1. Vehicle Dynamics Dr. Tahsean Ali Albadry Najaf Technical College

2. References *M.Khuvakh, Motor Vehicle Engine, MIR Publications *Hamilton H. Mobie, Mechanisms and Dynamics Machinery, John Willey & Sons *P. W. Kett, Motor Vehicle Science, part2, Chapman& Hall

3. Crank Gear Mechanism

4. Definitions • Crank Gear :- Is a group of mechanical parts which convert the thermal energy into mechanical energy , it is consist of piston and its accessories , connecting rod , and crank shaft

5. Piston Travel:- is the distance which piston traveled from the TDC due to crank angle • Central crank gear :- The crank gear in which the axis of the cylinder intersects that of the crank shaft.

6. Offset Crank Gear The crank gear in which the axis of the cylinder was shifted with an offset (e) from the center of the crank shaft.

7. Harmonic Motion • When the distance, velocity, and acceleration, of any particle change it’s magnitude periodically from zero to maximum to zero, this motion called harmonic motion

8. Dynamic & Kinematic calculation • It is important to make two kind of calculation:- • 1- Dynamic Calculations:- Which used the magnitude and the nature of the change in the gas forces of gas pressure . • 2- Kinematic Calculations :- The aim of it to determine the travel, velocity, and acceleration of piston,

9. Crank mechanism notations φ β

10. Φangle of crank travel counted from the cylinder axis in the direction of clockwise crank shaft rotation, Φ =0 at TDC, Φ=180 at BDC. • β angle between the connecting rod and the cylinder axis. • ω angular velocity of crank shaft rotation. • S =2R = piston stroke. • R = crank radius. • L = connecting rod length. • λ = R/L ratio of crank radius to connecting rod length (λ = 0.25 – 0.3).

11. Calculation of piston travel( xp) L+R

12. XP R +L) - [R cosφ+L cosβ) • XP= R(1 - cos φ ) +L(1 - cos β) • sinβ = sin φ= • L sin β = R sin φsinβ =sin φ • cos β = • XP= R(1 - cos φ ) +L(1 -) (1

13. XP= R(1 - cos φ ) +L(1 -) But 2 sin2φ = 1-cos 2φ OR sin2φ = (½)(1-cos 2φ) XP = R(1 - cos φ ) +L() XP = R[(1 - cos φ ) +()] XP = R[(1 - cos φ ) +]

14. Piston Velocity Equation XP = R[(1 - cos φ ) + ] Vp = = Vp =R [ sin φ + sin2φ ] When φ =90 Vp= R The harmonic of the second order R sin2φwhich take in account the finite length of the connecting rod the shifts maximum piston velocityVp max towards TDC with the given accuracy Vp max = R [ 1+ ] and φVp max =90-57.3o

15. The Piston Mean Velocity • The mean velocity of the piston has a practical importance among the parameters characterizing engine design Vp.m= Where n is the crank shaft speed, rpm The mean velocity of the piston in motor vehicle engine is limited by conditions of adequate charging of the cylinder and reliable operation of piston group. The velocity ratio. =(1+)

16. Piston Acceleration • The piston acceleration can be obtained by taken the derivative of piston velocity • ap= = • ap=R2 [ cosφ + cos2φ ] The piston acceleration involves two parts:- the first order harmonic a and the second order a : a= R2 [ cosφ] a =R2cos2φ ] The acceleration of the piston reaches its maximum absolute value at TDC i.e. atφ =0

17. Offset Crank Gear • An offset crank gear is one in which the cylinder axis dose not intersect the crankshaft but is displaced with respect to it by the distance e. The crank gear is additionally characterized by the magnitude of the relative offset K= e/R (which usually ranges between 0.05 and 0.15)

18. The parameters of offset crank gear offset crank gear piston travel:- XP = R[(1 - cosφ ) +(1 – cos2φ ) - ksinφ] Piston velocity:- Vp=R [ sin φ + sin2φ - kφ ] Piston acceleration:- ap =R2 [ cosφ + cos2φ + kφ]

19. The travel, velocity, and acceleration diagrams

20. Xp Xp Piston Travel diagram

21. vp vp Piston Velocity Diagram

22. ap ap Piston Acceleration Diagram

23. Problems 1- Calculate the piston travel, velocity, and acceleration for an engine with R= 50 mm and lrod=170 mm using the equation and graphical representation and compare the results (take45o increment ,N= 3000 r.p.m). 2- An engine having λ =0.25 rotates at 3000 r.p.m, the connecting rod length is 140 mm. Determine the piston velocity and acceleration at a) TDC. b)When engine crank lies at 30o from TDC. c) At BDC. 3- Show how the ratio of crank radius to the connecting rod length affects the piston velocity( take the engine data in the problem (1) and three values of λ ) plot the result.

24. Force Acting on A Crank Gear • To determine the loads on engine bearing we must analyze all the forces acting a crank gear. • The forces divided into three categories:- 1- The forces of gas pressure in the cylinder (Fg). 2- The forces of inertia of the moving parts, that forces divided in turn into two forces, the inertia of reciprocating parts (Fi) and the rotating parts(FR). 3- the friction forces (Ff)

25. GAS PRESSURE FORCES • INDICATOR DIAGRAM The plot of the engine cycle on P-V coordinates is often called indicator diagram PRESSURE VOLUME TDC BDC

26. Work is the output of any heat engine. • In IC engine the work is generated by the gasses in the combustion chamber. • Work is the result of force acting through a distance. F = Force P = Pressure in combustion chamber. Ap= Area against which the pressure acts. x = Distance piston travelled.

27. The gas pressure force is a function of a distance (piston travel) cylinder Fg piston

28. Fg cylinder Fg piston Fg’

29. INERTIA FORCES • To determine the inertia forces we must estimate the masses of moving parts. • The moving parts divided into three groups:- 1- Reciprocating parts (Piston, Rings, Piston pin ) 2- Rotating parts (crank shaft) 3- Complex plan parallel motion (connecting rod group )

30. REDUCED SYSTEM • The crank gear is a complicated group of different members. • These members moved in a different shapes to perform the final function. • To determine the forces acts the crank gear we must reduces these members to two groups one of them have a linear motion, and the other is rotates around a fixed axis. • The new shape called the reduced system.

31. REDUCED SYSTEM

32. Reciprocating parts • Reciprocating parts include the following parts:- • Piston • Rings • Piston pin • The piston mass to be lumped on the axis of piston pin axis it designated by (mP) mp

33. ROTATING PARTS • The mass of crank shaft This mass reduced to the crank radius (R) and designated by (mcr). • The mass of crank pin and balance mass (web) was lumped on the axis of crank pin, and designated (mcp).

34. REDUCING THE CRANK SHAFT mcp mcr web R R +cg ρ

35. The mass of the crank web mcwhave a center of gravity lies in the point denoted (cg) which located at (ρ ) distance from the crank shaft center. This mass will reduced to the radius (R):- mcwρ The reduced mass of crank is to be :-

36. COMPLEX PLAN PARALLEL MOTION (CONNECTING ROD) • The connecting rod reduced into two masses:- 1- The mass lumped on the piston pin axis(mrodpp) 2- The mass lumped on the crank pin axis(mrodcp)

37. CALCULATING THE MASSES OF UNSYMETRICAL BODY my = m mz z mz = m cg y my

38. REDUCING THE CONNECTING ROD MASS mrodpp Lrodpp Lrod m Cg mrod Lrodcp mrodcp

39. Cont.

40. The conditions of reduced system • There are three conditions observed to obtain dynamically equivalent system:- 1- A constant total mass:-

41. 2- A constant position of center of gravity of the System:-

42. 3- A constant moment of inertia with respect to the center of gravity:- I )

43. A mi=mp+mrodpp β Lrod ω mR=mcr+mrodcp ϕ B R

44. Now the entire crank gear replaced by a system of two masses connected by a rigid weightless links :- • The first at point (A) • The second at point (B) • In V-engine

45. In the above REDUCED SYSTEM there are only two forces

46. Analyzing the forces of the crank gear 1- The Reciprocating force induced by mi :- Fi= - mi a = - mi R ω2 (cosϕ + λ cos2ϕ) it may represented as the sum of two forces of inertia:- FiI = - mi R ω2 (cosϕ) = Z cosϕ FiII = - mi R ω2λ (cos2ϕ) = Z λcos2ϕ When Z = - mi R ω2

47. F”g Y Forces of gas pressure and forces of inertia of the reciprocating and rotating masses acting in a crank gear Fi A X mi Fg FR ω mR ϕ F’i B F’g

48. F”g Fi Fg FR F’i F’g

49. Fg FR

50. 2- The centrifugal force of rotating masses of crank gear:- FR = - mR R ω2 • FR is always directed along the crank radius. • It is constant in magnitude. • It is applied at center B of crank pin. • It is rotates together with the crank and not being balanced. • It is transmitted to the engine supports through the shaft bearing and the crank case. • It is resolved into two components:-