • Because of the square root in the equation for the variable t , it becomes complex when (A2 + B2) < C2 • If this happens, the mechanism cannot be assembled for the specified value of r1 .

50. Analytical Solution when r1 is known • Knowing θ2 , the closure equations can be solved for θ3. As in the previous cases, it is essential that the signs of the numerator and denominator be maintained to determine the quadrant in which the angle θ3 lies.