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Mission :BAJA SAE ASIA 2010
Team : BLACK MAMBA
Institution: N.I.T.Rourkela, Orissa
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Mission :BAJA SAE ASIA 2010
Team : BLACK MAMBA
Institution: N.I.T.Rourkela, Orissa
Visit us at :
www.saenitr.in
Tube Size:
Outer diameter=1 inch. , Inner diameter=0.74 inch.
Material selected:
1018 carbon steel because of following reasons:
 High tensile strength.
 Good weldability.
 Low cost.
 Availability
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Element
Carbon(C)
Manganese(Mn)
Phosphorous (P)
Sulphur (S)
Iron (Fe)
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Weight (%)
0.2
0.9
0.04(max)
0.05(max)
Remaining is iron
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Conditions
Treatment
Properties
Density (×1000 kg/m3)
7.7-8.30
T (°C)
25
Poisson's Ratio
0.27-0.3
25
190-210
25
Elastic Modulus (GPa)
Tensile Strength (Mpa)
634
Yield Strength (Mpa)
386
Elongation (%)
27
Reduction in Area (%)
48
197
Hardness (HB)
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25
carburized at 925°C, cooled, reheated
to 775°C, water quenched, tempered at
175°C, core properties (round bar (1932 mm))
25
carburized at 925°C, cooled, reheated
to 775°C, water quenched, tempered at
175°C, core properties (round bar (1932 mm))
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A measure of whether a steel can be easily welded is to determine the
carbon equivalent (Cequiv) of the steel.
For 1018 Carbon Steel :
Cequiv=0.2+(0.9/6)=0.35 (Easily weldable)
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Tungsten Inert Gas (TIG) welding is preferred for welding the parts of rollcage.
 The TIG (Tungsten Inert Gas) welding process generates heat from an electric arc
maintained between a non consumable tungsten electrode and the part being welded.
 Gas is fed through the torch to shield the electrode and molten weld pool to prevent
oxidization.
 Filler rod , same as that of base metal is added to the weld pool separately.
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•Superior quality welds
• Welds can be made with or without filler metal
• Precise control of welding variables (heat)
• Free of spatter
• Low distortion
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The vehicle collides with a rigid wall at 60kmph (sixty kilometers per hour) and
comes to a standstill in 0.2 seconds. We analyze the impact from four directions.
This value of deceleration coupled with the full loaded mass of the vehicle which
is approximately 350 kg (three hundred fifty kilograms) amounts to forces of
approximately 29 kN (twenty nine kilo Newton).
The calculations are as follows :
acceleration in meters per second squared=(60)*(5/18)*(1/0.2)
=83.33
Force is mass times acceleration. So force in Newton
=350X83.33
=29166N
~29kN
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•The load is applied on the two
horizontal front extremity
members while keeping the
rear plane fixed .Each force is
having a magnitude of 14.9kN
and directed inwards along
the X-axis. The massive force
no doubt bends the bars
substantially, but it doesn’t
deform
the
driver’s
compartment much.
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Max Stress= 4.56 x 108 pa.
Min Stress= 0
•
A sideways impact is on that
poses maximum threat to the
driver’s safety. This has the
potential tendency of crushing the
compartment space. We have
applied a load of 29 kN at the side
bracing upper member while
keeping the other side fixed. Once
again no doubt the deformation is
present but the driver has a good
chance of staying unhurt.
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Max Stress= 2.64 x 109 pa.
Min Stress= 0
• Similar to the front impact case,
here also we have two 14.5 kN
forces doing the damage. The
forces are applied on the
horizontal members inwards
along the X-axis while the front
extremity plane is kept fixed at
its position. This does much less
deformation because of the
extensive frame work at the
drivers back resting plane.
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Max Stress= 9.21 x 108 pa.
Min stress= 0
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• The applied load for analysis is
3500N. From the Von-Mises
diagrams we see that the long
and mostly unsupported member
has bent quite some amount but
not enough to render the driver
hurt. So all the analysis confirm
the safety of the rollcage. The
material employed for analysis is
structural steel but the one to be
employed for actual fabrication is
the much tougher and lighter
alloy steel which thus guarantees
better results.
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Max Stress= 2.36 x 108 pa.
Min stress= 0
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Basic Suspension System
The primary purpose of a suspension system is to support the
weight of the vehicle and give a smooth ride.
It is desirable that it should also:
 allow rapid cornering without body roll when the car leans to one side.
 keep tires in firm contact with the road at all times and conditions.
 prevent body squat (tilting down at rear) when accelerating.
 prevent body dive (tilting down at front) when braking.
 allow front wheels to turn for steering.
 keep the wheels vertical and in correct alignment at all times.
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15 of 15
Double wishbone suspension

Double wishbone suspension is an independent suspension
design using two parallel wishbone-shaped arms to locate the
wheel.

Each wishbone (or arm) has two mounting positions and
attached to the knuckle via ball joints.

The shock absorber and coil spring mount to the wishbones to
control vertical movement.

Double-wishbone designs allow to control the motion of the
wheel throughout suspension travel, controlling such parameters
as camber angle, caster angle, toe pattern, roll center height,
scrub radius, scuff and many more.
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Why Double Wishbone ?
 It is fairly easy to work out the effect of moving each
joint, so one can tune the kinematics of the suspension
easily and optimize wheel motion.
 It decreases the cambering effect.
 It checks the body vibrations.
Material to be used: MS tube
Outer Diameter
: 1 inch
Wall Thickness
: 12
SWG
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Ackermann steering geometry
• The steering arm makes 74° with
the axis parallel to wheel axis . It
is calculated by the formula
• tanθ=wheel base/ (track width/2)
Or,θ=74°
Wheel Parameters
Diameter :21 inch
Rim Diameter :10 inch
Tread width :8 inch
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Roll center
The roll center is the point about
which the vehicle rolls when
cornering and is different for front
and rear part. The front roll centre is
usually lower than the rear, thereby
transferring weight to the rear during
cornering.
In our design we have parallel
wishbone system which results in
the roll center appearing very near
to the ground which is beneficial for
stability during cornering.
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Track rod length=62-(2*14.61)(2*8)=16.78 inch
Track rod offset from center of
wheels=3*sin74° ,where 74° is the
Ackerman angle.
Tie-rod length=14.32-(3*cos74°)=13.49
inch{Taking length of steering arm=3 inch
making 74° with axle}
Calculation of scrub radius:
Caster trail=10.5*tan10°=1.85 inch
Due to KPI of 10°, shift = 10.5*sec 10°*tan
10° =1.88 inch
Zoom
view
Caster trail
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•It consists of steering wheel , steering shaft , rack and pinion gear
box with gear ratio 18:1.
•Selected over recirculating ball steering system for its lesser
complexity of design.
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Design calculation
FOR THE DESIGN OF AXLES AND TIE ROD WE WILL USE MILD STEEL AND THE ENTIRE CALCULATION
IS BASED ON THE FOLLOWING DATA:
Material Property
Magnitude
Modulus of Elasticity
200GPa
Tensile Strength
455MPa
Yield Strength (tension)
250MPa
Poisson's Ratio
0.29
The maximum weight of the vehicle as per the specification is: 230Kg.
Taking factor of safety into account while design let the load be: 250Kg.
The aim is for the car to weigh approximately 250 kilograms and it is expected to generate
about one times the force of gravity in cornering. As stated previously these forces
were increased to give a suitable factor of safety. It was determined that a cornering
force of two times the force of gravity and a weight force of 250 kilograms would be
suitable.
After applying the weight force in the centre of the tyre and the cornering force across
the tread surface of the tyre, and working out the moments around the wheel bearing
centres, it was possible to work out the forces on the top and bottom suspension
mounting points to be approximately 2500N and 7500N …..(2500+2500*2) respectively.
STEERING RATIO:
It is defined as the angle made by the steering wheel for each 1deg movement of wheel.It has been found that
car requires 580deg of steer wheel for 35deg movement of wheel .Thus the steering ratio is: 580/35 or
16.5:1. It should be within 12 to 20 .So the design is safe.
Minimum Turn Radius( as pe
Track (T)= 48 inch
WHEEL BASE (W)=87 inch
θ=35deg.
R1= T/2 + W/sin θ =175.68 inch=446.23 cm=4.5m(approx)
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Determining the rack load
R=steering wheel radius: 300mm
r=pinion pitch circle radius: 5 mm
T= number of pinion teeth:5
P=linear or circular pitch: 2*pi*r/t:6.28 mm
E=input steering effort= 2*20= 40N
W= output rack load:
If the pinion makes one revolution;
input steering wheel movement Xi = 2*pi*R
output rack movement Xo = 2*pi*r
movement ratio(MR): 2*pi*R/(2*pi*r)= R/r=60.
Assuming no friction:
MR= W/E.
W=E*MR= 40*60=2400.
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Analytically finding the shear stress through the bolt for the bottom
spherical rod end.
Using the following values:
F = 7500N (as found previously for the bottom of the front upright);
d = 10mm (diameter);
Finding the area of the bolt:
A= pi*d2/4= 78.5mm2.
Finding the shear stress in the bolt:
σ=F/A= 7500/78.5= 95.54MPa.
This is much lower than the maximum permissible shear stress(455MPa) for high
tensile steel.
Analytically finding the stress in the tie-rods.( data taken from maruti alto)
Using the following values:
W = 2400N (as found previously for the rack output load);
d = 15mm (inside diameter of tie-rod);
D = 18mm (outside diameter of tie-rod);
L = 335mm (length);
FOS = 2 (factor of safety); and
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E = 200GPa (modulus of elasticity).
From Euler's formula:
W =π2 *EI/L2
I= FOS*W*L2/( π2 * E) =2.73*10-4 m4
Also I=pi*( D4-d4)/ 64 = 2.73*10-4 m4
Putting d=15mm we get Dmin= 4√50.62 = 15.000004.
But we have D= 18 mm .Thus our design is safe.
Finding stress at the base of front upright
F = 7500N (as found previously for the bottom of the front upright);
L = 80mm (length);
D = 40mm (outer diameter);
d = 30mm (inner diameter); and
y = 20mm (centroid).
Find the moment about the base of the upright:
M = F ¤ L = 7500 ¤ 80 = 600000Nmm
Find the second moment of inertia:
I= π( D4- d4)/64 =85904.9 mm4.
Find the stress at the base of the upright:
σ=M*y/I = 600000*20/85904.9= 139.6 MPa.
This value is lower than the 250MPa . So the design is safe.
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