Puncture & Skidding Management P14

8/12/2019 Puncture & Skidding Management P14

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DEPARTMENT OF AUTOMOBILE ENGINEERING

7TH SEMESTER

PROJECT TITLE : PUNCTURE AND SKIDDING

MANAGEMENT FOR TWO WHEELER

Under The Guidance of: Prof. A.V.TUMBAL

Project Batch No. : 14

Team Members:

ANIL S KUMBAR 2BV08AU002

MOHAN P 2BV08AU023

RAMESH B PATIL 2BV08AU065

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CONTENTS Objective

Literature survey

Specifications of vehicle used

Schematic representation of front wheel arrangementand working of front wheel arrangement.

Schematic representation of rear wheel arrangement

Components to be designed

Design calculations secondary rear axle, Rear verticalmember,Front vertical member.

Front and rear wheel arrangement models

Reference

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LITERATURE SURVEY Skidding is one of the major causes of accidents.

Arising from a lack of anticipation and knowledge of skid

prevention. This resulting in dangerously excessive speeds relative to

prevailing conditions.

The arrangement consists of extended arms attached

with small wheels on both sides of the front wheel to actas supports in the case of skidding.

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Punctureis also a major problem faced by the riders

but in contrast to skidding it does not affect life.

To push the two-wheeler weighing above 100kg to the

nearest garage is a no mere task. An arrangement

similar to the supporting wheels in a bicycle can

help us to move without much effort.

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SPECIFICATIONS OF VEHICLE USED Model: Yamaha RX 100

Power : 11.50 HP(8.4 kW) @ 7500 RPM

Max RPM: 5800

Torque: 8.60Nm @6500rpm

Gear Box: 4-Speed.

Transmission type, final drive: Chain

Dry weight: 99.0 kg

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SCHEMATIC REPRESENTATION OF

FRONT WHEEL ARRANGEMENT

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WORKING OF FRONT WHEEL

ARRANGEMENT In case of skidding when the driver applies the front

brakes to maximum potential the arrangement is

provided in the front brake lever by which hook is

triggered because of that side wheels will come down to

certain extent so that it prevents skidding.

In case of puncture the driver has to lift the front wheelabove the ground and the circular pipe is lowered from its

usual position and the hook arrangement is fixed in the

lowest position which is meant for puncture.

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SCHEMATIC REPRESENTATION OF

REAR WHEEL ARRANGEMENT

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CALCULATIONS FOR SECONDARY

REARAXLE In order to design the shaft, the forces due to the weight of the vehicle,

driver , bearings, sprockets and various other components is considered

along with factor of safety.

Torque, T = 9550*103*(P/n2)

T= 9550*103*(8.4/2000)

T= 41594.076 N-mm

due to the presence of Sprocket, T= Ft*r

41594.076 = Ft*(56/2)

Ft=1485.50 N

Ref: Design Data Hand Book by Mahadevan

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Free body diagram of secondary rear axle

Taking moments about A,

RE*500 = (1000*456.5) + (1485.50*150) + (1000*43.5)RE= 1445.65N

Taking moments about E,

RA*500 = (1000*456.5) + (1485.50*350) + (1000*43.5)

RA= 2039.85 N

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1000N 1485.50N 1000N

43.5 106.5 306.5 43.5

500

A B C D E


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Bending moments:

Bending moment at A = 0

Bending moment at B = (2039.85*43.5) = 88733.475 N-mm

Bending moment at C = (2039.85*150)(1000*106.5) = 199477.5 N-mm

Bending moment at D = (1445.65*43.5) =62885.775N-mm

Bending moment at E = 0

Hence, Max. bending moment = Mb= 199477.5 N-mm

Mb= 199477.5 N-mm

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The shaft is designed on maximum shear stress theory

Therefore, Diameter of the shaft,

D = {(16/(*ed))[(KbMb)2+ (KtT)

2]1/2}1/3

Therefore, Diameter of the shaft, For suddenly applied loads and minor shocks, Kb= 1.5 and Kt= 1.0

Yield strength of Mild Steel = 320MPa

y = y/2 = 320/2 = 160MPa

ed

= y/FOS =

y/2 = 160/2 = 80MPa

D = {(16/(*80))[(1.5* 199477.5)2+ (1*41594.076)2]1/2}1/3

Hence the diameter of the shaft is found to be D=26.76mm

According to standard dimensions, D= 30mm

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CALCULATIONS FOR REAR

VERTICAL MEMBER Wcr= (c*A*k

2)/ (k2+L2)

Wcr=crippling load=W*factor of safety

W=load acting on the vertical memberk= least radius of gyration of square section

L= equivalent length=(l/2)

c= compressive yield stress

A= cross sectional area of rod

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DRAWINGS OF REAR VERTICAL

MEMBER

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CALCULATIONS FOR FRONT

VERTICAL MEMBER Wcr= (c*A*k

2)/ (k2+L2)

Wcr=crippling load=W*factor of safety

W=load acting on the circular pipek= least radius of gyration of circular section

L= equivalent length=(l/2)

c= compressive yield stress

A= cross sectional area of rod

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DRAWINGS OF FRONT VERTICAL

MEMBER

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FRONT WHEEL ARRANGEMENT

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REAR WHEEL ARRANGEMENT

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REFERENCE A text book of machine design by R.S.Khurmi and

J.K. Gupta

Design data handbook by K. Mahadevan and K

Balaveerareddy

http//wikipidea.org

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Thank you

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Puncture & Skidding Management P14