Transcript Document

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
𝐿
sin φ =
• ∴ L sin β = R sin φ
• cos β = 1 −
𝑅
( sin
𝐿
ℎ
𝑅
𝑅
𝐿
∴ sin β = sin φ
φ)2
• XP = R(1 - cos φ ) +L(1 - 1 −
(1±𝐵2)½
=1±
−
2
𝐵
𝐵4 ±
2×4
𝑅
( sin
𝐿
φ)2 )
6 …
1 ×±3𝐵…
2×4×6
XP = R(1 - cos φ ) +L(1 -(1 − (
1 𝑅
) ( sin
2
𝐿
But 2 sin2φ = 1-cos 2φ
OR sin2φ = (½)(1-cos 2φ)
∴ XP = R(1 - cos φ )
𝑅2
+L(2𝐿2 (sin
φ)2 )
XP = R[(1 - cos φ ) +(
𝑅
(½)(1−cos
2𝐿
2φ)
XP = R[(1 - cos φ ) +
λ
(1−cos
4
2φ)
]
)]
φ)2 )
Piston Velocity Equation
XP = R[(1 - cos φ ) +
Vp =
𝑑𝑋𝑝
𝑑𝑡
=
𝑑𝑋𝑝
𝑑𝜑
×
Vp =R𝜔 [ sin φ +
𝑑𝜑
𝑑𝑡
𝜆
sin
2
λ
(1−cos
4
2φ)
]
2φ ]
When φ =90 Vp= R𝜔
𝜆
2
The harmonic of the second order R𝜔 sin2φ which take
in account the finite length of the connecting rod the
shifts maximum piston velocity Vp max2 towards TDC with
𝜆
the given accuracy Vp max = R𝜔 [ 1+ ] and
2
o
φ Vp max =90- 𝜆57.3
The Piston Mean Velocity
• The mean velocity of the piston has a
practical importance among the parameters
characterizing engine design
𝑆𝑛
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 𝜋 𝜆2
= (1+ ) ≅ 1.63
2
Vp.m 2
Piston Acceleration
• The piston acceleration can be obtained by taken the
derivative of piston velocity
• ap =
𝑑𝑣𝑝
𝑑𝑡
=
𝑑𝑣𝑝
𝑑𝜑
×
𝑑𝜑
𝑑𝑡
• ap =R𝜔2 [ cosφ + 𝜆 cos2φ ]
The piston acceleration involves two parts:- the first order
harmonic aΙ and the second order aΙΙ :
aΙ= R𝜔2 [ cosφ]
aΙΙ =R𝜔2 𝜆 [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 =R𝜔 [ sin φ +
𝜆
sin
2
2φ - k λ𝑐𝑜sφ ]
Piston acceleration:ap =R𝜔2 [ cosφ + 𝜆 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 = 𝐹𝑑𝑥 = 𝑃𝐴𝑃 𝑑𝑥
F = Force
P = Pressure in combustion chamber.
Ap = Area against which the pressure acts.
x = Distance piston travelled.
𝑑𝑣 = 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑥
𝑏𝑢𝑡 𝑑𝑣 = 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑φ
∴ 𝑑𝑥 = 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓(𝑑φ)
• The gas pressure force is a function of a distance
(piston travel)
•
𝑃𝑔 = 𝑓 𝑥
𝑁/𝑚2
𝐹𝑔 = 𝑃𝑔 × 𝐴𝑝
cylinder
Fg
piston
𝐴𝑝 = 𝑃𝑖𝑠𝑡𝑜𝑛 𝐴𝑟𝑒𝑎
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 𝑅 ω2
∴ mcwR =
𝜌
mcw
𝑅
The reduced mass of crank is to be :mcr = 𝑚𝑐𝑝 + 2mcwR = 𝑚 𝑐𝑝 +
𝜌
2 mcw
𝑅
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
𝑧
𝑦+𝑧
mz = m
𝑦
𝑦+𝑧
z
mz
cg
y
my
REDUCING THE
CONNECTING ROD MASS
Lrodpp
Lrodcp
Lrod
mrodpp
Cg mrod
m
mrodcp
Cont.
• 𝑚𝑟𝑜𝑑𝑝𝑝 × 𝑙𝑟𝑜𝑑 = 𝑚 𝑟𝑜𝑑 × 𝑙 𝑟𝑜𝑑𝑐𝑝
• 𝑚𝑟𝑜𝑑𝑝𝑝 =
𝑙𝑟𝑜𝑑𝑐𝑝
𝑚𝑟𝑜𝑑 ×
𝑙𝑟𝑜𝑑
• 𝑚𝑟𝑜𝑑𝑐𝑝 × 𝑙 𝑟𝑜𝑑 = 𝑚 𝑟𝑜𝑑 × 𝑙 𝑟𝑜𝑑𝑝𝑝
• 𝑚𝑟𝑜𝑑𝑐𝑝 = 𝑚 𝑟𝑜𝑑 ×
𝑙𝑟𝑜𝑑𝑝𝑝
𝑙𝑟𝑜𝑑
The conditions of reduced system
• There are three conditions observed to obtain
dynamically equivalent system:1- A constant total mass:𝑚𝑟𝑜𝑑𝑝𝑝 + 𝑚 𝑟𝑜𝑑𝑐𝑝 = 𝑚 𝑟𝑜𝑑
2- A constant position of center of gravity of the
System:-
𝑚𝑟𝑜𝑑𝑐𝑝 × 𝑙 𝑟𝑜𝑑𝑐𝑝 − 𝑚 𝑟𝑜𝑑𝑝𝑝 × 𝑙 𝑟𝑜𝑑𝑝𝑝 = 0
3- A constant moment of inertia with
respect to the center of gravity:I 𝑟𝑒𝑑 = 𝑚𝑟𝑜𝑑𝑐𝑝 × 𝑙2𝑟𝑜𝑑𝑐𝑝 + 𝑚 𝑟𝑜𝑑𝑝𝑝 × 𝑙2𝑟𝑜𝑑𝑝𝑝
𝐼𝑟𝑒𝑑 = (𝑚𝑟𝑜𝑑 ×
𝑙𝑟𝑜𝑑𝑝𝑝
𝑙𝑟𝑜𝑑
𝑙2
𝑟𝑜𝑑𝑐𝑝)
𝐼𝑟𝑒𝑑 = 𝑙 𝑟𝑜𝑑𝑝𝑝 × 𝑙𝑟𝑜𝑑𝑐𝑝 × 𝑚𝑟𝑜𝑑
+ (𝑚𝑟𝑜𝑑 ×
𝑙𝑟𝑜𝑑𝑐𝑝 2
𝑙 𝑟𝑜𝑑𝑝𝑝)
𝑙𝑟𝑜𝑑
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)
𝑚𝑖 = 𝑚 𝑝 + 𝑚 𝑟𝑜𝑑𝑝𝑝
• The second at point (B)
𝑚 𝑅 = 𝑚 𝑐𝑟 + 𝑚 𝑟𝑜𝑑𝑐𝑝
• In V-engine
𝑚𝑅 = 𝑚𝑐𝑟 + 2 𝑚 𝑟𝑜𝑑𝑐𝑝
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
𝐶𝑂𝑆𝛽
“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(∅+𝛽)
N = K cos(∅ + 𝛽) = F 𝐶𝑂𝑆𝛽
• The second is the force tangential
to the crank radius circle :Ft = K sin(∅ + 𝛽 ) = F
sin(∅+𝛽)
𝐶𝑂𝑆𝛽
h= R
𝑠𝑖𝑛𝛽
sin(∅+𝛽)
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 × 𝑅 = F
sin(∅+𝛽)
𝐶𝑂𝑆𝛽
𝑅
• T = F (sinφ + tanβ cosφ) R
• T ≅ F (sinφ
𝜆
+ 2sin2φ)
R
TILTING MOMENT
Mtilt = - Q’ ×h = - F tanβ h =- F tanβ
sin(∅+𝛽)
𝑆𝑖𝑛𝛽
𝑅
= - 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
𝒓
(mB)R =mB
𝑹
MR =FR * a
a
FB
FR
r
R
MR =MB
FB
b
FR
MB =FB * b
THE END