Automobile Class 2

8/3/2019 Automobile Class 2

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2

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 inEquilibrium then :-

(a) The resultant of all the forces acting on it is

zero.

And

(b) The resultant of all the couples and themoments 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 W L 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 mmfrom rear wheel axis. Determine the load shared by front and

rear wheels.

1

* 983.943*0.575

459.971.230

W b

R NL= = =

20.575

*[1 ] 983.943*[1 ] 523.9731.230

b R 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 ofCG, the above two wheeler was kept on an inclined

surface at 15 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

Taking moment about A,

2*cos * *sin * * *cos R L W H W a + =

2*sin * * *cos *cos *W H W a R L =

* si nW

2[ * ]* cot

R H a L

W

=

2*cot * *cot

R H a L

W =

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 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 H =

2*

*cos *cos * *sinW a W

R HL L

=

Now Rewriting Equation no. (2)

Thus limiting value of may be obtained as:

2 ** cos 0 * cos * * sinW a WR HL L

= =

( * cos * sin ) 0W

a HL

= =

0W

L

*cos *sina H =

0.655

tan 2.6440.2477L

a

H = = =

lim] 69.28 =

[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 10cms 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 NW= 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*1.208 0.65371667.7

R La mW

= = =

1.208 0.6537 0.5543b L a m = = =

W

A

B

R2

R1

LCOS

x

aCOSh=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 r =

2* *cos * *cos *( )*sin R L W a W h r =

2( ) * co t * * co t

Rh r a L

W =

Dividing by W*sin

2

( * ) * cot

R

H r a LW = +

863.280.225 (0.6537 *1.208) *cot

1667.7= +

4.7585 =

cot 4.7585 12.0385 =

0.566H m =

Now we know that Limiting value of slope,

0.6537

tan 0.5666La

H ==

tan 1.153L =

49.08L =

0.100

sin 0.082781.208

h

L = = =Now


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Weight Distribution in aWeight 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 3 R 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, iftotal 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

2 2 [ ]

W a x

R l c=

24031.91 0.900 0.160

[ ] 449.024 45.7722 1.524 0.435

R N kg= = =


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3

2

[ ]W x a

Rc l

=

4031.91 0.160 0.900[ ]

2 0.435 1.524= +

1932.025 196.944 N kg= =

1 2 3 R W R R=

4031.91 449.024 1932.025

1650.86 168.28 N 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 xRl 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

W

A

B

R2,3

R1

x

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 r =

* *cos *( )*sinW a W H r =

Dividing by sinW

2, 3( ) * cot * * cot

R H r a L

W

=

2, 3[ * ]* cot

R H r a L

W = +

Now0.11

sin 0.068751.6

h

L= = =

3.942 =

2305.350.125 [0.9035 *1.6]*14.511

4169.25H = +

0.3977H m =

0.9035tan 2.272

0.3977

a

H = = =

66.24L =


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Weight Distribution in a Four-wheeled

Vehicle:

L

a b

CG

W

R3,4R1,2

R2

R1 R3

R4B

c

d

x

B

d c

R1,3 R2,4

y = B/2 -x

H

y B/2 +

xW

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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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 W L

+ =

1 2( ) *b

R R W L

+ =

2 4( ) *d

R R W

B

+ =

1 3( ) *c

R R W B

+ =

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 cR WL B

=

2 * *b d

R WL B

=

3 * *a c

R WL B

=

4 * *

a d

R W L B=

1 2460.7R N=

4 2345.8N=

2 1804.5R N=

3 3198.9R N=


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Example:2

A four wheeled vehicle has a wheel base of 2.36m 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 mB=1.32 m

W=790 kg = 7749.9 N

c = 0.735

1 2( ) 0.48* 3719.952 R R W N + = =

2

By= +

0.5852

Bd y m= =

3 4 1 2( ) ( ) 4029.948 R R W R R N + = + =


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=

2071.337 211.145 N kgf = =

2 * *b dR WL B

=

1.1328 0.5857749.9* *

2.36 1.32=

1648.615N=

168.055kgf=


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3 * *

a c

R W L B=

1.2272 0.7357749.9* *

2.36 1.32=

2243.948N=

228.741kgf=

4 * *a dR WL B

=

1.2272 0.5857749.9* *

2.36 1.32=

1785.9997N=

182.059kgf=


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1st check: W=R =790 kgf

2nd check:

3rd check:

------------------------------------------------------------

3 4 *a

R R W L

+ =

410.8 410.8=

2 4( ) *d

R R W B

+ =

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

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

a R R W

L

b R R W

L

d R R W

B

c R R W

B

=

+ =

+ =

=

=

=

=

+ =

+ =

+ =

+ =

+ =

+ =


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cot 14.511

0.9035tan 2.272

0.3977

66.24L

a

H

=

= = =

=

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

a b

L

W

CG

A B

R2R1


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W

A

B

R2

R1

LCOS

x

aCOSh=0.10 m

R2

R1

W


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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 in 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

R H r a L

W

= +

= = =

=

=

=

= +

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