automobile(mechanics of road vehicles)
TRANSCRIPT
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AUTOMOBILE ENGINEERINGAUTOMOBILE ENGINEERING
DYNAMICS OF ROAD VEHICLE
-By Dr. S A Channiwala
Example 1 :
A two wheeler has a mass of 265 kg. It has awheelbase of 1.24 m & ratio of x/y=1.2. Ifmanufacture claims the vehicle to achieve theacceleration of 3 m/s2 & height of CG is 0.7 m.Determine:
i. Effort required on the rear wheel
ii. Reactions R1&R2
iii. Load transfer
iv. Time required to achieve a speed of 80 km/hr
v. If the tyre of radius is 0.325 m, what is the theoretical power required?
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x y
L
mg
CG
H
Fi=ma
a=3 m/s2
A B
R2R1
P
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VEHICLE MECHANICS
1.1 Laws of Equilibrium : If a body is in
Equilibrium then :-
(a) The resultant of all the forces acting on it is
zero.
And
(b) The resultant of all the couples and the
moments of all the forces taken about
any axis whatsoever is zero.
In the most general case, the laws of
equilibrium will enable us to write down
six equations,
Three by equating to zero the force along
the three axis
Three by taking moment about the
three axis
But the majority of problems do not require
the formulation of all six equation.
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1.2 Weight/Force Distribution in Two wheeler :-
a b
L
R2R1
CGFront Rear
W
x
x
∑ Fy = 0, R1 + R2 = W ....... (i)
Taking moment about x-x,
……………(ii)
……………(iii)
Substituting (iii) in (i),
……………(iv)2*
*[1 ]W b b
R W WL L
= − = −
1 * *R L W b=
1*W b
RL
∴ =
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Example 2 :
A two wheeler has an unladden weight of 100.3 kg and a
wheel base of 1230mm. If the CG is displaced by 575 mm
from rear wheel axis. Determine the load shared by front and
rear wheels.
1* 983.943*0.575
459.971.230
W bR N
L= = =
20.575
*[1 ] 983.943*[1 ] 523.9731.230
bR W N
L= − = − =
Soln. :
W= 100.3 * 9.81 = 983.943 N
L = 1.230 m, b = 0.575 m
we know that:
Example 3 :
During experimental measurement of the height of
CG, the above two wheeler was kept on an inclined
surface at 15 ⁰ and the load distribution at front &
rear wheels was found to be 52.3 kg & 48 kg
respectively. Determine the height of CG of this
vehicle.
The mean tyre radius is 0.225 m. Also state the
limiting value of slope on which the vehicle will
be stable.
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R2
R1
W
θ
θ
Taking moment about ‘A’,
2*cos * *sin * * *cosR L W H W aθ θ θ+ =
2*sin * * *cos *cos *W H W a R Lθ θ θ∴ = −
* s inW θ
2[ * ]* cot
RH a L
Wθ= −
2*cot * *cot
RH a L
Wθ θ= −
48*9.81[0.655 *1.23]* cot15 0.2477
983.943H m∴ = − =
………….. (1)
…………….. (2)
……………….. (3)
Ans.
Dividing by
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Now R2COS θ is a reaction at rear wheel.
When this reaction tends to zero, the
vehicle becomes unstable & may over-
turn about pt. A.
2*cos * * *cos *sin *R L W a W Hθ θ θ= −
2*
*cos *cos * *sinW a W
R HL L
θ θ θ= −
Now Rewriting Equation no. (2)
Thus limiting value of ‘θ’ may be obtained as:
2*
* cos 0 * cos * * sinW a W
R HL L
θ θ θ= = −
( * cos * sin ) 0W
a HL
θ θ= − =
0W
L≠
*cos *sina Hθ θ=
0.655tan 2.644
0.2477L
a
Hθ = = =
lim] 69.28θ = °
∴
∴
∴
∴
[Note: In practice, this is
limited by road adhesion]
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• These equations very clearly signifies two
facts:
(i) For better stability of vehicle, the CG
disposition must as further as possible
from front axle
(ii) CG must be as lower as possible.
Example 4 :
A bike driver takes his bike to weigh bridge
for determination of its correct CG, he
intends to take part in a bike race, when
front wheel alone enters the weigh bridge,
the load indicated is 78 kg when both
wheels are taken to weighbridge, the
indicated load is 170 kg.
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Now he takes out front wheel from the bridge
on level road & raises the rear wheel by 10
cms on weighbridge. The load indicated is 88
kg. If wheel base is 1.208 m & tyre radius is
225 mm.
Determine
(i) CG Position.
(ii) Limiting value of slope on which vehicle
will be stable.
a b
L
W
CG
A B
R2R1
Case : 1
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Data:Case : 1
R1 = 78 kg = 765.18 N
W= 170 kg = 1667.7 N
∑ Fy = 0; W = R1 + R2
R2 = 902.52 NNow taking moments about ‘A’,
R2*L = W*a
∴
2* 902.52*1.2080.6537
1667.7
R La m
W∴ = = =
1.208 0.6537 0.5543b L a m∴ = − = − =
WW
A
B
R2
R1
θ
θ
LCOSθ
x
aCOSθh=0.10 m
Case : 2
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Selecting AB as a moment arm and taking moment
about point ‘A’,
2* *cos *{ *cos ( )*sin }R L W a h rθ θ θ= − −
2* *cos * *cos *( )*sinR L W a W h rθ θ θ= − −
2( ) * co t * * co t
Rh r a L
Wθ θ− = −
Dividing by W*sinθ
2( * ) * cot
RH r a L
Wθ∴ = + −
863.280.225 (0.6537 *1.208) *cot
1667.7θ= + −
4.7585θ∴ =
cot 4.7585 12.0385∴ =
0.566H m∴ =
Now we know that Limiting value of slope,
0.6537tan
0.5666L
a
Hθ ==
tan 1.153Lθ =
49.08Lθ °=
0.100sin 0.08278
1.208
h
Lθ = = =Now
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Weight Distribution in a Weight Distribution in a
ThreeThree--Wheeled VehicleWheeled Vehicle
a b
L
W
CG
A B
R2R3R1
C
C
R3
R2
H
R1
x
x
z
z
y yB=2c
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∑ Fy = 0; W = R1 + R2 + R3 ……………… (i)
Now taking moments about x-x,
R1*L = W*b ……………… (ii)
Now taking moments about y-y,
R3*c = W*x +R2*c
R3 - R2 = ……………… (iii)
Now taking moments about z-z,
R2*L+ R3*L = W*a
R2 + R3 = ……………… (iv) *W a
L
*W x
c
1 2 3R W R R= − −
2 [ ]2
W a xR
l c= −
(iii) + (iv)
…………………. (v)
(iv) - (iii)
………………… (vi)
……………… (vii)
3 [ ]2
W x aR
c l= −
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Example 5:
A three-wheeler has a wheelbase of
1524 mm and its CG is 900 mm behind
front wheel axle and 160 mm from the
longitudinal axis of vehicle on near
side. The track of rear wheel is 870
mm. Determine the wheel loads, if
total weight of the vehicle is 411 kg.
Sol.Given Data:
L=1.524 m
a=0.900 m
X=0.160 m, c= = 0.435 m
W=411 x9.81=4031.91 N
we know that,
0.870
2
22
[ ]W a x
Rl c
= −
24031.91 0.900 0.160
[ ] 449.024 45.7722 1.524 0.435
R N kg= − = =
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[ ]W x a
Rc l
= −
4031.91 0.160 0.900[ ]
2 0.435 1.524= +
1932.025 196.944N kg= =
1 2 3R W R R= − −
4031.91 449.024 1932.025
1650.86 168.28N kg
= − −
= =
Example 6:
• It is required to determine the CG of a threewheeler experimentally when one of the nearsiderear wheel is taken to weigh bridge, the loadrecorded is 160 kg while when opposite rearwheel is taken to weigh bridge the load is foundto be 80 kg. When both the rear wheels weretaken on weighbridge and raised by 110mm theload recorded was 235 kg.
Determine (i) CG position
(ii) limiting slope for stability.
The wheelbase and rear wheel track for this vehicleare 1.6m and 1m respectively while total weightof the vehicle is 425kg. The wheel diameter is 250mm.
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Solution:
• Given data:
= 160 kg = 1569.6 N
D = 0.250 m
W = 425 kg
L = 1.6 m
C =1/2 = 0.5 m
Raised condition: [R2 + R3]=235 kg=2305.35 N
3R
• For a three wheeler we know that :
22
[ ]W a x
Rl c
= −
4169.25784.8 [ ]
2 1.6 0.5
a x= −
4169.251569.6 [ ]
2 1.6 0.5
a x= +
32
[ ]W a x
Rl c
= +
784.8 = 1302.891*a – 4169.25*x …………. (1)
And
1569.6=1302.891*a + 4169.25*x …………. (3)
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Adding (1) + (3)
2354.4 = 2605.782*a
a=0.9035 m
b= L-a = 1.6 - 0.9035 = 0.6965 m
(3) - (1)
784.8 = 8338.5 * x
x=0.0941 m
WW
A
B
R2,3
R1
θ
θx
h
h
r
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Selecting A-B as moment arm & taking moments
about ‘A’,
2, 3* *cos *{ *cos ( )*sin }R L W a H rθ θ θ= − −
* *cos *( )*sinW a W H rθ θ= − −
Dividing by sinW θ
2, 3( ) *cot * * cot
RH r a L
Wθ θ− = −
2, 3[ * ]* cot
RH r a L
Wθ∴ = + −
Now0.11
sin 0.068751.6
h
Lθ = = =
3.942θ∴ =
2305.350.125 [0.9035 *1.6]*14.511
4169.25H∴ = + −
0.3977H m∴ =
0.9035tan 2.272
0.3977
a
Hθ∴ = = =
66.24Lθ∴ = °
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Weight Distribution in a Four-wheeled
Vehicle:
LL
aa bb
CGCG
WW
R3,4R1,2
R2
R1 R3
R4
B c
d
x
B
dd cc
R1,3 R2,4
y = B/2 -x
H
y B/2 +
x WW
• Data Given:
W=1000 kg
L=2.3 m
B=1.3 m
a=1.3 m
b=1.0 m
x= 0.1 m
c=0.75 m
d=0.55 m
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Taking moment about ‘A’,
3 4( ) * *R R L W a+ =
1 3( )* *R R B W c+ =
2 4( )* *R R B W d+ =
Taking moment about ‘C’,
Taking moment about ‘D’,
1 2( )* *R R L W b+ =
Taking moment about ‘B’,
3 4( ) *a
R R WL
+ =
1 2( ) *b
R R WL
+ =
2 4( ) *d
R R WB
+ =
1 3( ) *c
R R WB
+ =
From (i) to (iv),
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From method of Superimposition:
• From (vi) to (viii)
• From (vi) to (vii)
• From (v) to (vii)
• From (vi) to (vii)
1 * *b c
R WL B
=
2 * *b d
R WL B
=
3 * *a c
R WL B
=
4 * *a d
R WL B
=
1 2460.7R N=
4 2345.8R N=
2 1804.5R N=
3 3198.9R N=
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Example:2
A four wheeled vehicle has a wheel base of 2.36
m and mean track of 1.32 m. It is designed in
such a manner that front axle shares 48% of
vehicle load. The CG of this vehicle is displaced
by 0.075 m from longitudinal axis away from
driver side. Determine load distribution if
vehicle load is 790 kg under unladden
condition.
[Fiet-UNO]
L
a b
CG
W
R3,4R1,2
R2
R1 R3
R4
B = 1.32mC = 0.735m
d = 0.585m0.075m
B
d cR1,3 R2,4
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• Given data:
L=2.36 m
B=1.32 m
W=790 kg = 7749.9 N
c = 0.735
1 2( ) 0.48* 3719.952R R W N+ = =
2
By= +
0.5852
Bd y m= − =
3 4 1 2( ) ( ) 4029.948R R W R R N+ = − + =
Taking moment about ‘A’,
3 4( ) * *R R L W a+ =
3 4( )* 4029.948*2.361.2272
7749.9
R R La m
W
+∴ = = =
2.36 1.2272 1.1328b L a m= − = − =
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Now we know that:
1 * *b c
R WL B
=
1.1328 0.7357749.9* *
2.36 1.32=
2071.337 211.145N kgf= =
2 * *b d
R WL B
=
1.1328 0.5857749.9* *
2.36 1.32=
1648.615N=
168.055kgf=
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3 * *a c
R WL B
=
1.2272 0.7357749.9* *
2.36 1.32=
2243.948N=
228.741kgf=
4 * *a d
R WL B
=
1.2272 0.5857749.9* *
2.36 1.32=
1785.9997 N=
182.059kgf=
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1st check: W=∑R =790 kgf
2nd check:
3rd check:
------------------------------------------------------------
3 4 *a
R R WL
+ =
410.8 410.8=
2 4( ) *d
R R WB
+ =
350.114 350.114=
Example 3:• A four wheeled vehicle stands on a weighbridge
such that only front wheels are on the bridge.
The load recorded is 614 kgf. The vehicle is now
taken on weighbridge & the load was found to
be 1180 kg. The vehicle then was kept on a
weighbridge in such a way that its rear wheels
were on the weighbridge raised by 10 cms.
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R2
R1
W
θ
θ
Case : 2
3 4
1 2
1
2
3
4
2 4
1 3
3 4
1 2
2 4
1 3
1 2 3
( )* *
( )* *
2460.7
1804.5
3198.9
2345.8
( )* *
( )* *
( ) *
( ) *
( ) *
( ) *
R W R R
R R L W a
R R L W b
R N
R N
R N
R N
R R B W d
R R B W c
aR R W
L
bR R W
L
dR R W
B
cR R W
B
= − −
+ =
+ =
=
=
=
=
+ =
+ =
+ =
+ =
+ =
+ =
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cot 14.511
0.9035tan 2.272
0.3977
66.24L
a
H
θ
θ
θ
=
∴ = = =
= °
x y
L
mg
CG
H
Fi=ma
a=3 m/s2
A B
R2R1
P
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x y
L
mg
CG
H
Fi=ma
a b
L
R2R1
CGFront Rear
W
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R2
R1
W
θ
θ
a b
L
W
CG
A B
R2R1
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WW
A
B
R2
R1
θ
θ
LCOSθ
x
aCOSθh=0.10 m
R2
R1
W
θ
θ
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a b
L
W
CG
A B
R2R3R1
C
C
R3
R2
H
R1
x
x
z
z
y yB=2c
8 6 3 .2 80 .2 2 5 ( 0 .6 5 3 7 * 1 .2 0 8 ) * c o t
1 6 6 7 .7
0 .1 0 0s i n 0 .0 8 2 7 8
1 .2 0 8
4 .7 5 8 5
c o t 4 .7 5 8 5 1 2 .0 3 8 5
0 .5 6 6
2( * ) * c o t
h
L
H m
RH r a L
W
θ
θ
θ
θ
= + −
= = =
∴ =
∴ =
∴ =
∴ = + −