BAJA SAE, IITK MOTOR MOTORSPORTS SPORTS INDIAN INSTITUTE OF TECHNOLOGY, KANPUR POWERTRAIN 2014-2015
Faculty Advisor – Dr. Avinash Kumar Agarwal Submitted By-
Bhanu Chaturvedi Sansit Patnaik
Purwaj Tiwari Somesh Patel
Ravi Kothari Tushar Agarwal garwal
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 run 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 engine to the wheels. It allows 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 a as s outline the design parameters and discuss the challenges and pitfalls of the system.
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CONTENT OVERVIEW……………………………………………… ………………………………………………3 ……………………………………………… ENGINE………………………………………………….. …………………………………………………..3-10 ………………………………………………….. • • •
TOP SPEED……………………………………………… ………………………………………………3-6 ACCELERATION………………………………………...6 ………………………………………...6-9 HILL CLIMB……………………………………………….9 ……………………………………………….9
CVT……………………………………………………….11 ……………………………………………………….11 GEAR BOX………………………………………………12 ………………………………………………12-38 ………………………………………………12 • • • •
•
OBJECTIVE……………………………………………..12 ……………………………………………..12 GEARS…………………………………………………..12 …………………………………………………..12-17 SHAFT…………………………………………………...18 …………………………………………………...18-26 ANALYSIS………………………………………………27 ………………………………………………27-35 1. Static Structural Structural………………………..28 2. Fatigue…………………………………30 …………………………………30 3. Random Vibration Vibration…………………….32 4. Oil Flow ………………………………..34 EFFICIENCY…………………………………………….36 …………………………………………….36-38
CV JOINT………………………………………………… …………………………………………………39 ………………………………………………… ……………………………………………………..39-41 …………………………………………………….. HUBS…………………………………………………….. • • • •
OBJECTIVE……………………………………………. …………………………………………….39 FRONT HUB…………………………………………… ……………………………………………39 REAR HUB…………………………………………….. ……………………………………………..40 STEEL SLEEVE……………………………………….. ………………………………………..41
REFRENCES……………………………………………..42 ……………………………………………..42
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OVERVIEW Powertrain 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. Component
Name of the Product
Engine
Briggs and Stratton 10HP Intek OHV
Transmission
Continuously Variable Transmission, CVTech
Gear Box
Custom
CV Shafts
Rzeppa
Hub
Al hubs with 4340 splines
Rims
Polaris Sportsman 400 Front Rims – 12X7
Tires
Carlisle AT 489 - 23X8X12
Powertrain layout
3
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
19.65Nm /3200 RPM
Displacement
305 cc
Weight
50.4lbs
Bore
3.12in
Stroke
2.44in
Fuel
Gasoline
Spark plug
RC12YC
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: 0.43 Gearbox Reduction:11.5 Wheel Diameter: 23” (= 0.542 m m) Effective wheel radius*:11.5’’-1’’ 1’’ (=10.5’’) (*Compression of tire due to the weight of the vehicle)
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Therefore, theoretical top speed: =
.
.
.
=19.8 m/s or 71.3 kmph
However the theoretical top speed will be less because of the following factors: • • •
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 = 250 9.8 0.06 = 147N
DRAG FORCE: Drag force on a vehicle ehicle is equal to = (ρ v2 Cd A)/2 Where,
ρ (density of air) =1.2kg/m3
Cd (drag coeff.) =1.15
The drag force on B 16 will be mainly due to frontal area (A) of:
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1 Fire wall 2 Tires CALCULATION OF AREA: Total tal area of firewall = 0.857 m2 Area of two tires=0.227m2 Total frontal area = 1.084m^2 m^2
Coefficient of v2 = 0.9 At top speed,Power Power delivered at the wheels = Rate of work done by dissipating forces 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 = 227.5 N
ACCELERATION Maximum Acceleration The maximum acceleration of the car is determined by the maximum torque delivered at the wheels. F= (Tengine
Efficiency
Gearbox reduction CVT reduction) / Effective ffective radius of wheel
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T= (18.4 0.8 11.5 3) =507.8 .8 Nm Note:: 17.9 Nm is the minimum torque generated which peaks to 18.6 Nm. The above graph is fairly linear between the required range therefore mean value of torque is taken. Force on the car= T/ radius dius of the wheel =
.
.
.
.
N= 1940 N
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 nature of performance for a CVT.
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The CVT is tuned to engage at optimum RPM ((i.e. 2800) in order to exploit the maximum torque 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 =
=
.
= 278N
=
.
= -105.2……….. ((-k)
F–f=m
!
!
F = kv+ c; c = Fmax – f; -kv + c = ma -kv + c = mv #x =
!
!"
$ !
% &
8
Integrating the above equations, %" $
% &
= v + log ( %
)
Now, substituting the values of m m, c, and k; 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 θ = W cosθ
730
When the buggy is on the verge of toppling, N2 = 0
9
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°
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CVT (Continuous Variable T 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 transmission that offers a fixed number of gear ratios. CVTech was used as cvt for or this season.
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. This leads to better fuel efficiency. • 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 allows ows the vehicle to better handle obstacles, rough terrain and accelerate out of corners giving the vehicle an advantage in endurance event.
•
Better maneuverability as the driver doesn’t have shift 5+1 gears and hence less possibility of errors while driving
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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 Normal ormal stress, Shear stress and Torsion with Structural Rigidity R by maintaining taining a factor of safety of more than 22. 4) Compact 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 can be sacrificed as they were not required on the BAJA track. As this is our second season in BAJA, reverse comes with risk of failure during competition. As compared to 16.5kg DANA, our gear box weighs only 6kg, giving a reduction in weight of 10.5 kg hands down.
GEARBOX SPECIFICATION Type
Forward reduction
Reduction
11.5
Material
AL 6061-T6/ AISI 9310
Weight Weight*
6kg (*expected)
GEARS The gear calculations were based on theory and fformulae ormulae mentioned in the Shigley Handbook of 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 gear (T1), T1), say the smaller gear. Now using the relation given below we can determine number of teeth on other gear gear, T2.
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So we got number of teeth on both the gears, but one should als also check for interference, if gear 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 shown in following figure.
A pair of gear teeth under interference
For minimum interference, the pinion should have a minimum number of teeth specified by following relation.
Where aw represents addendum of tooth. For 20 degree pressure angle (which which is normally taken by designers) aw = 1 m and bw = 1.2 m. Module m, and pitch circle diameter Pd are 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 and T2 is shown below.
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Flow chart to determine number of teeth on each gear
THE LEWIS BENDING EQUATION UATION
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KV
KO
KM
Y
Pinion
3.26
1.25
1.6
0.43
Intermediate 1
3.26
1.25
1.6
0.43
Intermediate 2
1.75
1.25
1.6
0.43
Gear
1.75
1.25
1.6
0.302
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 between them. Such collisions have impact in long run. If a pair of gears failed at 500 lb. tangential load at zero velocity and at 250 lb. at velocity V1, then a velocity factor, designated Kv, of 2 was specified for the gears at velocity V1. Then another, identical, pair of gears run running at a pitch-line velocity V1 could be assumed to have a load equal to twice the tangential or transmitted load. This effect is incorporated ed in dynamic loading factor, Kv value of which is a function of pitch line velocity.
Kv =
&
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At root of the gear there could be fatigue failure due to stress concentration effect. Effect of which is incorporated corporated in a factor called Kf value of which is more than 1. There will be factors to check for overload (Ko) and load d distribution on gear tooth (Km). While incorporating all these factors Lewis stre strength equation will be modified like this
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 b, so we have obtained all the output parameters required for gear design. But such a gear does not guarantee a peaceful operation 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 minimum nimum weight over a range of gear ratio.
RESULT Gear material=AISI 9310 Gear Module = 2 Pressure Angle = 20 degrees Number of teeth
Face Width
Max stress
Pinion
18
0.5’’
1352Mpa
Intermediate 1
66
0.5’’
657Mpa
Intermediate 2
22
1’’
1595Mpa
Gear
66
1’’
1677Mpa
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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 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 not no arise. This problem can be dealt with by Case Hardening (case depth =0.9mm)
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SHAFTS
Material of shaft is same as gears=AISI 9310 For high bending strength and low weight 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+ Pinion is embedded on first shaft because of small diameter and the ssecond shaft does not rotate. And therefore it does not requir require splines, this configurations saves two bearings and lot of weight. Final shaft has the max OD due to heavy bending and torsion moment it has to transfer. A manual calculation also takes in account all stress concentration factors due to o steps and grooves.. Stress concentration due to grooves is almost 3 times higher than steps, so considering factor of safety grooves, retaining ring are avoided. Module of splines =1.25
Weight
Length
Deflection(mm)
Max stress
Shaft 1
272 g
129mm
0.0042
28Mpa
Shaft 2
143 g
76mm
0.0011
33.3Mpa
Shaft 3
478 g
173 mm
0.035
331Mpa
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Shaft 1 4000 3500
84.5, 3354.69
97, 3354.69
Shear Force in N
3000 2500 2000 1500 1000
103, 565.75
500 0
-500 0 0, -700 -1000
20
40
60
80
103, -565.75 565.75 112.2, -565.75 100 120
84.5, -700 Length (mm)
19
20
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Schematic representation of gear assembly
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SPLINE CALCULATION
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 *
+
-
+
.
.
* 34.375 34.375 mm ~ 34 mm
Where, P = 1/m
* 30.625 mm ~ 31 mm
* 1.9625 1.9625 mm ~ 2mm
Shear stress at pitch diameter ((, ) *
./0 /1
2-34 /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
Taking uniform shock, Ka = 1 For fixed splines, Km = 1 Therefore, ,
* 59.26 MPa
Taking Factor ctor Of Safety (FOS) = 2, ,
* 120 MPa
Taking factor of 0.5 for Kf and 2 for Ka,
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,
* 480 MPa
Taking 9310 Steel as material for manufacturing splined shaft, ,$
"
* 6789: ;<=8;; ∗ 0.577 = 550 MPa
Changing the initial parameters, s, 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 (Ss) *
./0
C / 2B4 5
Where, T = torque Ka = Spline application factor Dre = minor diameter Kf = Fatigue life factor
Therefore taking FOS = 1.5, Ss = 504.77 MPa
Changing some parameters, Minor diameter (Dre) = 31 mm Ss = 392.23 MPa
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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 (Sc) =
./1 /0
2-34 D/5
Where,
T = Torque Km = Load distribution factor Ka = Spline application factor D = Pitch diameter N = number of teeth Le = Max effective tive Spline length h = 0.9m (flat root) = m (fillet root) Therefore, Case 1: Taking N = 26, D = 32, Ka = 1, h=0.9m = 1.125 mm, Kf = 0.5 Sc = 14.29 MPa FOS = 3.7
Case 2: Taking N = 24, D = 30, Ka = 1, h=0.9m = 1.125 mm, Kf = 0.5 Sc = 16.51 FOS = 3.11
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Now, According to a book max. permissible ermissible Sc = 4000 psi for max shear stress = 40,000 psi, psi using unitary analogue, For max. Shear stress = 530 MPa , max. Sc = 53 MPa Hence in the both the cases Sc is in safe limit.
Rack profile for 37.5 37.5⁰ pressure angle spline
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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 conducted nducted in a virtually on the software ANSYS. The objective behind these analysi analysis was to determine the durability and efficiency of the gearbox. Reasons for failure of gearbox1. Deformation of casing 2. Heating of gears 3. Churning of gears due to shock loading
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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 applied according to results drawn from hand calculations. MAX STRESS= 116MPA
MINIMUM FACTOR OF SAFETY=2.4
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MAXIMUM ELASTIC DEFORMATION=0.26MM
DESIGN EVOLUTION OF GEAR BOX Iteration of changes in gearbox casing is solely based on reducing deformation and weight.
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FATIGUE It has been statistically estimated in the industry that 50 50-90% of structural failure is due to 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 designed product. After designing the gearbox on Solidworks 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 beari bearings ngs on which shafts are loaded. These forces have been calculated previously in the calculations.
Scale of deformation =536: 1
Maximum deformation=0.04mm
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Minimum fatigue life=2718hrs
Fatigue Safety factor 0.63
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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 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 applications, however, 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 gearbox via chassis, etc. These vibrations show drasticc resul results on the body, hence it is necessary to ensure that these vibration 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 PSD accelerations to which the design will be subjected.
32
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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 following1. Ensure that the gears maintain structural integrity (producing minimum friction for ensuring that gear teeth do not get damaged). 2. Avoid oil over-heating. 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 load-independent spin power losses 4) Assessment of wall effects on gear housing
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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)
Max oil pressure(103Pa)
Reach(out of 10)
10mm
9.5
1.5
2
20mm
9.42
4.16
5
30mm
9.36
5.5
9
40mm
9.28
7.824
9.5
Oil level increase comes at a cost of 1) Weight 2) Viscous drag
Streamline velocity curve for 10 mm oil level
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EFFICIENCY OF GEARBOX In calculating efficiency of gearbox various losses are accounted accounted1)Tooth Tooth engagement loss 2)Oil churning loss 3)Bearing loss 4) Seal frictional loss Total Power Loss (Pt) = Lt + Lch + Lb + Ls Where, Lt = Power loss at tooth engagement (hp) 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 Where, W = Total power Z1 = Number of pinion tooth ψ = Helix Angle V = Terminal velocity
1ST ENGAGEMENT: Lt = 8.2E
.
I
.
.
J MN
= 0.6984 hp
2ND ENGAGEMENT: Lt = 8.2E
.
I
.
.
J MN
= 1.028 hp
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CALCULATIONS FOR LCH: Lch = OPQ E
R . J F &F
10 TU
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
1ST ENGAGEMENT: Lch = 0.009
Q
1.768 E
. &
R
J
. &
R
J
.
10 TU
= 5.7 x 10-3 hp
2ND ENGAGEMENT: Lch = 0.009
Q
0.482 E
.
10 TU
= 1.586 x 10-3 hp
CALCULATIONS FOR LB: Lb = 5.23
10
\]^ :_ `U
Where, F = radial load on bearing (N) fb= coefficient of friction d= shaft diameter (m) n = shaft speed (rpm)
1ST SHAFT BEARINGS: Left F = 3665 N
Right F = 581.60 N
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fb = 0.01 (worst case)
fb = 0.01 (worst case)
d = 22 mm
d = 22 mm
n==
n=
rpm
rpm
Lb = 8.33 x 10-3 hp
Lb = 0.053 hp
LAST SHAFT BEARINGS: Left
Right
F = 6371 N
F = 3280 N
fb = 0.01 (worst case)
fb = 0.01 (worst case)
d = 25 mm
d = 25 mm
n=
n=
rpm
Lb = 9.533 x 10-3 hp
rpm
Lb = 4.767 x 10-3 hp
CALCULATIONS FOR LS:
a * ]N =Ts = Seal torque (Nm) ] * bc
d ^ B
e
/
g= angular velocity of shaft f = seal friction pr = radial lip load (N) r = radius of shaft (m)
Therefore, Ls = 0.0250 hp If we take, Engine power = 8.2 hp Total power loss = 1.9hp, hence hence, Efficiency of gearbox = 77%
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CV JOINT Constant-velocity 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
RZEPPA JOINT It is a 6 ball continuous velocity joint giving 22° of working angle with 50mm of plunging. It is heavier than tripod joint and difficult to maintain in long run. As our past experience with tripod wasn’t good so this season Rzeppa are used.
PLUNGING DATA 1)Right ight hand shaft requires max. Plunge=1in 2) Left hand shaft requires max. Plunge=1.02in
HUBS The wheel hub is that part of car on which the wheels and the brake rotor are mounted. It must be able to o absorb loads from braking and cornering cornering, and allow the wheel to spin freely. It must accommodate bearings which support supports the spindle and CVshaft.
DESIGN OBJECTIVES 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 with without compromising on the strength.
FRONT HUB Front hub are designed by considering braking torque, cornering forces and extreme cases of free fall. After analyzing forces on damper, brake rotor and tire; the following cases for force application was considered ered for analysis analysis1)Remote cornering force of 1250N. 2)Remotecurb force of 5000N. 3)Braking torque of 249Nm.
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Red colour depicts the area where FOS is below 1.5
Greenish area shows stress of 200MPa
Material used for analysis is Al 6061 and weight of hub is 530gram.
REAR HUB As compared to last season steel hub this year Al 6061 6061-T6 T6 hubs were designed. For splines another steel sleeve embedded with splines was press fit in Aluminium hub.
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
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Area below FOS of 1.5 is shown in red and the total weight is 440 g.
Green area depicts stress of 230Mpa
STEEL SLEEVE Aluminium is not reliable for splines, especially with low module so instead of manufacturing 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 friction on coefficient between Aluminium and steel µ is 0.61 and calculated interference is 0.01mm for the torque transfer ansfer of 350NmTtotal weight of sleeve and hub is 800g which is 30% of last year steel hub.
Press fit analysis shows max stress generated is on the edge which is around 1000MPa. 1000MPa
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REFRENCES • • • • • • •
• • • • • • •
www.briggsandstratton briggsandstratton.com/engines-racing/racing-engines/.../model--20 students.sae.org/cds/bajasae bajasae/ nptel.ac.in/courses/IIT .ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/2_7.pd nptel.ac.in/courses/IIT .ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/2_17.pdf
Shigley's higley's mechanical engineering design www.ansys.com/Products/Simulation+Technology/.../ .com/Products/Simulation+Technology/.../ANSYS+CFX
press fit-intro intro to mechanics of solids solids-crandall spline-machinary machinary handbook shaft-mechanics of solid – crandal
predicting fatigue life life-RKHolman and PK liaw Machinery’s handbook Peterson stress concentration book http://learntoengineer.com/beam http://www.amesweb.info/StressConcentrationFactor/StressConcentration Factors.aspx
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