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Vehicle Dynamics Dr. Tahsean Ali Albadry Najaf Technical College 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 Crank Gear Mechanism 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 • 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. 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. 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 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, Crank mechanism notations β φ • R = crank radius. Calculation of piston travel( xp) L+R • XP = ( R +L) - [R cos φ+L cos β) • XP = R(1 - cos φ ) +L(1 - cos β) • sin β = ℎ sin φ = • ∴ L sin β = R sin φ • cos β = 1 − ( sin ℎ ∴ sin β = sin φ φ)2 • XP = R(1 - cos φ ) +L(1 - 1 − (1±2)½ =1± − 2 4 ± 2×4 ( sin φ)2 ) 6 … 1 ×±3… 2×4×6 XP = R(1 - cos φ ) +L(1 -(1 − ( 1 ) ( sin 2 But 2 sin2φ = 1-cos 2φ OR sin2φ = (½)(1-cos 2φ) ∴ XP = R(1 - cos φ ) 2 +L(22 (sin φ)2 ) XP = R[(1 - cos φ ) +( (½)(1−cos 2 2φ) XP = R[(1 - cos φ ) + λ (1−cos 4 2φ) ] )] φ)2 ) Piston Velocity Equation XP = R[(1 - cos φ ) + Vp = = × Vp =R [ sin φ + sin 2 λ (1−cos 4 2φ) ] 2φ ] When φ =90 Vp= R 2 The harmonic of the second order R sin2φ which take in account the finite length of the connecting rod the shifts maximum piston velocity Vp max2 towards TDC with the given accuracy Vp max = R [ 1+ ] and 2 o φ Vp max =90- 57.3 The Piston Mean Velocity • The mean velocity of the piston has a practical importance among the parameters characterizing engine design Vp.m =30 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. Vp.max 2 = (1+ ) ≅ 1.63 2 Vp.m 2 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 aΙ and the second order aΙΙ : aΙ= R2 [ cosφ] aΙΙ =R2 [cos2φ ] The acceleration of the piston reaches its maximum absolute value at TDC i.e. at φ =0 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) The parameters of offset crank gear offset crank gear piston travel:XP = R[(1 - cosφ ) + λ (1 4 – cos2φ ) - k λ sinφ] Piston velocity:Vp =R [ sin φ + sin 2 2φ - k λsφ ] Piston acceleration:ap =R2 [ cosφ + cos2φ + k λsinφ ] The travel, velocity, and acceleration diagrams Xp Ι XpΙΙ Piston Travel diagram vp Ι ×××× × ×× ×× × × p × × × × × × ××× ×× × ×××××× × ×× × ×× × × × × × × ×× ×× ××× ××××× × ×× × × × × × × × × ×× ×× ××× v ΙΙ Piston Velocity Diagram ×× ×× × × × ap Ι apΙΙ × ×× × × × × × × × × × × ×× ××× × ×× ××× × × × × ×× ×× × × × × ××× ×××××× ××× × ×× × × × × × × × × ×× ×× ××× Piston Acceleration Diagram 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 (take 45o 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. 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) GAS PRESSURE FORCES PRESSURE • INDICATOR DIAGRAM The plot of the engine cycle on P-V coordinates is often called indicator diagram TDC BDC VOLUME • 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. W = = F = Force P = Pressure in combustion chamber. Ap = Area against which the pressure acts. x = Distance piston travelled. = = φ ∴ = (φ) • The gas pressure force is a function of a distance (piston travel) • = /2 = × cylinder Fg piston = Fg cylinder Fg piston Fg’ 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 ) 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. REDUCED SYSTEM Reciprocating parts • - Reciprocating parts include the following parts:Piston Rings Piston pin mp - The piston mass to be lumped on the axis of piston pin axis it designated by (mP) 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). REDUCING THE CRANK SHAFT mcp web R ρ +cg mcr R • The mass of the crank web mcw have 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):= 2 mcw ρ ω mcwR ω2 ∴ mcwR = mcw The reduced mass of crank is to be :mcr = + 2mcwR = + 2 mcw 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) CALCULATING THE MASSES OF UNSYMETRICAL BODY my = m + mz = m + z mz cg y my REDUCING THE CONNECTING ROD MASS Lrodpp Lrodcp Lrod mrodpp Cg mrod m mrodcp Cont. • × = × • = × • × = × • = × The conditions of reduced system • There are three conditions observed to obtain dynamically equivalent system:1- A constant total mass: + = 2- A constant position of center of gravity of the System:- × − × = 0 3- A constant moment of inertia with respect to the center of gravity:I = × 2 + × 2 = ( × 2 ) = × × + ( × 2 ) A mi=mp+mrodpp β ω ϕ B mR=mcr+mrodcp • 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 = + 2 In the above REDUCED SYSTEM there are only two forces 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:Fi I = - mi R ω2 (cosϕ ) = Z cosϕ Fi II = - mi R ω2λ (cos2ϕ) = Z λcos2ϕ When Z = - mi R ω2 F”g Forces of gas pressure and forces of inertia of the reciprocating and rotating masses acting in a crank gear Fi Fg Y A X mi FR ω F’i ϕ F’g B mR F”g Fi Fg F’i F’g FR Fg FR 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:- 2 • FRx = - mRRω sinϕ 2 • FRy = - mRRω cosϕ FRy FR ϕ FRx Engine torque calculation • The initial force acting on the piston is:- F = Fg + Fi ↓ F can resolved into two component 1- The lateral force perpendicular to the cylinder axis. Q=F tanβ Q Piston pin mp β F K 2- The other force is directed along the connecting rod axis:K = F 1 “K” can be transferred to center of crank pin and in turn resolved into two components:The first is directed along the crank radius to the center of crank shaft cos(∅+) N = K cos(∅ + ) = F • The second is the force tangential to the crank radius circle :Ft = K sin(∅ + ) = F sin(∅+) h= R sin(∅+) Q β F ω F’ K ϕ Q’ h Q Q’ ω THE TORQUE • The forces :- Ft = Ft’ = Ft” • The couple of Ft , Ft” called the TORQUE(T) • T = Ft × = F sin(∅+) • T = F (sinφ + tanβ cosφ) R • T ≅ F (sinφ + 2sin2φ) R TILTING MOMENT Mtilt = - Q’ ×h = - F tanβ h =- F tanβ sin(∅+) = - Ft R = -T * The torque T is imparted to the driving through the wheel transmission • The tilting moment is taken by the engine supports trough the stationary parts of crank gear and balanced by the reactive moment. • The values of the calculated torque are used to plot the curves showing how these forces depend on the angle of crank shaft travel. ENGINE BALANCE Engine Balancing • An engine is said to be balanced if forces constant in magnitude and direction are transmitted to its supports during stable operation conditions. The cases of unbalance:1- The periodic change of the inertia force(Fi) 2- the variation of total torque ∑ T along the operation of multi cylinder engine. • The forces of inertia of the rotating elements of an engine crank gear are balanced so arranging the crank or counterweights to insure the following two conditions :1- The center of gravity of the reduced system of the shaft should be on the axis of rotation. That is the (( )) 1-the The sum of the moments of the centrifugal forces of inertia of the revolving masses should equal zero with respect to any point of shaft axis. That is the (( )) FR mR =mcr + mcw R r cg mB FB FB R R 2mBR (mB)R =mB MR =FR * a a FB FR r R MR =MB FB b FR MB =FB * b THE END