e-Baja Design Report

Car No. #E135
BAJA SAE INDIA DESIGN REPORT
Kinshuk Srivastava, Rohit raj, Shivnath karmakar
1) Abstract:
The purpose of the BAJA SAE-India for Electrically Driven
vehicles is basically to manufacture a simple electric ATV for
recreational purposes which is aesthetically and
ergonomically sound as well as pleasing. As there is a
ubiquitous need to conserve fossil fuels and reduce carbon
emissions, the development for electric vehicles is increasing
exponentially. Thus, being our first year into the event, we
had to work our way to the complete buggy from bare
minimum. The event organizers provided teams with an
official electric kit common for each team. The kit included a
motor, gearbox along with other peripherals required to run
along the electric kit. The chassis design was kept as simple
as it could be to surround the driver, but also comply with
each and every rule set by the SAE. The rear of the chassis
was designed to incorporate the Electric Powertrain along
with other ancillaries. FEM analysis was performed for the roll
cage to make sure the structural integrity of the roll cage also
to find the ultimate fracture point.
The front and rear suspension geometries were evaluated in
lotus suspension analyzer and correspondingly the design of
hub and uprights were finalized. The front and rear spring
setup was calculated and manufactured. The team worked
diligently working day and night to manufacture the required
materials, also was in touch with 3rd
party manufactures to
procure the require components on time. The team strictly
followed the planned project plan to timely finish the project,
considering the late arrival of the electric kit.
The team successfully completed the project on time with
activities like manufacturing, parts procurement, and
assembly, and it is looking forward for the event.
2) Introduction:
The goal is to design and build a single-seat, all terrain, sporting
vehicle which is to be a prototype for a reliable, maintainable,
ergonomic and economic production vehicle which serves a
recreational user market, sized at 4000 units per year.
The vehicle should aspire to market-leading performance in terms
of speed, handling, ride and ruggedness in off-road
conditions.
3) Roll Cage:
Objectives:
● The roll cage/chassis houses the various sub-systems of the
vehicle and is the main supporting structure to which all the
other components are attached.
● It should ensure driver safety in cases such as crash or rollover.
● It should be stiff enough to compensate for the load transfer
during cornering and other dynamic scenarios.
Roll Cage Design Methodology:
Key aspects taken into consideration while
designing the roll cage were
● Driver safety
● Suspension loading points
● Drive train integration
● Structural Rigidity
● Ergonomics and Aesthetics
Material Selection:
A study was done to determine the ideal chassis
material taking into consideration factors like tensile
strength, bending stiffness etc. AISI 4130 was chosen
for its high strength-to-weight ratio and also high Tensile
strength and Yield strength. The material further
undergoes heat treatment which improves its fatigue life
and machinability.
Bending Strength and Bending Stiffness for both AISI 4130
and AISI 1018 were calculated to figure out which would be
the better choice for chassis fabrication.
STANDARD PARAMETERS
● Yield strength of AISI 4130(Sy) =460Mpa
● Modulus of elasticity for AISI 4130 E=200Gpa
● Yield strength of AISI 1018(Sy) =365Mpa
● Modulus of elasticity for AISI 1018 E=205Gpa
● Bending stiffness = E*I= The bending stiffness of a material
is proportional to the modulus of rigidity (E) and second
moment of inertia of material of cross section (I).
● Bending strength = (Sy * I)/C
● Moment of inertia of material cross section(I)= [π/4(R^4-r^4)]
Primary member(X) material AISI (4130) -
Outer diameter (D) =31.75mm
Inner diameter (d) =D-t=31.75-3=28.75
Outer radius(R) =15.87
Inner radius(r) =14.37
Ix= [π/4(15.87^4-14.37^4] = 16329.156mm^4
Ex = Modulus of elasticity which is = 200GPa
C= Distance from neutral axis to extreme fiber. =31.75/2=15.875mm
Bending Strength calculation of the “X” = (Sy*I)/ C=
(460*16329.156)/15.875=473159N-mm
Bending Stiffness calculation of the “X” = Ex*Ix
=200*16329.156=3265831.2N-mm^2
Secondary member(Y) material AISI (4130)-
Iy= [π/4(12.7^4-11.2^4] = 8073.32mm^4
Ey = Modulus of elasticity which is = 200GPa
C= Distance from neutral axis to extreme fiber.=25.4/2=12.7mm
Bending Strength calculation of the “Y” = (Sy *I)/ C=
(460*8073.32)/12.7=292419.46N-mm

Bending Stiffness calculation of the “Y” = Ey*Iy
=200*8073.32=1614664N-m^2
Primary member(X) material AISI (1018) -
Ix= [π/4(15.87^4-14.37^4] = 16329.156mm^4
Ex = Modulus of elasticity which is = 205GPa
C= Distance from neutral axis to extreme fiber. =31.75/2=15.875mm
Yield strength of AISI 1018(Sy*) =365Mpa
Bending Strength calculation of the “X” = (Sy*I)/ C=
(365*16329.156)/15.875=375442.012N-mm
Bending Stiffness calculation of the “X” = Ex Ix
=205*16329.156=3347476N-mm^2
Secondary member(Y) material AISI (1018)-
Inner radius(r)=11.2
Iy=[π/4(12.7^4-11.2^4] = 8073.32mm^4
Ey = Modulus of elasticity which is = 205GPa
C= Distance from neutral axis to extreme fiber. = 25.4/2=12.7mm
Yield strength of AISI 1018(Sy*) =365Mpa
Bending Strength calculation of the “Y” = (Sy*I)/ C=
(365*8073.32)/12.7=232028.48N-mm
Bending Stiffness calculation of the “Y” = Ey*Iy
=205*8073.32=1655030N-mm^2
Table 1: A comparative study has been shown to highlight the
differences between AISI 4130 and AISI 1018 (a material quite
commonly used for frame design).
Property AISI 1018 AISI 4130
Yield Strength
(MPa) 370 460
Tensile Strength
(MPa) 430 670
Strength-to-
Weight Ratio
(kN-m/kg)
54-60 71-130
Bending Strength
(N-mm) 375442.01 473159
Bending
Stiffness (N-mm2
) 3347476 3265831.2
Elongation (%) 15.0 20.5
Welding:
MIG welding was chosen for its precise heat control which lowers
the heat affected zone which in turn allows the material to retain
its original material properties. It also eliminates slag and spatter
due to the inert gas that shields the weld, creating a clean weld
bead.
Finite Element Analysis:
Analysis of the chassis was done on ANSYS 15 using the
co-ordinates obtained from SOLIDWORKS 2017.
Various scenarios such as front impact, side impact etc. was
taken into consideration. An example of force calculation method
and justification has been shown.
Case: Front Impact
The front impact scenario considers two vehicles travelling
towards each other having a head-on collision and then coming
to a stop. In an event scenario, the vehicles top-out at 40 km/hr,
so this was taken as the maximum speed and both vehicles
weighed 300 kg (laden weight of our vehicle). Impact time was
set at 0.2 seconds after cross-referencing with various research
papers.
Maximum Speed: 40km/hr (11.11 m/sec) Impact Time:
0.2 sec
Mass of vehicle: 300 kg
Force= Mass * (Change in velocity/Impact time)
= M*(Vrelative/∆t)
= 300 * ((22.22-0)/0.2)
= 33330 N
Similarly, other forces were calculated and the FEA analysis
was completed.
● Rear Impact: Stationary vehicle hit in the rear at 40 km/hr
by another vehicle.
● Side Impact: Stationary vehicle hit in the side members at
40 km/hr by another vehicle.
● Torsional Analysis: Maximum load transfer at current
turning radius.
● Roll Over: Running Vehicle topples over from an
obstruction.
Table2: Results of Finite Element Analysis
Analysis
Type
Max
Stress(N
)
Max.
Deformation(mm)
Front
Impact
368.17 1.48
Rear
Impact
229.14 1.94
Side
Impact
297.28 1.79

4) Suspension:
Abstract:
The main idea behind this section of the design report is to deal with
the suspension design and calculations. As the ATV stands for an
All-Terrain vehicle, the ATV has to be designed factoring in its most
fundamental function of versatility. Because of the immense
versatility of the ATV, suspension forms the basic constraint for
handling. From designing, analysis, and tuning, suspension setup
requires tremendous iterations for the required configuration.
Suspension characteristics like camber, caster angles, roll centers
and center of gravity forms a few of the many variables for tuning
the geometries. As the ATV undergoes huge amount of shocks in
rough terrains, stability as well as driver comfort is considered
important considering top notch performance. The suspension setup
includes a pair of spring and dampers along with the suspension
geometries of linkages. The spring parameters were calculated from
scratch to support our Electric Buggy. Along with that, the
suspension geometries were plotted in the lotus suspension
software which was tuned for our requirement. The arms were
modeled in Solidworks and underwent the Ansys Analysis.
Introduction:
An ATV as the name suggests is capable of maneuvering all type of
terrains. The same is achieved by the immense low pressures of the
tires and correct ratio of sprung to the un-spring mass for the
accurate momentum to be controlled by the suspension setup. The
ATV market especially for E-ATV is up and running and if not
constant, increasing on a yoy basis. The ATV’s have plethora of use
worldwide and suspension systems are the first mechanism to
undergo the rigorous beating. Because of such requirement and so
many variables to play with, suspension is just another compromise
with the most important requirement being able to stick the 4 wheels
on the ground most of the time providing the traction.
Objective:
The main objective for the suspension of any ATV is to minimize the
momentum to be controlled by the suspension springs, followed by
maintaining maximum amount of traction at all 4 wheels specially
the two driving wheels. Suspension has to be designed such as to
isolate the ATV from road shocks and to provide the road holding
throughout the full range of power transmission. Driver comfort and
road grip thus forms the ulterior motive independent suspension
system.
Thus, this year we decided to use such a system which would make
our vehicle to perform even in the worst terrains and provide agility
while cornering or overcoming obstacles.
The suspension setup is discussed not only considering the
mechanism of springs and dampers, but also must include the
linkages that connect the vehicle and wheels controlling the motion
of wheels under a bump and droop. Under the event of rolling, huge
amount of weight transfer changes the orientation of the outside
wheels thus reducing the traction. To reduce the same camber curve
should be on spot to increase the traction and thus tire life. Besides
the aforementioned dynamic parameters, other factors to be
considered are cost, weight, space and packaging,
manufacturability, etc.
Design Evaluation:
• As our vehicle is for off-road purpose vehicle control is a very
crucial factor. Stabilizing the vehicle and a better road grip at
deep droops, high bumps and steep corners is a very sensitive
job.
• We begin our calculations for the suspension system by
assuming vehicle track width, wheel base and ride height (static
condition). As this is our first time and observing track nature
we considered our front track width at 51’’ while the rear is set
to 48’’ so that the tires could be more evenly loaded and could
produce higher braking force and improve the cornering.
Making ride smoother as it gives more total suspension travel.
• Moving on to the next step where we decided the suspension
geometry for the vehicle. For the front we choose unequal and
non-parallel double wishbone setup with damper mounting on
lower arm. For the rear we’re sticking with the trailing arm
geometry setup with camber links.
Design Process:
Double wishbone suspension setup being the most fundamental
setup is the most utilized geometry. The setup has a control arm
which helps the engineer to tune different characteristics of the
double wishbone geometry while maintaining the road holding and
traction. The double wishbone, with its optimum un sprung weight
fits in perfect with the aim of gaining camber.
In the rear, trailing arm was used due to its ease of installation and
proper damper mounting points, while maintaining a good
installation ratio. The camber links help in modifying the camber
characteristics in the corners. The trailing allows the wheel to move
up & down to deal with bump. The propensity of camber change is
minimum with this geometry. The camber angle changes only when
the trailing arm rolls for the same degree as the car body which is
desirable. Semi-trailing geometry was not opted as when the wheel
moves up and down, camber angle changes like the double
wishbone setup which is highly undesirable.
Next, roll centers were calculated and verified. The biggest effects
that roll center location has on a car’s handling have to do with how
the car responds to steering input in a corner. It defines how friendly
is the car in the event of rolling along with the control over car’s
balance.
Figure 1. Dynamic Functioning of Roll Couple and Roll Center.
Higher roll centers result in higher jacking force which can be
detrimental. Secondly it also affects the roll moments. Jacking force
is the effect due to lateral force acting on roll moment or connecting
the tire contact patch with roll center, the vertical component
determines the jacking force. If the roll center is too low below the
center of gravity, the roll couple formed by this distance would be too
long and thus creating a more leverage for cg on suspension
through the center of gravity eventually leading to more tendency to
roll. Thus, our vehicle has a front roll center of 241 mm and rear roll
center of 360 mm. This data we achieved not only reduced jacking
force but also the effect of lateral forces like wheel travel. As our
front roll center is lower we can get better braking but at the same
time we lose acceleration. It also helps in good steering control.
Now taking all these assumptions and values, we achieve the spring
rates and roll rates which help in finding of spring stiffness Front
spring stiffness was found to be 31.05 N/mm and the rear spring
stiffness was found to be 34.52 N/mm
This result was obtained for the vehicle subjected to a cornering
force of 2G and a bump of 1.5G.
The forces considered for hub and upright analysis broadly are
classified as bump force, braking/ acceleration, jacking, lateral load
transfer and longitudinal load transfer.
We are employing customized springs and dampers, which offer
more range of suspension movement and allows us a wider tuning
envelope through the suspension range than its counterparts. Even
the tuning characteristics of the conventional springs and dampers
is comparatively less than that of fox air dampers, the team decided
to go with the former option. The spring rates for both front and rear
geometries were calculated separately using the moment equation
and considering the angle correction factor.
As weight is added, the spring rate becomes stiffer to increase
stability without the ill effects of bouncing, and it prevents the ride

height from dropping significantly. Coil springs also allow for a
greater amount of up and down wheel travel, which is a huge
advantage on rutted and rocky off-road trails.
Thus, after finalizing the trackwidth and wheelbase, the geometry of
the front and rear suspensions was plotted and tuned in Lotus
Suspension Analysis Software which helped in further evaluation of
the suspension. Procuring hard points from the Lotus Software lead
to the further development of the roll cage altogether.
The following images represents the working on Lotus software.
Figure 2. 3D view for Front Double Wishbone Setup
Figure 3. Front view for the Double Wishbone Setup
Figure 4. Front view for 3D Bump full static and dynamic
Figure 5. Front view for 3D Roll full static and dynamic
Figure 6. 3D view for Rear Trailing Arm Setup
Figure 7. Front View for Trailing Arm Setup
Figure 8. 3D view for Bump full static and dynamic- Rear Trailing Setup
Figure 9. 3D view for Roll Static and dynamic- Rear Trailing Setup

Components Used in Suspension: -
Ball Joints Maruti 800
Rose joints MTR Rod Ends
Trailing Arm AISI 4130 Chromyl
Molybdenum Steel
(OD: 25.4mm, Thickness:
1.20mm)
Shock Absorber Customized Spring &
Damper
Double Wishbone Arms AISI 4130 Chromyl
Molybdenum Steel
(OD: 25.4mm, Thickness:
1.20mm)
Hub(customized) 7075-T6
Upright(customized) 7075-T6
Bushing Used Nylon bush
Static Data: -
Camber angle (in degrees) -2
Caster angle (in degrees) 2
Kingpin inclination (in
degrees)
6
Scrub radius (in mm) 45.29
Toe angle (in degrees) 2
C.G height (in inch) 18.03
Ground Clearance (in inch) 13.5
Front Rear
Spring Stiffness (in N/mm) 31.05 34.52
Spring Travel (in inch) 4.5 5.5
Ride frequency (in Hz) 1.208 1.283
Motion Ratio 0.578 0.833
Installation Ratio 0.501 0.742
Roll Center Height (in mm) 241 360
Given data:
Weight at front axle= W1+W2= Wf= 178.57lbs (38+43=81 kg)
Weight at rear axle= W3+W4= Wr= 416.67lbs (92+97=189 kg)
Total weight= Wf + Wr= 595 lbs (270 kg)
Track width front (tf)= 51” (1295.4 mm)
Track width rear (tr)= 48” (1219 mm)
d= (tf-tr)/2= 1.5” (38.1 mm)
Wheelbase (l)= 60” (1524mm)
1) Centre of Gravity location:
a) Horizontal position:
Figure 10. Dynamic Functioning of Roll Couple and Roll Center
Taking moment, the rear axle,
Wf.l=W.b
b= = (81/270). (1524) =457.2 mm
𝑊𝑓
𝑊
. 𝑙
a=l-b= 1524 – 457.2 = 1066.8mm
a=1066.8 mm
b=457.2mm
Then, taking moment about X-X1 axis:
W2.(tf-d) – W1.d + W3.tr = W.y’
y’= .(tf-d) – .(d) + .(tr)
𝑊2
𝑊
𝑊1
𝑊
𝑊3
𝑊
y’=598.07 mm
Now, y”=y’- ( )
𝑡𝑟
2
y”= .(tf-d) – .(d) + .(tr) – ( )
𝑊2
𝑊
𝑊1
𝑊
𝑊3
𝑊
𝑡𝑟
2
y” = -11.43 mm
b) Vertical Position:
Figure 11. Dynamic Functioning of Roll Couple and Roll Center
Angle of the used ramp: 14 degrees.
Radius of front tire = 11.5” = 292.1 mm
Radius of rear tire = 12.5” = 317.5 mm
The front wheel loads were measured again and were found to be
41.38 kg on the front left tire and 46.2kg on the front right tire.
Thus, Wf= 87.58 kg
Now,
l1= l cosƟ
Taking moment about O:
Wf.l1 = W.b1
Wf.l .cosƟ = W.b1
b1 = ( ).l.cosƟ
𝑊𝑓
𝑊
Then,
= cosƟ
𝑏
𝑏+𝑐
c= ( ) – b
𝑏1
𝑐𝑜𝑠θ
c= ( ).l - b
𝑊𝑓
𝑊
Using tan Ɵ = ;
𝐶
ℎ1

h1= ( )
(𝑊𝑓.𝑙 – 𝑊.𝑏)
𝑊.𝑡𝑎𝑛Ɵ
=
87.58 × 1524 – 270 × 457.2
( )
270×tan𝑡𝑎𝑛 14
( )
= 149 mm = 5.93”
Then, CG height= h1+ radius of loaded tire
Now, Rf = 11.5” = 292.1 mm
Rr = 12.5” = 317.5 mm
Rcg = Rf. ( ) + Rr. ( )
𝑏
𝑙
𝑎
𝑙
= 292.1 ( ) + 317.5( )
457.2
1524
1066.8
1524
= 309.53 mm
= 12.1”
Thus, CG height = h1+ Rcg= (309.53 +149) mm
= 18.03” = 458.49 mm
Load calculations:
a) Lateral load transfer:
Lateral load transfer is given by the formula:
∆W = Lateral acceleration (per g) × weight (lb) × C.G. Height
(inches)
Track width (inches)
Now, maximum lateral acc. =
(𝑉𝑚𝑎𝑥)2
𝑅.𝑔
Then, Vmax= 40 kmph = 11.11 m/s
R= Turning radius= 3.13 m
g= 9.81 ms-2
Then, max lateral acc.= 4.5 ms-2
Thus, max lateral load transfer=
∆W= = 975.611lb =442.52kg
(4.5×595.215×18.03)
49.5
b) Lateral load transfer gain sensitivities:
Now, roll sensitivity to lateral acceleration is given by:
KƟ= =
Ɵ
𝐴𝑦
−𝑊.𝐻
(𝐾𝑓 + 𝐾𝑟)
Then, lateral load transfer equations for the front and rear axles,
expressed in terms of gains to Ay, are given by:
= [ + ]
∆𝑊𝑓
𝐴𝑦
𝑊
𝑡𝑓
(𝐻×𝐾𝑓)
(𝐾𝑓+ 𝐾𝑟)
𝑏×𝑧𝑟𝑓
𝑙
= [ + ]
∆𝑊𝑟
𝐴𝑦
𝑊
𝑡𝑟
(𝐻×𝐾𝑟)
(𝐾𝑓+ 𝐾𝑟)
𝑎×𝑧𝑟𝑟
𝑙
Where, Kf= front roll rate
Kr = rear roll rate
= front and rear roll center heights
𝑍𝑟𝑓
𝑍𝑟𝑟
c) Banking of road:
Now, centripetal force, Fα= W.Aα
Where, Aα= V2
/ Rg
Resolving along y and z axes, we have:
Fy = -Fαcosα+ Wsinα
Fz = Fαsinα + Wcosα
Since, Fα= W. Aα
Fy = W (-Aαcosα + sinα)
Fz = W (Aαsinα + cosα)
Thus, changed weight due to banking is
W’ = W(Aαsinα + cosα)
W’ = 595.215[-0.74.sin(-10) + cos(-10)] = 662 lb = 300.592 kg
hen, effective rear and front axle weights are:
Wf’=W’. = 662. ( ) =198.6 lb = 90.083 kg
𝑏
𝑙
457
1524
Wr’=W’. = 662. ( ) = 463.4 lb = 210.194 kg
𝑎
𝑙
7
10
Lateral load transfer
∆Wf = (Aαcosα - sinα) ( ) [H + ]
𝑊
𝑡𝑓
. 𝐾𝑓
(𝐾𝑓+𝐾𝑟)
𝑏.𝑧𝑟𝑓
𝑙
∆Wr = (Aαcosα - sinα) ( ) [H + ]
𝑊
𝑡𝑟
. 𝐾𝑟
(𝐾𝑓+𝐾𝑟)
𝑏.𝑧𝑟𝑟
𝑙
hus;
Wf = (Aαcosα - sinα) ( ) [ H. + + ]
𝑊
𝑡𝑓
. 𝐾𝑒𝑓
(𝐾𝑒𝑓+𝐾𝑒𝑟)
𝑏.𝑧𝑟𝑓
𝑙
= -0.74 × × [ + ( ) × 0.790]
595.215
4.5
0.5337 × 2500
(2500 + 2000)
3
10
= - 12.92 lb
= - 5.86 kg
Wr = (Aαcosα - sinα) ( ) [ H. + + ]
𝑊
𝑡𝑟
. 𝐾𝑒𝑟
(𝐾𝑒𝑓+𝐾𝑒𝑟)
𝑏.𝑧𝑟𝑟
𝑙
= -0.74 × × [ + ( ) × 1.18]
595.215
4.5
0.5337 × 2000
(2500 + 2000)
7
10
= -30.14 lb
= -13.67kg
2) Roll rate/ ride rate/ roll angle calculations:
efore, the calculations are done, the following data is known/assumed:
Wf = 81 kg = 178.57 lb Wr = 189 kg =416.67 lb
W= 270kg=595.215lb
f== 54” = 4.5 ft. Tr= 54” =4.5ft l=
0” = 5 ft

h= C.G. Height= 18.03” H= C.G to roll axis = 0.5337 ft.=162mm
zrf = front roll centre = 0.790 ft. = 241mm
zrr= rear roll centre = 1.18 ft. = 360 mm
Vmax= 40 kmph = 36.45 ft/sec.
We now assume the roll rates,
KƟf = 2500 lb-ft/rad = 59.15 N-m/deg
KƟr = 2000 lb-ft/rad = 47.32 N-m/deg
First, the CG position is calculated:
Wf.l=W.b
b= 457.2mm = 1.5 ft.
a= l-b = 1066.8mm = 3.5ft
Now, suppose the car is going on a left banked turn of radius 100 ft. at
Vmax, at an angle of 10 degrees, then:
Aα=
𝑉2
𝑅𝑔
=
(36.45)2
−100 ×32.2
= - 0.412 g’s
Then, Ay= (Aαcosα – sinα) = - 0.412 .cos(-10) – sin(-10) = - 0.232
Thus, effective weight due to banking:
W’ = W(Aαsinα + cosα) = 595.215[-0.412.sin(-10) + cos(-10)] =
543.58 lb = 246.56 kg
Then, effective rear and front axle weights are:
Wf’=W’. = (543.58) . ( =198.6 lb = 90.083 kg
𝑏
𝑙
3
𝑙10
)
Wr’=W’. = (543.58) . ( ) = 463.4 lb = 210.194 kg
𝑎
𝑙
7
10
Thus, the roll gradient, is then given by
Ɵ/Ay = = = 0.0706 rad/g = 4.048deg/g
−𝑊.𝐻
(𝐾𝑓 + 𝐾𝑟)
− 595.215 × 0.5337
2500 + 2000)
Thus, Ɵ = body roll angle
= 4.048 × - 0.412
= -1.66deg
The front and rear weight transfers due to lateral accelerations:
Wf = (Aαcosα - sinα) ( ) [H + b. ]
𝑊
𝑡𝑓
. 𝐾Ɵ𝑓
𝐾Ɵ𝑓+𝐾Ɵ𝑟
𝑧𝑟𝑓
𝑙
= -0.412 × 595.215/4.5 × [ 0.5337 × 2500 / (2500 + 2000) + (3/10) ×
0.790] = - 12.92 lb = - 5.86 kg
Wr = (Aαcosα - sinα) ( ) [H + b. ]
𝑊
𝑡𝑟
. 𝐾Ɵ𝑟
𝐾Ɵ𝑓+𝐾Ɵ𝑟
𝑧𝑟𝑟
𝑙
= -0.412 × 595.215/4.5 × [ 0.5337 × 2000 / (2500 + 2000) + (7/10) ×
1.18]
= -30.14 lb = -13.67 kg
We observe that the load on the outside wheel increases and that on the
inside wheel decreases. Thus, the individual wheel loads can be given
by:
Front outside = (196.6/2)+ 12.92 = 111.02 lb = 50.35 kg
ront inside = (196.6/2) – 12.92 = 85.18 lb = 38.63 kg
Rear outside = (463.4/2) + 30.14 = 261.84 lb = 118.76 kg
Rear inside = (463.4/2) – 30.14 = 201.56 lb = 91.42 kg
Changes from static loads:
ront outside = 111.02 – 100.31 = 10.71 lb =4.85 kg
ront inside = 85.18 – 100.31 = -15.13lb = -6.86 kg
Rear outside = 261.84 – 208.33 = 53.51 lb = 24.27 kg
Rear inside = 201.56 – 208.33 = -6,77 lb = -3.07 kg
ow, we have 5’’ of jounce travel available. It is favourable to
ccommodate these load changes in 4’’ of jounce travel allowing 1’’ of
dditional travel due to acceleration or road bumps etc.
hus, we choose ride rates that are compatible with these wheel loads.
ow, ride rates required are the load changes divided by the ride travel.
hus,
KRF = 12.92/4 = 3.23 lb/in. = 565.65 N/m
KRR = 30.14/4 = 7.53 lb/in. = 1318.70 N/m
hen, ride frequencies can be calculated as:
wf = 1/2π × ( )1/2
= 0.595 Hz = 35.4 cpm
2 × 𝐾𝑅𝐹 × 12 × 32.2
𝑊𝑓
wr = 1/2π × ( )1/2
= 0.595Hz = 35.4 cpm
2 × 𝐾𝑅𝑅 × 12 × 32.2
𝑊𝑟
hen, roll rates can be calculated as:
KƟf =
(12 × 𝐾𝑅𝐹 × 𝑡𝑓 2 )
2
KƟr =
(12 × 𝐾𝑅𝑅 × 𝑡𝑟2 )
2
hus,
Ɵf = / 2 = 784.89 lb-ft/rad
12×3. 23×(4. 5)2
Ɵr = = 1829.79 lb-ft/rad
12×7. 53×(4. 5)2 / 2
3) Frequency Calculations
) Sprung mass natural frequency:
𝑓𝑛𝑓 =
1
2π
× (
𝐾𝑟𝑓
𝑀𝑓
)1/2
𝑓𝑛𝑟 =
1
2π
× (
𝐾𝑟𝑟
𝑀𝑟
)1/2
Where, fn = Body bounce frequency
Kr= Ride rate
M= Sprung mass
Using the above-mentioned values;
fnf = 3.47 Hz
fnr = 2.516 Hz
Installation ratios and motion ratios:
i) For front arms:
Distance of ball joint from hinge(suspension) point = 530.8 mm =
20.9”
Distance of shocker mounting point from hinge (suspension) point =
307.3 mm = 12.1”

Thus, motion ratio = 307.3/530.8 = 0.578
Then, Installation ratio = Motion ratio × cos (angle of shocker
mounting with vertical)
= 0.578 × cos(30)
= 0.501
ii) For rear arms:
Distance of upright mountings from hinge(suspension) point = 487.6
mm = 19.2”
Distance of shocker mounting point from hinge (suspension) point =
406.6 mm = 16”
Thus, motion ratio = 406.4/487.6 = 0.833
Then, Installation ratio = Motion ratio × cos (angle of shocker
mounting with vertical)
= 0.833 × cos(27)
= 0.742
Roll Camber and roll change:
Now,
Roll camber = Wheel Camber Angle/ Chassis Roll angle.
= 2/3. 034
= 0.66
Then, Front view swing arm (FVSA); front = . (1-roll camber)
(𝑡𝑓/2)
= (0.647). (1-0.66)
= 0.221
Front camber roll change = = 77.59
𝑡𝑎𝑛 − 1(1/𝐹𝑉𝑆𝐴𝑓)
Front view swing arm (FVSA); rear = (1-roll camber)
(𝑡𝑓/2).
= (0.609). (1-0.66)
= 0.207
Rear camber roll change = tan-1
( ) = 78.14
1/𝐹𝑉𝑆𝐴𝑟
Body roll stiffness:
For the purpose of the calculation, the standard radius of 100 ft
(30.48 m) is taken;
M = Centrifugal moment about roll center + Transverse
displacement moment
Centrifugal moment = F.a
Force = 270. (11.12
) / 30.48 = 1093.39 N
= 𝑚𝑣2/𝑟
a= distance of roll centre from cg
a) For front:
i) Front transverse displacement moment:
W= Unsprung Weight = 140 kg
a= 1.002 ft = 0.305m
Ɵ= 3 deg
M1 = W.a.Ɵ
= 242 × 0.305 ×1.66
= 122.52 Nm
ii) Front centrifugal moment:
M2 = F.a
= 2440.98 × 0.305
= 744.498 Nm
iii) Total M = 122.52 + 744.498 = 867.022 Nm
iv) Front Roll stiffness=Roll couple/Chassis roll
angle= =290.866Nm/deg
872. 6/1. 66
b) For rear:
v) Rear transverse displacement moment:
W= Unsprung Weight = 242 kg
a= 0.612 ft = 0.187 m
Ɵ= 1.66 deg
M1 = W.a.Ɵ
= 242 . × 0.187 × 1.66
= 75.12 Nm
vi) Rear centrifugal moment:
M2 = F.a
=2440.98 × 0.187
= 456.463 Nm
vii) Total M = 75.12 +456.463 = 531.58 Nm
viii) Rear Roll stiffness=Roll couple/Chassis roll angle=
=178.34Nm/deg
531. 58/3
Overturning Couple:
Overturning couple = F × h(CG height) = 2440.98 × 0.458 = 1117.96
Nm
Shaft Travel:
Now, Angle of rear shocker from the vertical = 27 deg
Maximum travel of rear shocker = 5.5” = 139.7 mm
Thus, maximum shocker travel in vertical direction = 139.7 × cos(27)
= 124.5mm
Then, maximum wheel travel = Maximum shocker travel/ Motion
ratio = = 149.4mm
124. 5/0. 833
Length of drive shaft = 18” = 457.2mm
Thus, maximum driveshaft travel angle = Ɵ(rad) =
𝑊ℎ𝑒𝑒𝑙 𝑡𝑟𝑎𝑣𝑒𝑙
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒 𝑠ℎ𝑎𝑓𝑡
=
= 22.5 deg
149. 4 / 457. 2
Now, the driveshaft is already at an angle of 36 degrees from the
horizontal. It was verified that the maximum angle that the driveshaft
could travel before its clamp opened was 42 degrees.
Thus, with our current configuration, the maximum angle the drive
shaft can travel is 22.5 degrees, which is 17.3 degrees below the
horizontal and is much less than 42 degrees.
Hence, the shaft can easily accommodate the entire wheel travel in
a bump condition without detaching from its boot
Trailing arm calculation (dimensions):
We have a governing equation:
e and d can be interpreted from the diagram
Fz = vertical reaction on the ground on each wheel
Fx = tractive effort on the ground
Therefore, d = = 0.4476 m
From geometry:
1 = tan-1
( ) = tan-1
() = 9.66
2 = tan-1
( ) = tan-1
() = 21.66
Again, from the geometry we can calculate the forces acting on the
controlling arm (P1& P2)
P1 = 506.45 N
P2 = 2397.79 N

SPRING DESIGN CALCULATION
d= wire diameter of spring (mm)
Di = inside diameter of spring coil (mm)
Do= outside diameter of spring coil (mm)
D= Mean coil diameter (mm)
D =
( 𝐷𝑖 + 𝐷𝑜 )
2
Spring index (c) = =
𝑀𝑒𝑎𝑛 𝑐𝑜𝑖𝑙 𝑑𝑎𝑖𝑚𝑒𝑡𝑒𝑟
𝑊𝑖𝑟𝑒 𝑑𝑎𝑖𝑚𝑒𝑡𝑒𝑟
𝐷 / 𝑑
Low value of C = High sharpness of curvature
C < 3 (Hard to manufacture)
C > 15 (Lead to Buckling)
Preferred Range = 4 ≤ C ≤ 12
6 ≤ C ≤ 9 (Close Tolerance Spring)
We assume that, C = 7
Shear Stress, τ = 700 N/mm2
Deflection δ: δF = 21mm, δR = 40mm
Wahl factor = + ( )
(4𝑐 − 1)
(4𝑐−4)
0.615
𝐶
Shear Stress, τ =
𝐾 (8𝑃𝐶)
(π𝑑2)
Coil diameter = d = ( )
8𝐾𝑃𝐶
(π τ)1/2
= ( )1/2
1.212 𝑥 8 𝑥 330𝑥9.81 𝑥 7
π 700
= 10mm
Mean diameter = D = d x c = 10 x 7 = 70mm
δ =
(8𝑃𝐷3𝑁)
(𝐺𝑑4)
N =
(δ 𝐺𝑑4)
(8𝑃𝐷3)
So, 𝑁𝑓 = (δ𝑓 𝐺𝑑4) / (8 𝑃𝑓 𝐷)
= (21 x 116000 x 104
) / (8 x 810 x 703
)
= 10.95 = 11
NR = (40 x 116000 x 104
) / ( 8 x 810 x 703
)
= 8.94 = 9
Total number of coil in front spring = 11+2 = 13
Total number of coil in rear spring = 9+2 = 11
Free length, L = 235mm
Pitch, P = L / ( Nt – 1)
Pf = 235 / (13 – 1) = 19.58mm
Pr = 235 / (11-1 ) = 23.5mm
Stiffness of front spring:
Total force o rear spring X total length of A arm = force on damper X
distance of damper form
end of arm
120x9.81x 17’’ =f x6’’
f =3335.4
F =f/correction factor
=3335.4/cos(20) =3549.45
Stiffness = F/spring travel
=3549.45/(4.5 x25.4) =31.05 N/mm
Stiffness of rear spring:
Total force of rear spring X total length of trailing arm = force on
damper X distance of damper
form end of arm
231x9.81x 28’’ =f x14’’
f =4532.22
F =f/correction factor
=4532.22/cos(20) =4823.08
Stiffness = F/spring travel
=4823.08/ (5.5 x25.4) =34.52 N/mm
Figure 11. Front Spring
Figure 11. Rear Spring
Thus, after calculations and initial iterations the double
wishbone was fabricated as follows-

For the same, the lotus iteration graphs are as follows-
Figure 11. Anti-Dive% vs Droop-Bump
Figure 11. Camber Angle vs Droop-Bump
Figure 11. Caster Angle vs Droop-Bump
Figure 11. Damper Ratio vs Droop-Bump

Figure 11. Damper Travel vs Droop-Bump
For the Trailing Arms, the graphs are as follows
Figure 11. Anti-Dive% vs Droop-Bump
Figure 11. Camber Angle vs Droop-Bump
5) Brakes:
Objective:
The objective of the braking system is to either slow the vehicle
down or completely bring it to a halt to protect the vehicle and the
driver from accidents, by converting the kinetic energy of the vehicle
into heat dissipation via the brake rotors.
SAE Baja shares the above-mentioned design objective along with
locking of all four wheels when brakes are applied.
Introduction:
A F/R split hydraulic braking system was used along with
custom SS410 rotors coupled with Wilwood PS-1 brake calipers
on all four wheels.
The choice of material for the rotors was based on how good
the heat dissipation capacity of the material was, SS410 was
thus chosen, Multiple key points were taken into consideration
while designing the rotors:
1. The rotor should be light weight to minimize unsprung
mass.
2. Rotor should be so designed to maximize heat dissipation;
thus, a drilled design was chosen.
3. The rotors should be designed to minimize crack formation
in them while their use.
4. The rotor should be large enough to maximize torque due
to clamping action.
All the above points were taken into consideration thus concluding
to this design:
The choice of caliper was based on the following points:
1. The caliper should be light to minimize the unsprung mass.
2. The caliper should be able to accommodate brake pads of
large dimension to maximize contact area between the
rotor and pad
3. The caliper should be compact to minimize packaging area
inside the rim.
Keeping all this in mind Wilwood PS-1 was chosen due to its small
size, ability to accommodate large brake pads, and lightweight
construction compared to its expensive counterpart Wilwood
GP-200.
The brake lines throughout the braking circuit are made from steel
braided brake lines, this was done to avoid the following:
1. Minimize the loss of pressure in the circuit due leakage.
2. Bending of metal brake lines causes change of bore diameter
thus changing line pressure.
A tandem master cylinder from Bosch was chosen to generate
brake pressure for the entire circuit. A tandem setup was used
rather than a pair of single piston master cylinders to avoid
complexity in mounting the master cylinder, more over the former
would be more compact. The pressure generation would be assisted
by a roof mounted pedal with a pedal ratio of 4.5:1. A lightweight
pedal was designed to favor driver ergonomics along with
maximizing force on the master cylinder.

Design Criteria:
1. The major consideration during designing of the system was
providing enough Braking force which would counter the torque
due to weight transfer and torque supplied by the transmission.
2. Along with above mentioned criteria, 4-wheel lock on
application of the brakes was a necessity.
Keeping in mind the aforementioned points, the braking system was
designed.
Table-2 Design parameters:
Master cylinder bore
diameter:
19.05mm.
Caliper Bore Diameter: 25.4 mm.
Pedal mounting Used: Roof mounted.
Weight of the car: 250 kg
Weight of the driver: 70 kg
Weight distribution of the
car:
65:35
Tire diameter: Front: 584.2 mm
Rear: 584.2 mm
Coefficient between tire
and road:
0.50
Coefficient between pad
and rotor:
0.45
Wheel Base: 1524 mm
C.G. Height: 457.962 mm
Number of piston in
each calliper:
2
Pad width: 7.62 mm
Certain parameters had to be assumed to assist in the designing of
the system:
1. 400 N force on the pedal.
2. Deceleration was assumed to be 1g according to the book
“Brake design and safety by Rudolf Limpert”.
3. Top speed of the vehicle is 45 kmph
The calculation was carried out in the following manner:
1. First the static weight on the front and rear
Wheels was calculated.
2. Then weight transfer was calculated which would occur during
braking.
3. This weight transfer is then added to the front static weight and
substracted from the rear static weight.
4. Master cylinder and caliper bore area was calculated.
5. Force on the master cylinder which the driver will apply is
calculated as this would be used to calculate the operating
pressure that is applied on the caliper pistons.
6. This pressure is then used to calculate the clamping force that
the caliper will apply on the rotor when the brakes are applied.
7. The required braking torque is then calculated by adding the
torque due to weight transfer and torque supplied by the
powertrain.
8. Using the value of clamping force and required braking torque
the brake rotor diameter was calculated.
9. A single stop temperature rise was calculated for the rotor to
check if adequate heat dissipation is occurring and brake fade
doesn’t occur.
Calculation:
Static front weight: (Total weight*35)/100=112kg(W )
𝑓
Static rear weight: (Total weight*65)/100=208kg(W )
𝑟
Actual weight transfer: (W ) (W*a) /b=943.32N.
𝑡
Dynamic front weight: (W +W )=2042.07N.
𝑓 𝑡
Dynamic rear weight: (W -W )=1097.16N. Master cylinder bore
𝑟 𝑡
area: (3.14*a ²)/4=284.87mm².
𝑚
Caliper bore area: (3.14*d ²)/4=506.45mm².
𝑐
Force on master cylinder: (force on pedal*pedal ratio) =1800N.
Pressure supplied by the master cylinder: Force on master
cylinder/master cylinder bore diameter=6.31 N/mm².
Operating pressure: 6.31N/mm². (Front and Rear)
Clamping force: Front: front operating pressure*caliper front
area*no of piston*coefficient of friction between rotor and pad =
6.31*506.45*0.45*2=2876.12N.
Rear: rear operating pressure*caliper rear area*coefficient of friction
between rotor and pad*no of pistons =
6.31*506.45*2*0.45=2876.12N.
Torque: Front: {(Dynamic front weight/2) *(tire diameter/2)
*coefficient of friction between road and tire + (0.44* Torque supplied
by the drive shaft)} *FOS (1.4) = {(2042.04/2) *(584.2/2) *0.50 +
20000} * 1.4 =236767.95Nmm.
Rear: {(Dynamic rear weight/2) *9.81*(tire diameter/2) *coefficient of
friction between road and tire + (Torque supplied by the drive shaft)}
*FOS (1.4) = {(1097.16/2) *(584.2/2) *0.50 + 45000} * 1.4
=175168.15Nmm.
Effective rotor diameter:
Front: 2*(front tire torque/front clamping force) = (236767.95/
2876.12) =164.64mm.
Rear: 2*(rear tire torque/rear clamping force) = (175168.15/2876.12)
=121.80mm.
Temperature rise during braking:
Stop Energy
The energy dissipated in a stop is the sum of energy from three
sources, kinetic, rotational.
Kinetic Energy

Assuming the stop is from the test speed down to zero then the
kinetic energy is given by: -
KE= =0.5*320*12.5² = 25000 J
0. 5 * 𝑀 * 𝑣²
Where:
KE=kinetic energy (Joules)
M=total vehicle mass (kg)
V=test speed (m/s)
Rotational Energy:
The rotational energy is the energy needed to slow rotating parts. It
varies for different vehicles and which gear is selected however
taking 3% of the kinetic energy is a reasonable assumption.
Potential Energy= 750 J
Braking Power:
To calculate the power, we need to know the brake on time:
T= =12.5/ (9.81) =1.27 s
𝑣/(𝑎 * 𝑔)
Where,
T=brake on time(s)
v= test speed (m/s)
a= deceleration (m/s²)
g= acceleration due to gravity (ms²)
Power:
P=E/t =25750/1.27=20275.59W
Where,
P=average power (Watts)
E=energy (Joules)
t=brakes on time (s)
This is the average power, the peak power at the onset of braking is
double this.
Dry Disc Temperature Rise
These calculations are based on that given in the following
reference:
“Brakes design and safety 2nd
edition” by “Rudolf Limpert”
Single Stop Temperature Rise
To approximate the temperature, rise of the disc an assumption as
to where the energy is going has to be made. Initially most of the
heating takes place in the disc, however this can then be rapidly
cooled by surrounding components and the air stream. The
calculation assumes 80% goes to the disc.
Heat flux into one side of the disc:
q=4P/ ( )=1770548.17W/m²
π * (𝐷
2
− 𝑑²)
Where,
q= heat flux (Watts/m²)
P=average power (Watts)
D=disc useable outside diameter (m)
d=disc useable inside diameter (m)
Single Stop Temperature Rise is
T = 0.527*q* / ( ) +T
𝑚𝑎𝑥 √𝑡 √(ƥ * 𝑐 * 𝑘) 𝑎𝑚𝑏
=97.76 ˚C
Where,
T = maximum disc temperature (°C)
𝑚𝑎𝑥
T =ambient temperature (°C)
𝑎𝑚𝑏
q=heat flux (Watts/m²)
t=brakes on time (s)
Typical Stainless-Steel Figures
ƥ= density of disc material (kg/m³) =7850kg/m³
c= brake disc specific heat capacity (J/kg/K) =530J/kg/K
k= brake disc thermal conductivity (W/ (m*K)) =50.2W/(m*k)
Actual Weight Transfer: 943.32 N
Dynamic Front Weight: Front:2042.04N
Rear:1097.16 N
Clamping Force: Front:2876.12N
Rear:2876.12N
Braking Torque: Front:236767.95Nmm
Rear:175168.15Nmm
Rotor Diameter: Front:164.64mm
Rear: 121.80mm.
Single stop temperature
rise:
97.76˚C
Table-2 Design parameters calculated:
Structural Analysis of the pedal:
A structural analysis of the pedal was done to ensure it undergoes
minimum deformation and minimum stress generation.
In our analysis we found that the pedal under a force of 400N had a
maximum deformation of--- and maximum stress of ---. The stress
was found to be much less than the yield strength of the material
thus ensuring no plastic deformation.
Figure-3 Pedal deformation
Thermal Analysis of the rotor:
A thermal analysis of the brake rotor was done to ensure minimal
heat generation occurs during the application of brakes in a single
stop, this is used to check the probability of occurrence of brake
fade.
In our analysis we found that the maximum temperature generated
when the brakes were applied was 96°C. This is almost equal to our
calculated value of 97.76°C thus further fortifying our design

method.
Figure-4 Thermal analysis
References:
1.Brake design and safety- “Rudolf Limpert”
2.Brakes 2000- “David Barton”
3. Brake Handbook- “Fred Puhn”

6) Steering:
Introduction:
Steering system is basically a combined assembly of steering
wheel, gears, linkage and other components, which controls the
direction of a vehicle in motion. First and the primary function of the
steering system is to achieve angular motion of front wheel to
negotiate a turn. Secondary functions of steering system are It helps
to achieve directional stability of the vehicles when going straight
ahead, to achieve perfect rolling motion on the road, to gain straight
ahead recovery after a turn, to reduce the wear on tyre. So for a
good steering system the steering system must be very accurate
and easy to handle, effort required to steer should be minimal and
the most important is the steering mechanism should provide
directional stability, that is the vehicles should have a tendency to
return back to its straight ahead position. In our steering mechanism
we use Ackermann mechanism.
Material used:
Steering: 6063-T6 Al-Alloy (AA 7075 T6 is an aluminum alloy,
with magnesium and silicon as its alloying elements. Its
composition is maintained by The Aluminum Association. It has
good mechanical properties and is heat treatable and weldable. It
has a density of 2.68g/cm^3 (0.0975 lb/cubic inch)
Tie -rod, Mild steel:(Mild steel, also known as
plain-carbon steel, is the most common form of steel
because its low price and it provides material properties
that are acceptable for many applications. Low-carbon
steel contains approximately 0.05–0.15% carbon which
makes it malleable and ductile. Mild steel has a low
tensile strength, but it is cheap and easy to form;
surface hardness can be increased through carburizing.
Parameters used:
Wheelbase “b” (the distance between the front and rear wheels) of
the vehicle is taken as 60inches.
Front Trackwidth “a” (distance between the front wheels) of the
vehicle is taken as 54inches.
Distance between the two kingpins “c” is taken as 46inches.
Parameters assumed:
Maximum turning angle of the inner front wheel is
assumed as 38 degrees.
Calculated parameters:
For perfect rolling conditions (for all wheels): - COT ∅ − COT 𝜃 = 𝑐/𝑏.
According to Ackermann Steering Mechanism: - SIN (𝛼 + 𝜃) + SIN
(𝛼 − ∅) = 2 SIN 𝛼.
For Perfect(true) Ackermann,
𝛼 = TAN−1
(C/2b)
For over true Ackermann,
First, we take the required condition for turning angle Example: -
take the inner turning angle (𝜃) to its maximum (38°) Then we put it
in the formula
cot ∅ − cot θ= 𝑐/𝑏
And calculate the value of ∅ = 𝟐𝟔.0𝟒°
Then we put it in the equation
SIN (𝛼 + 𝜃) + SIN (𝛼 − ∅) = 2 SIN 𝛼
Then the value of 𝛼 = 27.25° is calculated
Turning radius of the vehicle from common Centre to the outer front
wheel: -
ROF=[2b/sinɸ-(a-c)] =132.44” (3.36m)
Horizontal Components of R: -
R1=[c/2+b/tanθ] = 99.79”
Turning radius of vehicles from common center to the cg point is
R= =102.17”
𝑅1
2
+ 𝑏2
2
( )
Minimum turning radius
(Rmin)=R1-A/2=74.29”
Maximum turning radius
( )=144.25” (3.66M)
(𝑅𝑚𝑖𝑛 + 𝑎)
2
+ (𝑏 + 𝑔)
2
Where, g = radius of tyre(11.5")
Rack travel from Centre to lock= T=2.25"
Length of knuckle arm= T/θ= 86.60MM
Particulars Current Year
Inside Turning Angle at Rolling 38 degrees.
Outside Turning Angle at Rolling 26.04 degrees.
Ackerman Angle 27.25 degrees
Turning Radius (m) 2.59 m
Steering Ratio 8.10
Steering ratio: It the ratio between the angle turned by
the steering wheel and to the corresponding angle
turned by the wheels.
Angle turned by pinion from Centre to
lock, β= (270 deg.)
Steering ratio: (2 )/ ( )] =8.43,
×β θ + ∅
Where, (𝜃 + ∅) is the total lock to lock angle turned by the wheels
Weight on each front tyres: - Total weight of the
vehicles =W(320kg)
Distance of Centre of gravity from the Centre of rear
tyres along wheelbase=(0.563m.),
Total weight shared by front wheels = W =
×
𝑒
𝑏
110.82kg
Maximum cornering force: - The maximum lateral force
(F) transmitted by the rack to the tie rods to steer the
knuckle arms and turn the wheels will be equal to the
maximum cornering force. (Without slipping of tyres).
Where, is the coefficient of friction between ground and
rubber. (approx. 0.85)
F=𝜇 ×G=924.07N
Maximum effort on steering wheel: -Torque transmitted (T) by pinion
and hence by steering wheel,
T= F× PINION DIAMETER(18MM.) =5.54NM
Now, Effort on steering wheel (Z) = tangential force on steering
wheel
Z=T/steering wheel diameter =13.64N
Torsion in the steering column: -

The steering column subjected is to torsion moment
and hence the maximum shear stress developed =
,
τ 𝑚𝑎𝑥
( ) =
𝑇×𝑟
( )
𝐽
= 31. 72𝑀𝑃𝐴
Analysis of tie rod:
1. Axial compressive force (reaction to the steering effort)
=13.64N
2. Bump force (acting perpendicular to the axial force).
STATIC STABILITY FACTOR= / (2 )=1.45
𝑎 ×ℎ
a= track width
h= height of CG from the ground.
Figure-1 Deformation
Figure-2 Stress
Figure-4 Toe vs Rack travel
Figure-4 Camber vs rack travel

6) Powertrain:
Introduction
The drivetrain is the power house of the vehicle, which provides
controlled application of the power. When the vehicle is stationary, it
requires much more inertia than usual to overcome friction. This
generally helps when the vehicle is climbing a hill or employing a
reverse gear. The engine torque supplied is not sufficient enough for
the same, thus something else must be employed to make the
engine torque sufficient enough. Thus, this is where the
transmission system comes in where it provides various torque
multipliers and magnifies the engine torque as required as well as
controls the speed of the vehicle.
The general layout of the powertrain consists of several
components, where the power delivery generally starts from the
Engine/Motor of the vehicle from where the power is delivered to the
gearbox of the vehicle which is used to give a suitable torque or
RPM to the vehicle. The outputs by the gearbox is then transferred
to the wheels with the help of propeller shaft and a differential thus
helping the vehicle move in the desired direction i.e. forward or
reverse.
Type of powertrain used
The wheels of the vehicle can be powered by either mechanical or
electrical transmission. A conventional generally consists of an IC
engine powering the vehicle while in an electrical vehicle a motor
which feeds on the battery is used to give drive to the wheels.
With the boom of electric vehicle all over the vehicle all over the
world, aiming to reduce the damage to the environment we have
decided to make a switch to electrical drivetrain.
Why electric powertrain?
•The first and foremost reason to opt for the electric drivetrain is to
contribute to the reduction of the carbon emission by the vehicles
and also to keep up with the changing dynamics of the automobile
industry.
•As compared to IC engine the electric powertrain provides much
higher starting torque
•The efficiency is claimed to be better.
•Cheap to maintain and run.
Powertrain (e-kit) parts and specification
Parts
▪ Motor
▪ Controller
▪ Main contactor
▪ Accelerator
▪ Gearbox
▪ Battery pack
Specifications
1. Motor
⮚ Max. Power – 4.5 KW
⮚ Max RPM - 4500
⮚ Nominal Current - 93 Amp
⮚ Peak current – 200 Amp
⮚ Nominal torque – 10 N-m
⮚ Peak torque - 38 N-m
⮚ Weight – 11 KG
2. Battery
⮚ Li-ion - 48 V / 110 A-h
⮚ Nominal voltage – 50 V
⮚ Cut off voltage – 40 V
⮚ Charging current – 15 Am
3. Gearbox
⮚ 2 stage reduction
⮚ Gear ratio- 1:10
Electrical circuit
Gearbox Assembly
Calculations (pertaining to battery)
Battery capacity (KWh): -
Assumption 1: - The efficiency of the entire system was considered
to be 80%.
Assumption 2: - The peukert effect for Li-ion battery was taken to be
1.0
Actual capacity = Voltage x Charge x efficiency x peukert effect
= (48 V x 110 A x 0.8 x 1.0)/1000
=4.224 KWh
Battery consumption (Wh/min): -
Assumption 3: - The top speed of the vehicle was considered to be
28 MPH (i.e 45 KMPH) at 48V and 94 Amp rated current
Consumption = (V x I/) Speed of the vehicle
= (48 V x 94 Amp)/ 28 MPH
= 161.14 Wh/min
~ 162 Wh/min
Distance travelled before the battery drains (Km): -

Distance = Actual capacity / Battery consumption
= (4.224 KWh) / (162 Wh/min)
= 26.074 miles
~ 26 miles
~ 41 Km
Ideal horsepower by the motor (HP): -
HP = (V x I)/ 746
= (48 V x 94 A)/ 746
= 6.04 HP
~ 6 HP
Battery charging time
Assumption 4: - The charging current should be 10% of the Ah
rating of battery.
Ideal current requirement for charging for 110Ah Battery = 110 x
(10/100)
=11 Amperes.
considering the losses due to charging we take 15 Amp as the
charging current
Charging time for 110Ah battery = 110 / 15
= 7.33 hr
~ 8 hr (ideal case)
Assumption 5: - for practical purpose taking 40% losses (in
case of battery charging)
=110x (40 / 100) =44 Ah
110 + 44 = 154 Ah (110 Ah + Losses)
Therefore,
Final charging time = Ah/Charging Current
=154 / 15 = 10.266 hr
~11 hr
Calculations pertaining to Vehicle performance
Data we have
Max rpm of the BLDC motor-4500 rpm
Tyre dimension= 23*7*10
Assumption 6: - Max speed of the vehicle = 45 Km/h
= 12.5m/s
Radius of the wheel in metre=0.2921
Therefore,
Velocity of the vehicle = radius x ω
12.5=0.2921*ω
ω=42.955 rad/s
Now rpm,
ω = (2*π*N)/60
42.955 = (2*π*n)/60
N=391.98 rpm
overall reduction ratio at max speed
= 4500/391 .98=12
Therefore Final reduction calculated was = 1:12
Rolling Resistance:
This is the resisting force that opposes the rolling of the tires, which
is caused due to non-elastic effects at the tire-road surface.
Rr = K*W = 0.015*280*9.81 = 41.20 N
K=0.015 for loose unpaved road
W= Weight of Vehicle in Kg (With Driver) =310
Gradient Resistance:
When negotiating a slope, a component of weight acts against the
direction of motion which is proportional to the angle of inclination of
the road surface.
For the evaluation of the torque required to obtain the enough streng
th to climb the 30° incline
Rg = W*sin (α) = 280*9.81*sin (30) = 1373.4 N
Aerodynamic resistance:
Ra = 0.5*density* Af *Cd*(V/3.6) ²
Ra = K*A*(V²)
Af =frontal area,
Cd=co-eff. of aerodynamic resistance
Af = 0.88 m²
K= 0.032 (Average Car)
Speed Aerodynamic drag (N)
Max speed 70.83

Total Resistance:
For levelled road
Total Resistance = rolling resistance + aerodynamic resistance
For inclined road
Speed Total resistance
(levelled road)
Total resistance
(inclined road)
Max speed 112.03 1307.25
Total Resistance=rolling resistance+aerodynamic resistance+gradie
nt resistance
Traction
Te = mean MOTOR torque = 10 N-m
Torque on wheels (Tw) = Te*G*transmission eff.  [Trans eff. = 0.8]
Speed Torque on wheels (N-m)
Max speed 78.84
Tractive Effort
It is the net force available at wheels.
Tractive effort = Tw/r
r=radius of the wheel=D/2=0.2921m
Speed Tractive effort (N)
Max speed 269.03
Acceleration
(At maximum Speed)
a = (Road resistance- Tractive effort) / mass of vehicle
a = (1307.07-269.03) / 310
a = 3.34 m/sec²
Torque on Wheels
The team assumed the incline to be approximately 30 degrees.
Through the inspection of previous courses, as a group we felt this
would be the maximum angle in any hill climb we might encounter.
In order to complete the incline, the force on two wheels will need to
be greater than the component force of gravity along the incline,
which is G1 in the figure below:
Figure-1 FBD for Baja Buggies
G1 = G * sin α = 310* 9.81 * sin 30 = 1177.2
Force per wheel = 588.6
Torque per wheel = 588.6* 𝐷/2 = 588 .6 * 0.2921
= 171.93 Nm
Total torque (𝑇𝑡) = 343.86 Nm+00-l, k
From above we can assume that the minimum torque that needs to
be transferred to the wheels is 343.86 Nm.
Gradeability
The grade ability of a vehicle is expressed as a percentage (%). For

example, the figure 25% means that for a horizontal length of L =
100m, a height of H = 25m can be overcome.
P = [Fz / (9.81 * Gz)] * 100
Where,
P = Grade ability in %
Fz = Tractive force in N at mean engine torque
Gz = Overall combined mass in kg
Fr = Coefficient of rolling resistance = 0.015
Tractive effort at mean engine torque = (19.66 * 18.00 * 0.8) /
(0.2921) = 969.20 N
So, P = [(965.97 / 9.81 * 250)] * 100
=41 %
Ansys Report of Drive shaft
Figure-2 Deformation
The shaft was subjected to a torque of 45N-m. And this
analysis states a maximum deformation of 0.2mm with a
maximum principal stress of 146.56Mpa.
Figure-3 Stress

e-Baja Design Report

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