BAJA SAE, IITK MOTOR
INDIAN INSTITUTE OF
Faculty Advisor
Bhanu Chaturvedi Purwaj Tiwari Ravi Kothari
Sansit Patnaik Somesh Patel
BAJA SAE, IITK MOTORSPORTS
INDIAN INSTITUTE OF TECHNOLOGY,
KANPUR
POWERTRAIN
2014-2015
Faculty Advisor – Dr. Avinash Kumar Agarwal
Submitted By-
Bhanu Chaturvedi Purwaj Tiwari Ravi Kothari
Sansit Patnaik Somesh Patel Tushar Agarwal
SPORTS
TECHNOLOGY,
Dr. Avinash Kumar Agarwal
ABSTRACT
The purpose of the Baja SAE series is to test students against one
another in critical real life situations of engineering design and
management. One of the most influential pieces of this process is the
build and design of the vehicle. A vehicle cannot ru
train. The Powertrain for an off road Baja car is an extremely pivotal
part of the design process and the cars build as a whole. Without a
power train there is no way to transmit the power generated from the
engine to the wheels.
It allows the designers to be constrained to a given space, but still
have to produce maximum performance. The overall goal is to outline
the design and function of a not so typical power train for the Baja
SAE application and discuss its pros and cons as well a
design parameters and discuss the challenges and pitfalls of the
system.
The purpose of the Baja SAE series is to test students against one
another in critical real life situations of engineering design and
management. One of the most influential pieces of this process is the
build and design of the vehicle. A vehicle cannot run without a power
train. The Powertrain for an off road Baja car is an extremely pivotal
part of the design process and the cars build as a whole. Without a
power train there is no way to transmit the power generated from the
ows the designers to be constrained to a given space, but still
have to produce maximum performance. The overall goal is to outline
the design and function of a not so typical power train for the Baja
SAE application and discuss its pros and cons as well as outline the
design parameters and discuss the challenges and pitfalls of the
The purpose of the Baja SAE series is to test students against one
another in critical real life situations of engineering design and
management. One of the most influential pieces of this process is the
n without a power
train. The Powertrain for an off road Baja car is an extremely pivotal
part of the design process and the cars build as a whole. Without a
power train there is no way to transmit the power generated from the
ows the designers to be constrained to a given space, but still
have to produce maximum performance. The overall goal is to outline
the design and function of a not so typical power train for the Baja
s outline the
design parameters and discuss the challenges and pitfalls of the
CONTENT
OVERVIEW………
ENGINE………..
• TOP SPEED……… • ACCELERATION………...6 • HILL CLIMB……….9CVT……….11
GEAR BOX………12
• OBJECTIVE………..12 • GEARS………..12 • SHAFT………...18 • ANALYSIS………27 1. Static Structural 2. Fatigue………30 3. Random Vibration 4. Oil Flow ………..34 • EFFICIENCY……….36CV JOINT………
HUBS………..
• OBJECTIVE………. • FRONT HUB……… • REAR HUB……….. • STEEL SLEEVE………..REFRENCES………..42
………
………..
………3-6 ………...6-9 ……….9……….11
………12
………..12 ………..12-17 ………...18-26 ………27-35 Static Structural………..28 ………30 Vibration……….32 ………..34 ……….36-38………
………..
……….39 ………39 ………..40 ………..41………..42
………3
………..3-10
……….11
………12-38
………39
………..39-41
………..42
OVERVIEW
Powertrain is the mechanism that transmits the drive from the engine of a vehicle to its axle. The components that we’ve used in our design are listed in the following table.
Component Engine Transmission Gear Box CV Shafts Hub Rims Tires
train is the mechanism that transmits the drive from the engine of a vehicle to its axle. The components that we’ve used in our design are listed in the following table.
Name of the Product Briggs and Stratton 10HP Intek OHV Continuously Variable Transmission, CVTech
Custom Rzeppa
Al hubs with 4340 splines
Polaris Sportsman 400 Front Rims – 12X7 Carlisle AT 489 - 23X8X12
Powertrain layout
train is the mechanism that transmits the drive from the engine of a vehicle to its axle.
ENGINE
We are provided with Briggs & Stratton engine with fixed power output of 10hp. This power is to be transmitted to the wheels through the transmission system.
Torque Displacement Weight Bore Stroke Fuel Spark plug
TOP SPEED
Ratio Limited Top Speed
The top speed of B – 16 is determined by taking into account the following parameters which are:
• CVT Overdrive • Gearbox reduction • Wheel effective diameter • Powertrain efficiency
• Maximum achievable engine RPM • Drag force
The following are the calculations for the same: Max Engine RPM: 3600
CVTech CVT Overdrive reduction: Gearbox Reduction:11.5
Wheel Diameter: 23” (= 0.542 m Effective wheel radius*:11.5’’-1’’
(*Compression of tire due to the weight of the vehicle)
We are provided with Briggs & Stratton engine with fixed power output of 10hp. This power is to be transmitted to the wheels through the transmission system.
Torque 19.65Nm /3200RPM Displacement 305 cc Weight 50.4lbs 3.12in Stroke 2.44in Gasoline
Spark plug RC12YC
16 is determined by taking into account the following parameters which
Wheel effective diameter
Maximum achievable engine RPM
The following are the calculations for the same:
reduction: 0.43
m)
1’’ (=10.5’’)
of tire due to the weight of the vehicle)
We are provided with Briggs & Stratton engine with fixed power output of 10hp. This power is
Therefore, theoretical top speed: =
. . .
=19.8 m/s or 71.3 kmph
However the theoretical top speed will be • Rolling friction at tires
• Drag force
• Powertrain efficiency
Assumed powertrain efficiency = 80% • Efficiency of CVT and engine. • Losses due to gear box
Max power delivered at the wheels = 80% of Max power =746 10 0.8
= 5.97 kW
FRICTIONAL FORCE:
Coefficient of rolling friction between tire and sand=0.06 Therefore, value of rolling friction =
DRAG FORCE:
Drag force on a vehicle is equal to = (ρ Where,
Cd (drag coeff.) =1.15
The drag force on B 16 will be mainly due to frontal area speed:
However the theoretical top speed will be less because of the following factors:
Assumed powertrain efficiency = 80% engine.
Max power delivered at the wheels = 80% of Max power
Coefficient of rolling friction between tire and sand=0.06 friction = 250 9.8 0.06 = 147N
ehicle is equal to = (ρ v2 Cd A)/2
ρ (density of air) =1.2kg/m3
1 Fire wall 2 Tires
CALCULATION OF AREA:
Total area of firewall = 0.857 m Area of two tires=0.227m2 Total frontal area = 1.084m^2
Coefficient of v2 = 0.9
At top speed,Power delivered at the wheels 147 V + 0.9 V2 = 5970
V= 15.9 m/s or 57.2 kmph
Drag Force experienced by the car at max achievable top speed =
ACCELERATION
Maximum Acceleration
The maximum acceleration of the car wheels.
F= (Tengine Efficiency Gearbox reduction
tal area of firewall = 0.857 m2
m^2
Power delivered at the wheels = Rate of work done by dissipating forces
Drag Force experienced by the car at max achievable top speed = 227.5 N
acceleration of the car is determined by the maximum torque delivered at the
Gearbox reduction CVT reduction) / Effective radius of wheel = Rate of work done by dissipating forces
is determined by the maximum torque delivered at the
T= (18.4 0.8 11.5 3) =507.8 Nm
Note: 17.9 Nm is the minimum torque generated which peaks to
fairly linear between the required range therefore mean value of torque is taken. Force on the car= T/ radius of the wheel =
F - f = Ma
1940 – 250x9.8x0.06 = 250xa. Therefore, a = 7.1 m/s2
Top speed in acceleration event
The following graph has been taken as reference for the .8 Nm
: 17.9 Nm is the minimum torque generated which peaks to 18.6 Nm. The above graph linear between the required range therefore mean value of torque is taken.
dius of the wheel = .
. . .
N
= 1940 Nin acceleration event
The following graph has been taken as reference for the nature of performance for a CVT.
The above graph is linear between the required range therefore mean value of torque is taken.
The CVT is tuned to engage at optimum RPM (
from the engine. The ratio of the CVT varies linearly with the speed of the vehicle and hence the torque delivered at the wheels.
Minimum force at the wheels =
=
=
.=
-105.2………. (F – f = m
! !F = kv+ c;
c = F
max– f;
-kv + c = ma
-kv + c = mv
! !"#x =
$ !% &The CVT is tuned to engage at optimum RPM (i.e. 2800) in order to exploit the maximum torque ratio of the CVT varies linearly with the speed of the vehicle and hence the torque delivered at the wheels.
Minimum force at the wheels = . = 278N
. (-k)
2800) in order to exploit the maximum torque ratio of the CVT varies linearly with the speed of the vehicle and hence the
Integrating the above equations,
%"
$
= v +
%log (
% &
)
Now, substituting the values of m v = 13.3 m/s or 47.8 km/h TIME
-kv+c = m
! !#t =
'!( % &t = -
$ %log (
% &)
Putting limits for v from 0 to 13.3 (m/s) and substituting values for k, c, m We get t = 3.59s
HILL CLIMB
fs = fr1 +fr2 + W sin θ N1 + N2 = W cos θ
N1, N2 are normal reaction forces. hcg = 516.3 mm
h1 = 730mm h2 = 763mm
Balancing the moment on the contact point of rear wheels N1 (h1 + h2) + 516.3 W sinθ
When the buggy is on the verge equations,
Now, substituting the values of m, c, and k;
Putting limits for v from 0 to 13.3 (m/s) and substituting values for k, c, m
are normal reaction forces.
moment on the contact point of rear wheels θ = W cosθ 730
Tan ()) = . =) = 54.7° W sin θ + μ W cosθ= 1940 Sin θ+ μ Cos θ = 0.79
The value of μ i.e. rolling friction on loose sand is 0.2 Sin θ+ 0.2cosθ = 0.79; θ = 40°
CVT (Continuous Variable T
A CVT is a transmission that can change seamlessly through an infinite number of effective gear ratios between maximum and minimum values. This contrasts with other mechanical
transmission that offers a fixed number of gear
ADVANTAGES
The CVT works by allowing the motor to rev up to a desired engine speed before engaging the drive train and starting vehicle movement.
• Less components in comparison to Manual Transmission thus leading to reduction in weight of car.
• When the vehicle slows down or in case of increased load, the motor RPMs drops from maximum horsepower to maximum available torque. This all
handle obstacles, rough terrain and accelerate out of corners giving the vehicle an advantage in endurance
• Better maneuverability
possibility of errors while driving
(Continuous Variable Transmission)
A CVT is a transmission that can change seamlessly through an infinite number of effective gear ratios between maximum and minimum values. This contrasts with other mechanical
a fixed number of gear ratios. CVTech was used as cvt for this season.
The CVT works by allowing the motor to rev up to a desired engine speed before engaging the drive train and starting vehicle movement. This leads to better fuel efficiency.
Less components in comparison to Manual Transmission thus leading to reduction in
When the vehicle slows down or in case of increased load, the motor RPMs drops from horsepower to maximum available torque. This allows the vehicle to better obstacles, rough terrain and accelerate out of corners giving the vehicle an
endurance event.
Better maneuverability as the driver doesn’t have shift 5+1 gears and hence less possibility of errors while driving
A CVT is a transmission that can change seamlessly through an infinite number of effective gear ratios between maximum and minimum values. This contrasts with other mechanical
or this season.
The CVT works by allowing the motor to rev up to a desired engine speed before engaging the Less components in comparison to Manual Transmission thus leading to reduction in
When the vehicle slows down or in case of increased load, the motor RPMs drops from ows the vehicle to better obstacles, rough terrain and accelerate out of corners giving the vehicle an
GEARBOX
DESIGN OBJECTIVES
1) To minimize the weight in order to increase the power to weight ratio of the buggy. 2) Final output reduction of 11.5.
3) To efficiently withstand the N
maintaining a factor of safety of more than 2 4) Compact and elegant design which
The main reason for using custom
• After discussing with team drivers, we came to a conclusion that differential and reverse can be sacrificed as they were
• As this is our second season in • As compared to 16.5kg DANA, our
10.5 kg hands down. GEARBOX SPECIFICATION Type Reduction Material Weight
GEARS
The gear calculations were based on theory and f Machining and NEPTEL lectures on gear analysis. DETERMINATION OF GEAR RATIOS AND INTERFERENCE
To do this we have to assume number of teeth on one
the relation given below we can determine number of teeth on other gear
DESIGN OBJECTIVES
minimize the weight in order to increase the power to weight ratio of the buggy. output reduction of 11.5.
efficiently withstand the Normal stress, Shear stress and Torsion with Structural R taining a factor of safety of more than 2.
and elegant design which is tough at the same time.
The main reason for using custom gearbox instead of DANA Spicer H12 is:
After discussing with team drivers, we came to a conclusion that differential and reverse were not required on the BAJA track.
As this is our second season in BAJA, reverse comes with risk of failure during
DANA, our gear box weighs only 6kg, giving a reduction in weight of
Type Forward reduction
Reduction 11.5
Material AL 6061-T6/ AISI 9310
Weight* 6kg
The gear calculations were based on theory and formulae mentioned in the Shigley and NEPTEL lectures on gear analysis.
DETERMINATION OF GEAR RATIOS AND INTERFERENCE
To do this we have to assume number of teeth on one gear (T1), say the smaller gear. Now using the relation given below we can determine number of teeth on other gear, T2.
minimize the weight in order to increase the power to weight ratio of the buggy.
ormal stress, Shear stress and Torsion with Structural Rigidity by
After discussing with team drivers, we came to a conclusion that differential and reverse comes with risk of failure during competition. gear box weighs only 6kg, giving a reduction in weight of
(*expected)
ormulae mentioned in the Shigley Handbook of
So we got number of teeth on both the gears, but one should als
system has to have a smooth operation. Interference happens when gear teeth has got profile below base circle. This will result
shown in following figure.
A pair of gear
For minimum interference, the pinion should have a minimum number of teeth specified by following relation.
Where aw represents addendum
taken by designers) aw = 1 m and b defined as follows.
If this relation does not hold for a given case, then one has to increase number of teeth T1, and redo the calculation. The algorithm for deciding number of teeth T1
So we got number of teeth on both the gears, but one should also check for interference, system has to have a smooth operation. Interference happens when gear teeth has got profile below base circle. This will result high noise and material churning problem. This phenomenon is
A pair of gear teeth under interference
the pinion should have a minimum number of teeth specified by
addendum of tooth. For 20 degree pressure angle (which is normally = 1 m and bw = 1.2 m. Module m, and pitch circle diameter Pd are
If this relation does not hold for a given case, then one has to increase number of teeth T1, and redo the calculation. The algorithm for deciding number of teeth T1 and T2 is shown below.
interference, if gear system has to have a smooth operation. Interference happens when gear teeth has got profile
problem. This phenomenon is
the pinion should have a minimum number of teeth specified by
which is normally = 1.2 m. Module m, and pitch circle diameter Pd are
If this relation does not hold for a given case, then one has to increase number of teeth T1, and and T2 is shown below.
Flow chart to determine number of teeth on each gear
THE LEWIS BENDING EQUATION
Flow chart to determine number of teeth on each gear
KV Pinion 3.26 Intermediate 1 3.26 Intermediate 2 1.75 Gear 1.75
DYNAMIC EFFECTS
When a pair of gear rotates we often hear noise from this, this is due to collision happening between gear teeth due to small clearance in be
run.
If a pair of gears failed at 500 lb. then a velocity factor, designated another, identical, pair of gears run
load equal to twice the tangential or transmitted load. This effect is incorporated in dynamic loading factor, velocity.
K
v=
& KO KM Y 1.25 1.6 0.43 1.25 1.6 0.43 1.25 1.6 0.43 1.25 1.6 0.302When a pair of gear rotates we often hear noise from this, this is due to collision happening between gear teeth due to small clearance in between them. Such collisions have impact
lb. tangential load at zero velocity and at 250 lb. at velocity V a velocity factor, designated Kv, of 2 was specified for the gears at velocity V1. Then another, identical, pair of gears running at a pitch-line velocity V1 could be assumed to have a
to twice the tangential or transmitted load.
ed in dynamic loading factor, Kv value of which is a function of pitch line
0.43 0.43 0.43 0.302
When a pair of gear rotates we often hear noise from this, this is due to collision happening collisions have impact in long
at velocity V1, . Then could be assumed to have a
At root of the gear there could be fatigue failure due to stress concentration effect. which is incorporated in a factor called K
There will be factors to check for overload ( incorporating all these factors Lewis stre
The above equation can also be represented in an alternating form (AGMA Strength equation) like shown below
Where J is
Using above equation we can solve for value of parameters required for gear design.
unless it does not a have enough surface resistance.
All the above calculations have been done over a number of iterations using a code on MATLAB. The code was designed such that one could get the mi
RESULT
Gear material=AISI 9310 Gear Module = 2
Pressure Angle = 20 degrees
Number of teeth Pinion
Intermediate 1 Intermediate 2
Gear
At root of the gear there could be fatigue failure due to stress concentration effect. corporated in a factor called Kf value of which is more than 1.
for overload (Ko) and load distribution on gear tooth (K
incorporating all these factors Lewis strength equation will be modified like this
The above equation can also be represented in an alternating form (AGMA Strength equation)
Using above equation we can solve for value of b, so we have obtained all the output parameters required for gear design. But such a gear does not guarantee a peaceful unless it does not a have enough surface resistance.
All the above calculations have been done over a number of iterations using a code on MATLAB. The code was designed such that one could get the minimum weight over a range of gear ratio.
Number of teeth Face Width Max stress
18 0.5’’ 1352Mpa
66 0.5’’ 657Mpa
22 1’’ 1595Mpa
66 1’’ 1677Mpa
At root of the gear there could be fatigue failure due to stress concentration effect. Effect of
d distribution on gear tooth (Km). While
The above equation can also be represented in an alternating form (AGMA Strength equation)
so we have obtained all the output
peaceful operation
All the above calculations have been done over a number of iterations using a code on MATLAB. nimum weight over a range of gear ratio.
Max stress 1352Mpa
657Mpa 1595Mpa 1677Mpa
DESIGN FOR SURFACE RESISTANCE
Usually failure happens in gears due to lack of surface resistance, this is also known as pitting failure. Here when 2 mating surfaces come in contact under a specified load a contact stress is developed at contact area and surfaces get deformed. A simple case of contact stress
development is depicted below, where 2 cylinders come in contact under a load
Surface deformation and development of surface stress due to load applied
For a gear tooth problem one can
If contact stress developed in a gear interface is more than a critical value (specified by AGMA standard), then pitting failure occurs. So designer has to make sure that this condition does no arise. This problem can be dealt with by Case
DESIGN FOR SURFACE RESISTANCE
Usually failure happens in gears due to lack of surface resistance, this is also known as pitting failure. Here when 2 mating surfaces come in contact under a specified load a contact stress is
area and surfaces get deformed. A simple case of contact stress development is depicted below, where 2 cylinders come in contact under a load F.
Surface deformation and development of surface stress due to load applied
For a gear tooth problem one can determine contact stress as function of following parameters
If contact stress developed in a gear interface is more than a critical value (specified by AGMA standard), then pitting failure occurs. So designer has to make sure that this condition does no
be dealt with by Case Hardening (case depth =0.9mm)
Usually failure happens in gears due to lack of surface resistance, this is also known as pitting failure. Here when 2 mating surfaces come in contact under a specified load a contact stress is
area and surfaces get deformed. A simple case of contact stress
F.
Surface deformation and development of surface stress due to load applied
determine contact stress as function of following parameters
If contact stress developed in a gear interface is more than a critical value (specified by AGMA standard), then pitting failure occurs. So designer has to make sure that this condition does not
SHAFTS
Material of shaft is same as gears=AISI 9310 For high bending strength and low weight
created using matlab code which takes into account bending moment and factor of safety 2 Pinion is embedded on first shaft because of small diameter and the s
rotate. And therefore it does not requir
of weight. Final shaft has the max OD due to heavy bending and torsion moment it has to transfer.
A manual calculation also takes grooves. Stress concentration due to factor of safety grooves, retaining ring Module of splines =1.25
Weight
Shaft 1 272 g
Shaft 2 143 g
Shaft 3 478 g
Material of shaft is same as gears=AISI 9310
For high bending strength and low weight, all shafts are kept hollow. Dimensions of shafts were created using matlab code which takes into account bending moment and factor of safety 2
shaft because of small diameter and the second shaft does
And therefore it does not require splines, this configurations saves two bearings and lot shaft has the max OD due to heavy bending and torsion moment it has to
calculation also takes in account all stress concentration factors due to steps and . Stress concentration due to grooves is almost 3 times higher than steps, so considering
retaining ring are avoided.
Weight Length Deflection(mm)
272 g 129mm 0.0042
g 76mm 0.0011
478 g 173 mm 0.035
hollow. Dimensions of shafts were created using matlab code which takes into account bending moment and factor of safety 2+
shaft does not configurations saves two bearings and lot shaft has the max OD due to heavy bending and torsion moment it has to
o steps and is almost 3 times higher than steps, so considering
Max stress 28Mpa 33.3Mpa
0, -700 -1000 -500 0 500 1000 1500 2000 2500 3000 3500 4000 0 20 S h e a r F o rc e i n N 84.5, -700 84.5, 3354.69 97, 3354.69 103, 565.75 103, -565.75112.2, 40 60 80 100 Length (mm)
Shaft 1
97, 3354.69 103, 565.75 565.75112.2, -565.75 120Schematic representation of gear assembly
Schematic representation of gear assembly
Calculations were started using Michigan’s design report. Later Machinery’s Handbook was used as reference book for all calculations. To start with, metric module was used as standard for calculations. Following data was used as starting point for calculations; it is based on standards followed by most of the teams.
Metric module (m) = 1.25 Number of teeth (N) = 26 Pitch Diameter (PD) = 32.5 mm Now,
Max. Outer diameter (flat root tooth) Min. minor diameter (flat root tooth) Tooth thickness *
+* 1.9625 mm ~
Shear stress at pitch diameter (
Taking uniform shock, Ka = 1
For fixed splines, Km = 1 Therefore,
, * 59.26 MPa
Taking Factor Of Safety (FOS) = 2,
, * 120 MPa
Taking factor of 0.5 for Kf and 2 for K
SPLINE CALCULATION
Calculations were started using Michigan’s design report. Later Machinery’s Handbook was used reference book for all calculations. To start with, metric module was used as standard for calculations. Following data was used as starting point for calculations; it is based on standards
Pitch Diameter (PD) = 32.5 mm
Max. Outer diameter (flat root tooth) *-& .
+ * 34.375 mm ~ 34 mm Where, P = 1/m
Min. minor diameter (flat root tooth) *- .+ * 30.625 mm ~ 31 mm 1.9625 mm ~ 2mm
Shear stress at pitch diameter (, ) *2-3./0/1
4 /5 Where,
T = torque
Ka = Spline application factor Km = Load distribution factor
D = Pitch diameter
N = Number of tooth in actual contact Le = Max. Effective length of spline t = tooth thickness
Kf = Fatigue life factor
ctor Of Safety (FOS) = 2,
and 2 for Ka,
Calculations were started using Michigan’s design report. Later Machinery’s Handbook was used reference book for all calculations. To start with, metric module was used as standard for calculations. Following data was used as starting point for calculations; it is based on standards
34.375 mm ~ 34 mm Where, P = 1/m
= Spline application factor = Load distribution factor D = Pitch diameter
N = Number of tooth in actual contact = Max. Effective length of spline
t = tooth thickness = Fatigue life factor
, * 480 MPa
Taking 9310 Steel as material for manufacturing splined shaft,
,$ " * 6789: ;<=8;; ∗ 0.577
= 550 MPa
Changing the initial parameters, N = 24
PD = 30 mm
Major diameter = 31.5 mm Minor diameter (Dre) = 28.5 mm Shaft diameter = 27
Therefore taking FOS=2,
, * 557 MPa Sheer stress calculation:
Sheer stress under roots of external teeth (S
Therefore taking FOS = 1.5, Ss = 504.77 MPa
Changing some parameters, Minor diameter (Dre) = 31 mm Ss = 392.23 MPa
Taking 9310 Steel as material for manufacturing splined shaft,
s,
) = 28.5 mm
Sheer stress under roots of external teeth (Ss) * 2./0
B4C/5 Where,
T = torque
Ka = Spline application factor Dre = minor diameter Kf = Fatigue life factor
= Spline application factor = minor diameter = Fatigue life factor
Hence,
Taking stress concentration factor = 3 Ss exceeds ,$ " limit of 530 MPa
Calculation for Compressive stress:
Compressive stress on sides of spline tooth for fixed splines (S Therefore, Case 1: Taking N = 26, D = 32, Ka = 1, h=0.9m = 1.125 mm, K Sc = 14.29 MPa FOS = 3.7 Case 2: Taking N = 24, D = 30, Ka = 1, h=0.9m = 1.125 mm, K Sc = 16.51 FOS = 3.11
Taking stress concentration factor = 3 limit of 530 MPa
Calculation for Compressive stress:
Compressive stress on sides of spline tooth for fixed splines (Sc) =
./1/0
2-34D/5 Where,
T = Torque
Km = Load distribution factor Ka = Spline application factor D = Pitch diameter N = number of teeth Le = Max effective Spline length h = 0.9m (flat root) = m (fillet root) = 1, h=0.9m = 1.125 mm, Kf = 0.5 = 1, h=0.9m = 1.125 mm, Kf = 0.5 Where, T = Torque
= Load distribution factor = Spline application factor
D = Pitch diameter N = number of teeth
tive Spline length h = 0.9m (flat root)
Now,
According to a book max. permissible S unitary analogue,
For max. Shear stress = 530 MPa , max. S Hence in the both the cases Sc
Rack profile for 37.5
ermissible Sc = 4000 psi for max shear stress = 40,000 psi
stress = 530 MPa , max. Sc = 53 MPa is in safe limit.
Rack profile for 37.5⁰ pressure angle spline
ANALYSIS
The gearbox had been subjected to various life like situations that involved fatigue, oil flow and random vibrations. This set of analysis has been co
The objective behind these analysi gearbox.
Reasons for failure of gearbox- 1. Deformation of casing 2. Heating of gears
3. Churning of gears due to shock loading
The gearbox had been subjected to various life like situations that involved fatigue, oil flow and random vibrations. This set of analysis has been conducted in a virtually on the software ANSYS.
these analysis was to determine the durability and efficiency of the
of casing
of gears due to shock loading
The gearbox had been subjected to various life like situations that involved fatigue, oil flow and nducted in a virtually on the software ANSYS. was to determine the durability and efficiency of the
STATIC STRUCTURAL
As our design objective is to reduce weight and durability. After analysis, casing material is chosen to be Al 6061 T6 for its light weight and high strength properties .Casing has been design by keeping in mind the minimum deformation the shafts can bear. Forces were
according to results drawn from hand calculations. MAX STRESS= 116MPA
MINIMUM FACTOR OF SAFETY=2.4
As our design objective is to reduce weight and durability. After analysis, casing material is light weight and high strength properties .Casing has been design by keeping in mind the minimum deformation the shafts can bear. Forces were applied
to results drawn from hand calculations.
MINIMUM FACTOR OF SAFETY=2.4
As our design objective is to reduce weight and durability. After analysis, casing material is light weight and high strength properties .Casing has been design
MAXIMUM ELASTIC DEFORMATION=0.26MM
DESIGN EVOLUTION OF GEAR BOX
Iteration of changes in gearbox casing is solely based on reducing MAXIMUM ELASTIC DEFORMATION=0.26MM
DESIGN EVOLUTION OF GEAR BOX
FATIGUE
It has been statistically estimated in the industry that 50
fatigue. In order to overcome this catastrophic result one has to perform random and repeated loading and unloading operations on the prototype of the desi
gearbox on Solidworks we imported the model to ANSYS to
We started the process by setting the model according to the forces that our gearbox will experience when it is functioning at the maximum possible load. The various forces will be transferred to the casing through the beari
been calculated previously in the calculations.
It has been statistically estimated in the industry that 50-90% of structural failure is due to fatigue. In order to overcome this catastrophic result one has to perform random and repeated
unloading operations on the prototype of the designed product. After designing the we imported the model to ANSYS to proceed with further analysis. We started the process by setting the model according to the forces that our gearbox will experience when it is functioning at the maximum possible load. The various forces will be transferred to the casing through the bearings on which shafts are loaded. These forces have been calculated previously in the calculations.
Scale of deformation =536: 1
Maximum deformation=0.04mm
is due to fatigue. In order to overcome this catastrophic result one has to perform random and repeated
After designing the proceed with further analysis. We started the process by setting the model according to the forces that our gearbox will experience when it is functioning at the maximum possible load. The various forces will be
Minimum fatigue life=2718hrs Minimum fatigue life=2718hrs
RANDOM VIBRATIONS
Determining the fatigue life of parts under periodic, sinusoidal vibration is a fairly straight forward process in which damage content is calculated by multiplying the stress amplitude of each cycle from harmonic analysis with
field. The computation is relatively simple because the absolute value of the vibration is highly predictable at any point in time. Vibrations may be random in nature in a wide range of applications, however, such as vehicles traveling on rough roads or industrial equipment operating in the field where arbitrary loads may be encountered. In these cases, instantaneous vibration amplitudes are not highly predictable as the amplitude at any point in time is
related to that at any other point in time. The example of one such random vibration is forces experienced on our gearbox. These can be due to vibrations in engine, forces propagated on the gearbox via chassis, etc.
These vibrations show drastic resul
vibration frequencies do not coincide with the modal frequencies.
The least fundamental mode for random vibrations is
frequencies to which are design will be subjected. The following are the screen shots to the accelerations to which the design will be subjected.
Determining the fatigue life of parts under periodic, sinusoidal vibration is a fairly straight forward process in which damage content is calculated by multiplying the stress amplitude of each cycle from harmonic analysis with the number of cycles that the parts experience in the field. The computation is relatively simple because the absolute value of the vibration is highly predictable at any point in time. Vibrations may be random in nature in a wide range of
wever, such as vehicles traveling on rough roads or industrial equipment operating in the field where arbitrary loads may be encountered. In these cases, instantaneous vibration amplitudes are not highly predictable as the amplitude at any point in time is
related to that at any other point in time. The example of one such random vibration is forces experienced on our gearbox. These can be due to vibrations in engine, forces propagated on the
c results on the body, hence it is necessary to ensure that these frequencies do not coincide with the modal frequencies.
The least fundamental mode for random vibrations is 1011.6 Hz which is way more than the frequencies to which are design will be subjected. The following are the screen shots to the
to which the design will be subjected.
Determining the fatigue life of parts under periodic, sinusoidal vibration is a fairly straight forward process in which damage content is calculated by multiplying the stress amplitude of
the number of cycles that the parts experience in the field. The computation is relatively simple because the absolute value of the vibration is highly predictable at any point in time. Vibrations may be random in nature in a wide range of
wever, such as vehicles traveling on rough roads or industrial equipment operating in the field where arbitrary loads may be encountered. In these cases, instantaneous vibration amplitudes are not highly predictable as the amplitude at any point in time is not related to that at any other point in time. The example of one such random vibration is forces experienced on our gearbox. These can be due to vibrations in engine, forces propagated on the
to ensure that these
which is way more than the frequencies to which are design will be subjected. The following are the screen shots to the PSD
OIL FLOW ANALYSIS
The gearbox transfer rotational kinetic energy coming from engine through CVT to drive shaft. The gear assembly must be well lubricated in all conditions for the following
1. Ensure that the gears maintain structural integrity (producing minimum friction for ensuring that gear teeth do not get damaged).
2. Avoid oil over-heating.
3. Ensure minimal churning losses (losses generated by oil’s being stirred by rotating gears).
Main objective of CFD model prediction is the optimization of the oil flow:
1) To ensure, the oil reaches to every region of gear contact. 2) Reduce the friction between the gearwheels (pitting) 3) Minimization of load
4) Assessment of wall effects on gear housing
The gearbox transfer rotational kinetic energy coming from engine through CVT to drive shaft. gear assembly must be well lubricated in all conditions for the following-
that the gears maintain structural integrity (producing minimum friction for that gear teeth do not get damaged).
heating.
minimal churning losses (losses generated by oil’s being stirred by rotating
Main objective of CFD model prediction is the optimization of the oil flow:
ensure, the oil reaches to every region of gear contact. the friction between the gearwheels (pitting)
of load-independent spin power losses of wall effects on gear housing
The gearbox transfer rotational kinetic energy coming from engine through CVT to drive shaft.
that the gears maintain structural integrity (producing minimum friction for
ANSYS 16.0 and CFX as solver Gear Oil – SAE75W-140 Oil level -30mm
@ running temperature-60°C Optimum oil level =30mm
COMPARISON OF DIFFERENT OIL LEVELS Oil level Max oil velocity(m/s) 10mm 9.5
20mm 9.42 30mm 9.36 40mm 9.28
Oil level increase comes at a cost of 1) Weight
2) Viscous drag
Streamline velocity curve for 10 mm oil level COMPARISON OF DIFFERENT OIL LEVELS
Max oil velocity(m/s) Max oil pressure(103Pa) Reach(out of 10)
1.5 2
4.16 5
5.5 9
7.824 9.5
Oil level increase comes at a cost of
Streamline velocity curve for 10 mm oil level
EFFICIENCY OF GEARBOX
In calculating efficiency of gearbox various losses are accounted 1)Tooth engagement loss
2)Oil churning loss 3)Bearing loss 4) Seal frictional loss
Total Power Loss (Pt) = Lt + Lch + L
Lt = Power loss at tooth
Lch= Power loss due to churning (hp) Lb= Power loss at bearing (hp) Ls= Seal friction power loss (hp)
CALCULATIONS FOR LT: Lt = WEF G H. I &. J hp 1ST ENGAGEMENT: Lt = 8.2E . I .. J MN = 0.6984 hp 2ND ENGAGEMENT: Lt = 8.2E . I . . J MN = 1.028 hp
EFFICIENCY OF GEARBOX
calculating efficiency of gearbox various losses are accounted- Tooth engagement loss
+ Lb + Ls Where,
= Power loss at tooth engagement (hp) = Power loss due to churning (hp) = Power loss at bearing (hp) = Seal friction power loss (hp)
:
Where, W = Total power
Z1 = Number of pinion tooth ψ = Helix Angle
V = Terminal velocity = Number of pinion tooth
CALCULATIONS FOR LCH Lch = OPQ EF &FRJ . 10 TU 1ST ENGAGEMENT: Lch = 0.009 Q 1.768 E . & = 5.7 x 10-3 hp 2ND ENGAGEMENT: Lch = 0.009 Q 0.482 E . & = 1.586 x 10-3 hp CALCULATIONS FOR LB: Lb = 5.23 10 \]^:_ `U 1ST SHAFT BEARINGS: Left Right F = 3665 N F = 581.60 CH: Where, c = splash lubrication b = face width (mm) V = Terminal Velocity
µ = viscosity of oil at operating temperature Z1 and Z2 = number of teeth in engaging gears
RJ . 10 TU
R
J . 10 TU
:
Where,
F = radial load on bearing (N) fb= coefficient of friction
d= shaft diameter (m) n = shaft speed (rpm)
Left Right
F = 3665 N F = 581.60 N
µ = viscosity of oil at operating temperature = number of teeth in engaging gears
fb = 0.01 (worst case) f d = 22 mm n = = rpm
Lb = 0.053 hp L
LAST SHAFT BEARINGS:
Left Right F = 6371 N F = 3280 N fb = 0.01 (worst case) d = 25 mm d = 25 mm n = rpm Lb = 9.533 x 10 -3 hp L CALCULATIONS FOR LS: a * ]N =Ts = Seal torque (Nm) ] * b cd ^ Be / Therefore, Ls = 0.0250 hp If we take, Engine power = 8.2 hp
Total power loss = 1.9hp, hence
= 0.01 (worst case) fb = 0.01 (worst case) d = 22 mm d = 22 mm n = rpm = 0.053 hp Lb = 8.33 x 10 -3 hp Left Right F = 6371 N F = 3280 N
= 0.01 (worst case) fb = 0.01 (worst case) d = 25 mm d = 25 mm n = rpm hp Lb = 4.767 x 10 -3 hp : = Seal torque (Nm)
g= angular velocity of shaft f = seal friction
pr = radial lip load (N) r = radius of shaft (m)
hence, Efficiency of gearbox = 77%
CV JOINT
Constant-velocity joints (CV joints) allow a drive shaft to transmit power through a variable angle, at constant rotational speed, without an appreciable increase in friction.
CV joints are mainly used in front wheel
independent rear suspension typically use CV joints at the ends of the rear axle half shafts
RZEPPA JOINT
It is a 6 ball continuous velocity joint giving
heavier than tripod joint and difficult to maintain in long wasn’t good so this season Rzeppa
PLUNGING DATA
1)Right hand shaft requires max. 2) Left hand shaft requires max.
HUBS
The wheel hub is that part of car on which the wheels and the be able to absorb loads from braking and cornering
accommodate bearings which support
DESIGN OBJECTIVES
We had the following targets in mind while designing the hub: 1. To make it strong enough to bear the forces during
2. To bear the force exerted on the rotor by the 3. To reduce the weight of the hub with
FRONT HUB
Front hub are designed by considering braking
free fall. After analyzing forces on damper, brake rotor and application was considered for analysis
1)Remote cornering force of 1250N. 2)Remotecurb force of 5000N. 3)Braking torque of 249Nm.
velocity joints (CV joints) allow a drive shaft to transmit power through a variable angle, at constant rotational speed, without an appreciable increase in friction.
CV joints are mainly used in front wheel drive, and many modern rear wheel drive cars with independent rear suspension typically use CV joints at the ends of the rear axle half shafts
It is a 6 ball continuous velocity joint giving 22° of working angle with 50mm of plunging.
heavier than tripod joint and difficult to maintain in long run. As our past experience with tripod Rzeppa are used.
ight hand shaft requires max. Plunge=1in shaft requires max. Plunge=1.02in
hub is that part of car on which the wheels and the brake rotor are mounted. It must o absorb loads from braking and cornering, and allow the wheel to spin freely. It must
which supports the spindle and CVshaft.
We had the following targets in mind while designing the hub:
1. To make it strong enough to bear the forces during curb and cornering 2. To bear the force exerted on the rotor by the caliper
3. To reduce the weight of the hub without compromising on the strength.
Front hub are designed by considering braking torque, cornering forces and extreme cases of analyzing forces on damper, brake rotor and tire; the following cases for force
ered for analysis- cornering force of 1250N.
velocity joints (CV joints) allow a drive shaft to transmit power through a variable
drive, and many modern rear wheel drive cars with independent rear suspension typically use CV joints at the ends of the rear axle half shafts
plunging. It is our past experience with tripod
are mounted. It must freely. It must
forces and extreme cases of following cases for force
Material used for analysis is Al 6061 and
REAR HUB
As compared to last season steel hub this year Al 6061 another steel sleeve embedded with splines was press fit in
Red colour depicts the area where FOS is below 1.5
Material used for analysis is Al 6061 and weight of hub is 530gram.
As compared to last season steel hub this year Al 6061-T6 hubs were designed. For splines another steel sleeve embedded with splines was press fit in Aluminium hub.
Red colour depicts the area where
Greenish area shows stress of 200MPa
Initial hub design with 650 g of weight and requires large bearing as compare to its successor .This designed also failed due to excessive stress concentration on edges
T6 hubs were designed. For splines Greenish area shows stress of
Initial hub design with 650 g of requires large bearing as compare to its successor .This designed also failed due to excessive stress concentration on
STEEL SLEEVE
Aluminium is not reliable for splines,
whole hub of steel, a sleeve is press fit into it for accommodating splines. The interference for press fit was calculated using torque transfer and fricti
between Aluminium and steel µ is 0.61 and calculated interference is 0.01mm for the torque transfer of 350NmTtotal weight of sleeve and hub is 800g which is 30% of last year steel hub.
Area below FOS of 1.5 is shown in red and the total weight is 440 g.
splines, especially with low module so instead of manufacturing sleeve is press fit into it for accommodating splines.
The interference for press fit was calculated using torque transfer and friction coefficient µ is 0.61 and calculated interference is 0.01mm for the torque ansfer of 350NmTtotal weight of sleeve and hub is 800g which is 30% of last year steel hub. Area below FOS of 1.5 is shown in
and the total weight is 440 g. Green area depicts stress of 230Mpa
Press fit analysis shows max stress generated is on which is around 1000MPa so instead of manufacturing
on coefficient µ is 0.61 and calculated interference is 0.01mm for the torque ansfer of 350NmTtotal weight of sleeve and hub is 800g which is 30% of last year steel hub.
Green area depicts stress of
Press fit analysis shows max stress generated is on the edge which is around 1000MPa.
REFRENCES
•
www.briggsandstratton•
students.sae.org/cds/bajasae•
nptel.ac.in/courses/IIT•
nptel.ac.in/courses/IIT•
Shigley's mechanical engineering design•
www.ansys.com/Products/Simulation+Technology/.../• press fit-intro to mechanics of solids
•
spline-machinary handbook•
shaft-mechanics of solid –•
predicting fatigue life•
Machinery’s handbook•
Peterson stress concentration•
http://learntoengineer.com/beam•
http://www.amesweb.info/StressConcentrationFactor/StressConcentration Factors.aspx briggsandstratton.com/engines-racing/racing-engines/.../model-bajasae/ .ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/2_7.pd .ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/2_17.pdfhigley's mechanical engineering design
.com/Products/Simulation+Technology/.../ANSYS+CFX
intro to mechanics of solids-crandall
machinary handbook – crandal
predicting fatigue life-RKHolman and PK liaw Machinery’s handbook
Peterson stress concentration book http://learntoengineer.com/beam
http://www.amesweb.info/StressConcentrationFactor/StressConcentration
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