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    400 Commonwealth Drive, Warrendale, PA 15096-0001 U.S.A. Tel: (724) 776-4841 Fax: (724) 776-5760 Web: www.sae.or

    SAE TECHNICAL

    PAPER SERIES  2002-01-3308

    Design of Formula SAE

    Suspension Components

    Badih A. Jawad and Brian D. PolegaLawrence Technological University

    Reprinted From: Proceedings of the 2002 SAE MotorsportsEngineering Conference and Exhibition

    (P-382)

    Motorsports EngineeringConference & Exhibition

    Indianapolis, IndianaDecember 2-5, 2002

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    All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or

    transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise,without the prior written permission of SAE.

    For permission and licensing requests contact:

    SAE Permissions400 Commonwealth Drive

    Warrendale, PA 15096-0001-USAEmail: [email protected]: 724-772-4028

    Tel: 724-772-4891

    For multiple print copies contact:

    SAE Customer ServiceTel: 877-606-7323 (inside USA and Canada)

    Tel: 724-776-4970 (outside USA)Fax: 724-776-1615

    Email: [email protected]

    ISSN 0148-7191Copyright © 2002 SAE International

    Positions and opinions advanced in this paper are those of the author(s) and not necessarily those of SAE.The author is solely responsible for the content of the paper. A process is available by which discussions

    will be printed with the paper if it is published in SAE Transactions.

    Persons wishing to submit papers to be considered for presentation or publication by SAE should send themanuscript or a 300 word abstract of a proposed manuscript to: Secretary, Engineering Meetings Board, SAE.

    Printed in USA

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    ABSTRACT

    This paper is an introduction to the design of suspensioncomponents for a Formula SAE car. Formula SAE is astudent competition where college students conceive,design, fabricate, and compete with a small formula-styleopen wheel racing car. The suspension componentscovered in this paper include control arms, uprights,spindles, hubs, pullrods, and rockers. Key parameters

    in the design of these suspension components aresafety, durability and weight. The 2001 LawrenceTechnological University Formula SAE car will be usedas an example throughout this paper.

    OVERVIEW

    In designing suspension components for the2001 Lawrence Technological University Formula SAEvehicle, safety and durability were the top priorities. Inorder to ensure the safety of the suspension system, theloads acting on every component were extensivelystudied by utilizing strength of material calculations and

    Finite Element Analysis. Another measure used toensure the safety of the suspension system was thatevery suspension fastener was put into double shearand was either safety wired or secured with a locknut.

    Weight was another important consideration while designing and manufacturing the suspensionsystem on the 2001 Lawrence Technological UniversityFormula SAE car. In order to achieve a weight reductionover previous Lawrence Technological UniversityFormula SAE vehicles, Finite Element Analysis wasutilized to remove the maximum amount of material fromevery suspension component while maintaining a critical

    safety factor of two. Also, chrome moly or 7075-T6aluminum was used for every suspension componentdepending on which material was more practicalconsidering any compatibility and dimensionalrestrictions. These materials were chosen due to theirsuperior strength to weight ratios. The suspension loadpaths were also extensively studied to ensure that they were fed into the suspension system and frame in arobust manner.

    SUSPENSION DESIGN

    CONTROL ARMS

    The purpose of the control arms is to secure the wheel assembly to the chassis. Coupled with thesuspension geometry, the control arms play a key role indetermining the kinematics of the car such as camberrejection and roll stability. [1] The control arms also

    provide a means for tuning the suspension for specificcourses.

    Figure 1: Assembled Rear Control Arms

    In order to reduce weight, spherical bearings arestaked into the pivot points and ball joints of the lowecontrol arms, along with the rear upper control arm pivotpoints and front upper control arm ball joints instead ofusing traditional heavier rod ends at these locations. Inorder to stake the spherical bearings into the 4130 steetubing, a hollow aluminum insert is pressed into the

    tubing. Non-tempered aluminum was used for the inserdue to the fact that tempered aluminum is too hard toflatten, which causes the 4130 steel tubing to crackNext, the end of the tube is crushed flat and a hole ismilled through the flat end of the tubing. The bearing isthen firmly staked into the control arm.

    2002-01-3308

    Design of Formula SAE Suspension Components

    Badih A. Jawad and Brian D. PolegaLawrence Technological University

    Copyright © 2002 SAE International

     

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    Figure 2: Inserted Spherical Bearing

    Rod ends with 4130 steel threaded inserts areused where suspension adjustment is necessary fortuning. Rod ends are used on the front upper controlarm pivot points to allow for camber and casteradjustments in the front suspension. Also, rod ends areused on the rear upper control arm ball joints to allow forcamber adjustment in the rear.

    The appropriate size of 4130 steel tubing wasdetermined to be 19.05 mm OD x 0.71 mm wallthickness from performing buckling and bending stresscalculations. (See Appendix) This size of tubing isequivalent to 0.75 inch x 0.028 inch wall thickness, which is a common size of tubing in the United States.Using this size of tubing with a cornering load of 1.3 gand a braking force of 1.4 g, which were determinedfrom equipping previous Lawrence TechnologicalUniversity Formula SAE vehicles with data acquisition,the minimum safety factor in any of the control armssubjected to simultaneous braking and cornering wascalculated to be 1.7. [2] The cornering and brakingforces were calculated using a vehicle and driver weightof 2935.8 N. (See Appendix) These calculations werealso verified by performing Finite Element Analysis using

    COSMOS software.

    Figure 3: Control Arm FEA Results

     Also, an extensive load study performed on thecontrol arms resulted in the arms of the control armsbeing directly in line with chassis nodes, minimizing anybending moments acting on the frame. All of the controlarm fasteners are put into double shear and meet ANspecifications. Therefore, the control arms weredesigned to be lightweight, reliable, and safe.

    UPRIGHTS

    The main function of the uprights is to providean interface to connect the upper and lower ball joints

     with the spindle. It is crucial to minimize the weight othe uprights because they are unsprung mass, and theshocks have to control this weight in bump. [3]

    The uprights were manufactured from 7075-T6aluminum instead of 4130 steel, which is common onmany Formula SAE vehicles, as a weight reduction Also, the 2001 Lawrence Technological UniversityFormula SAE uprights were manufactured using CNCprocesses instead of cutting and welding tubing in orderto increase their strength.

    For safety purposes, the ball joint fasteners were put intodouble shear by passing them through the top web othe ball joint pocket and threading them into the lowe web of the ball joint pocket.

    Front Uprights

    To maximize the strength and minimize thedeflection of the front uprights, many design iterationsconsisting of various shapes were performed. FiniteElement Analysis was executed on each iteration incornering, braking, and a combination of cornering andbraking situations as a worse case scenario. Thesesituations are seen while performing typical maneuverson a Formula SAE course. To simulate a corneringforce of 1.3 g, the spindle hole was rigidly constrained while a load of 895.0 N was applied to the upper bal joint and a load of 2304.7 N was applied to the lower bal joint in the opposite direction. (See Appendix) Braking was simulated in Finite Element Analysis by rigidly

    constraining the spindle hole and applying a force o1121.1 N at each of the brake caliper mountinglocations.

    Preliminary Finite Element Analysis resultsshowed that a wide upright with pockets in it wasstronger than a traditional narrow upright without anypockets. It was also discovered through Finite Elemen Analysis that triangular pockets provided the greatesstrength with the least amount of deflection. FurtheFinite Element Analysis iterations resulted in an uprigh with optimized triangular pockets with the majority of themass centered around the spindle hole. The safety

    factor of the final front upright was 4.0 in cornering, 2.8in braking, and 2.5 for combined cornering and braking.

    Front Lower Control Arm

    Maximum Stress=817.43 MPa

    SF=1.68

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    Figure 4: Final Front Upright FEA Results

    Rear Uprights

    Many design iterations were performed on therear uprights to maximize their strength while minimizingtheir weight. Finite Element Analysis was performed onthe rear uprights in cornering, braking, and acombination of both of these situations. To simulatecornering on the rear uprights, the center hole wasrigidly constrained, while forces of 781.3 N were appliedto the upper ball joint and toe bar mounting location anda force of 2227.0 N was applied to the lower ball joint inthe opposite direction of the upper ball joint force. (See Appendix) To simulate braking, a force of 2224.1 N wasapplied to the upper and lower ball joints in thelongitudinal direction, while the center hole was fixed.

    Finite Element Analysis results showed thattriangular pockets proved to be the most effective meansto reduce the weight of the rear uprights without

    compromising strength. The resulting safety factors ofthe rear upright were 4.3 in cornering, 3.3 in braking, and3.1 in combined cornering and braking.

    Figure 5: Final Rear Upright FEA Results

    SPINDLES

    The spindles provide a base at which the wheeassembly rotates about. Its outer diameter is critical toprevent slop in the bearings, resulting in premature wear. Also, as learned from previous LawrenceTechnological University Formula SAE cars, thissuspension component is greatly prone to fatigue loadsthat originate from abnormalities in the road surface andfrom weight transfer due to lateral and longitudina

    accelerations. [4]

    The spindles were manufactured from 4340steel due to its excellent fatigue-resistant properties andstrength. Key dimensional considerations for thespindles included bearing and upright dimensionsbecause the spindle is pressed into the upright. In ordeto prevent the hub from sliding into the upright, a step was designed into the spindle 5.08 mm from the uprightThe step is the same thickness as the bearings’ width inthe hub so that the slight impact load caused bymovements of the hub and bearings would be distributedthrough the entire bearing. Another step was added to

    the free end of the spindle so that a washer and castlenut could be used to secure the hub to the spindle.

    Figure 6: Final Spindle Design

     An analysis of the spindle was performed usingmaximum braking and vertical forces. The maximumbraking force experienced by the 2001 LawrenceTechnological University Formula SAE car was found tobe 1.4 g, and the maximum cornering force on a tire wasfound to be 1.3 g. Using these forces and FiniteElement Analysis, the spindle's safety factor wasrevealed to be 25.8 in braking and 2.8 during corneringThe spindle's safety factor was determined to be 2.6 when braking and cornering loads were applied to thespindle simultaneously.

    Front Upright FEA Results

     Loading Case  Max Stress

    (MPa)

     Deflection

    (mm)

     Safety Factor 

    Cor ner ing 1 23. 35   .363   4

    Braking   179.66   .044   2.8

    Cornering and

    Braking

    205.10   .409   2.45

    Deflection

     

    Combined Cornering

    and Braking Stress

    CorneringStress

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      Figure 7: Spindle FEA Results

    For safety purposes, a backplate was welded tothe back of the spindle so that it could be bolted onto theback of the front uprights after it was pressed into theupright.

    HUBS

    The purpose of the hubs is to rotate the wheels

    and tires. It is critical to minimize the weight of the hubsbecause they are rotating and unsprung mass. [5] Also,to reduce friction due to the rotational motion of thehubs, the 2001 Lawrence Technological UniversityFormula SAE car utilizes bearings in the hubs.

    Front Hubs

    The front hubs are manufactured from 7075-T6aluminum due to its superior strength to weight ratio. Also, in order to minimize weight without compromising

    strength, the front hub utilizes a four-leaf clover patternto hold the lugs and brake rotor fasteners. The brakerotor and lug four-leaf clover patterns are offset 45

    0 from

    each other to allow for ease of pressing lugs into the huband installation of brake rotors. Also, a 10.16 mm highstep that is 19.69 mm wide is designed into the insidecenter of the front hub to keep the bearings apart fromeach other and to dissipate heat from the bearings.

     Another key function of the front hubs is to allowthe brake rotors to float freely. This is accommodatedthrough a slot in the brake rotor fingers. Floating rotorsprovide maximum stopping force by allowing both brake

    pads on the caliper to grip the rotor evenly. Also, theslot in the brake rotor fingers allow the brake rotorfasteners to be put into double shear as a safetyfeature. [6]

    Figure 8: Final Front Hub

    Finite Element Analysis was performed on thefront hubs to validate their design. A braking force of 1.4g applied to the brake rotor fastener holes resulted in asafety factor of 5.0. Also, a cornering force of 1.3 g wassimulated by applying a force of 4469.9 N on the bottomlug hole and 3024.2 N on the top lug hole in oppositedirections while constraining the brake rotor fingers from

    translating. (See Appendix) This loading resulted in asafety factor of 2.8. A safety factor of 2.7 was obtained when combining braking and cornering.

    Figure 9: Front Hub FEA Results

    Rear Hubs

    The main function of the rear hubs is to connecthe driveline to the wheels through the halfshafts andconstant velocity (CV) joints. The rear hubs aremanufactured from 4340 steel because they must becompatible with the outer CV joints that are splined toaccept a hub.

    The rear hubs also utilize a four-leaf clovedesign to minimize their weight. To reduce friction fromthe rotating hub, bearings are placed in between theupright and the shaft on the hub. As a safety feature, aspindle lock nut is applied on the end of the CV joint toprevent the hub from becoming disengaged from thedriveline.

    Finite Element Analysis proved that the reahub’s design was adequate for its application. A safetyfactor of 4.9 was obtained when a torque of 697.0 N-m

    Spindle FEA Results

     Lo ad ing Ca se  Ma x S tre ss

    (psi)

     Def lec tio n

    (in)

     Sa fet y F act or 

    Cornering   324.05   .150   2.8

    Braking 35.21 4.32 x 10 -4 25.8

    Cornering and

    Braking

    343.90 .155 2.63

    Cornering

    Deflection

     

    CorneringStress

    CorneringStress

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    originating from the driveline was applied to the hubshaft while restraining the hub from rotating. When acornering force of 1.3 g was simulated on the hub byplacing a load of 3024.2 N on the top lug hole and4469.9 N on the bottom lug hole in the opposite direction while preventing the shaft from rotating, the resultantsafety factor was 4.2. (See Appendix) When combininga cornering situation with the torque acting on the hub,the resultant safety factor was 3.8.

    Figure 10: Rear Hub FEA Results

    ENERGY MANAGEMENT MECHANISMS

    The energy management system consists of thepullrods, rockers, and shocks. The system’s purpose isto activate the inboard shocks, which greatly improvesthe handling qualities of the car. The geometry of thesuspension mechanism is critical because it determinesthe motion ratio, which is the ratio of vertical wheelmovement to shock displacement. This ratio is used todetermine the car’s natural frequency, which significantlyaffects the car’s handling qualities. Friction is kept to anabsolute minimum in the dampening system to allow formaximum efficiency of the shocks. [7] It is also critical

    that the pullrod, suspension mechanism, and shock arelocated in the same plane to eliminate bending momentson these suspension components.

    Figure 11: Placement of Pullrod and SuspensionMechanism

    Pullrods

    Pullrods are utilized on the 2001 LawrenceTechnological University Formula SAE car instead ofpushrods, which are typically used on Formula SAE

    vehicles, for a weight reduction. Smaller diameter tubingcan be used for pullrods because when the car hits abump, the pullrod pulls the suspension mechanism which puts it in tension. During rebound the pullrods aresubjected to an insignificant buckling load equivalent tothe weight of the wheel assembly. Therefore, bucklingdoes not have to be considered in the design of pullrodsfor a Formula SAE vehicle. However, a pushrod issubjected to a buckling load whenever the car goes intorebound, so buckling must be considered in the designof pushrods. Also, pullrods allow the rockers and shocksto be packaged towards the bottom of the car resulting ina lower center of gravity.

     A stress analysis of the pullrods using a 1.4 gtensile force resulted in a maximum stress of 118.89MPa. A safety factor of 2.6 is obtained from this stressFinite Element Analysis was also performed on thepullrods to verify the stress calculations and resulted in asafety factor of 2.5.

    Rockers

    Rockers, commonly called bellcranks, wereplaced on the 2001 Lawrence Technological UniversityFormula SAE car so that the shocks could be packagedinboard. This packaging scheme significantly reducesunsprung mass. Also, rockers allow the pullrod andshock displacements to be in different directions, whichaids tremendously in the packaging of suspensioncomponents. The rockers were designed to have a oneto one motion ratio, which means that the shocks travelsthe same amount that the wheel moves in the verticadirection. A one to one motion ratio was selectedbecause it allows the total range of shock displacemento be used, which improves the sensitivity of the

    suspension system.

    The rockers were manufactured from 7075-T6aluminum due to its superior weight to strength ratio. Toreduce friction in the energy management system, rollebearings were placed in between the two rocker platesand thrust bearings were incorporated into the bellcrankplates at the pivot points.

    Extensive Finite Element Analysis wasperformed on the rockers to minimize their weight. Asafety factor of 4.9 was obtained for the final design ofthe front rockers when a force of 1.3 g was applied to

    them at the pullrod attachment point while restraining itfrom translating. A safety factor of 4.5 was obtained fothe rear rocker when it was loaded in a similar manner.

    CorneringStress

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    Figure 12: Rocker FEA Results

    CONCLUSION

    This paper addressed essential designconsiderations for a small formula-style open wheelracing car. These considerations were addressedproperly because none of the suspension componentson the 2001 Lawrence Technological University FormulaSAE car failed during intensive driver’s training or at the2001 Formula SAE competition.

    ACKNOWLEDGMENTS

    The authors would like to thank the 2001 LawrenceTechnological University Formula SAE team for all oftheir hard work and effort put into the project.

    REFERENCES

    1. Smith, Carroll. Tune to Win. Fallbrock, CA: AeroPublishers, 1978.

    2. Hibbeler, R.C. Mechanics of Materials. 3rd ed. UpperSaddle River, NJ: Prentice Hall, 1994.

    3. David E. Woods and Badih A. Jawad. “Numerical Designof Racecar Suspension Parameters”, Wahington, D.C.:SAE International, 1999.

    4. 2000 LTU FSAE Team, “2000 Formula SAE Final Report”,Southfield, Michigan: Lawrence Technological University,2000.

    5. Puhn, Fred. How to Make your Car Handle. Los Angeles,CA: HPBooks, 1981.

    6. Smith, Carroll. Engineer to Win. Osceola, WI:Motorbooks International, 1984.7. William F. Milliken and Douglas L. Miliken. Race Car

    Vehicle Dynamics. Warrendale, PA: SAE International.

    Suspension Mechanism FEA Results

    Stress Displacement

      Rocker FEA Results

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    APPENDIX 

    SAMPLE CALCULATIONS

    TOTAL WEIGHT

    Target Vehicle Weight = 2224.11 N Average Driver Weight = 711.72 N

    Total Weight = 2224.11 + 711.72 = 2935.83 N

    CONTROL ARMS

    FRONT CONTROL ARM AXIAL FORCES 

    1445.7 N 4.40o  28.06

    15.45o FUCA 

    FPR 

     

    FLCA 

    556.0 N

     N  F 

     N  F 

     N  F 

     F 

     F  M 

     F  F 

     F  F 

     F  F 

     F  F 

     PR

    UCA

     LCA

    o LCA

    o LCA

    o LCA

    o PR

    oUCA y

    o LCA

    o PR

    o

    UCA z

    0.233,2

    6.208,17.259,2

    )1.4)(40.4sin()8.21)(0.556(

    )8.238)(40.4cos()381)(7.1445(0

    40.4cos06.28cos

    45.15cos7.14450

    40.4sin06.28sin

    45.15sin0.5560

    −=

    −=−=

    +−

    +==+

    +−

    +==→+

    −+

    −==↑+

    )

     

    REAR CONTROL ARM AXIAL FORCES 

    1445.7 N 4.13o  28.03

    14.30o R UCA 

    R PR  

    R LCA 

    556.0 N

     N  R

     N  R

     N  R

     R

     R M 

     R R

     R F 

     R R

     R F 

     PR

    UCA

     LCA

    o LCA

    o LCA

    o LCA

    o PR

    oUCA y

    o LCA

    o PR

    oUCA z

    0.088,2

    5.060,1

    3.267,2

    )29.2)(13.4sin()64.8)(0.556(

    )3.241)(13.4cos()381)(7.1445(0

    13.4cos03.28cos

    30.14cos7.14450

    13.4sin03.28sin

    30.14sin0.5560

    −=

    −=

    −=

    +−

    +==+

    +−

    +==→+

    −+

    −==↑+

    )

     

    BUCKLING CALCULATIONS 

    2

    2

     L

     EI  P

    cr 

    Π=  

    4121020.1723

    84.206

    m I 

    GPa E 

    ×=

    =

     

    ( ) ( )

     N  P

     P

     MAX 

     MAX 

    90.167,52

    0176.00191.04

    1058.1206  226

    =

      

       Π×=

     

    y

    z

    y

    z

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    Front Upper Control Arm Front Arm 

    ( (( )

    65.226.660,19

    90.167,52

    26.660,19

    423.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Arm

    ( )

    90.234.006,18

    90.167,52

    34.006,18

    442.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Front Lower Control Arm Front Arm

    ( )

    01.458.009,13

    90.167,52

    58.009,13520.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Arm

    ( )

    49.406.629,11

    90.167,52

    06.629,11

    550.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Upper Control Arm Front Arm

    ( )

    15.367.552,16

    90.167,52

    67.552,16

    461.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Arm

    ( )

    66.355.241,14

    90.167,52

    55.241,14

    497.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Lower Control Arm

    Front Arm

    ( )

    21.471.382,12

    90.167,52

    71.382,12

    533.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    Rear Arm

    ( )

    37.482.930,11

    90.167,52

    82.930,11

    543.0

    1020.17231084.206

    2

    1292

    ==

    =

    ××Π=

    SF 

     N  P

     P

    cr 

    cr 

     

    MOMENT CALCULATIONS

    M=Fd

    Bending Moment due to Vertical Force Front Upper Control Arm

    m N  M 

    d VF  M 

    −=

    ==

    61.93

    45.25cos413.045.25cos0.278))((   00 

    Front Lower Control Arm

    ( (m N  M 

    d VF  M 

    −=

    ==

    45.131

    40.14cos504.040.14cos0.278))((   00 

    Rear Upper Control Arm 

    m N  M 

    d VF  M 

    −=

    ==

    14.101

    30.24cos438.030.24cos0.278))((  00

     

    Rear Lower Control Arm

    m N  M 

    d VF  M 

    −=

    ==

    97.107

    13.14cos413.013.14cos0.278))((  00

     

    Bending Moment due to Cornering Force Front Upper Control Arm

    (   m N d CF  M    −===   86.94)117.0(45.15cos16.841))((   0 

    Front Lower Control Arm

    (   m N d CF  M    −===   37.92)041.0(40.4cos70.2259))((   0 

    Rear Upper Control Arm

    (    N d CF  M    −===   29.117)114.0(30.14cos79.1061))((   0 

    Rear Lower Control Arm

    m N d CF  M    −===   92.85)038.0(13.4cos81.2266))((   0

     

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    Shearing Moment due to Braking Force Front Upper Control Arm

    m N d  BF  M    −===   39.90)127.0)(72.711())((  

    Front Lower Control Arm

    m N d  BF  M    −===   39.90)127.0)(72.711())((  

    Rear Upper Control Arm

    m N d  BF  M    −===   39.90)127.0)(72.711())((  

    Rear Lower Control Arm 

    m N d  BF  M    −===   39.90)127.0)(72.711())((  

    STRESS CALCULATIONS 

     I 

     MyS   =  

    m y

    m I 

    3

    412

    10525.9

    1020.1723

    ×=

    ×=

     

    Bending Stress due to Vertical Force Front Upper Control Arm

    ( )(

    33.243.517

    58.1206

    43.517

    1020.1723

    10525.961.93

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Front Lower Control Arm

    ( )(

    66.159.726

    58.1206

    59.726

    1020.1723

    10525.945.131

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Upper Control Arm 

    ( )(

    16.205.559

    58.1206

    05.559

    1020.1723

    10525.914.101

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Lower Control Arm

    ( )(

    02.280.59658.1206

    80.596

    1020.1723

    10525.997.107

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Bending Stress due to Cornering Force 

    Front Upper Control Arm

    ( )

    30.234.524

    58.1206

    34.524

    1020.1723

    10525.986.94

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Front Lower Control Arm

    ( )(

    36.258.510

    58.1206

    58.510

    1020.1723

    10525.937.92

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Upper Control Arm

    ( )(

    86.132.648

    58.1206

    32.648

    1020.1723

    10525.921.117

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Lower Control Arm

    ( )(

    54.292.474

    58.1206

    92.474

    1020.1723

    10525.992.85

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Shearing Stress due to Braking Force

    Front Upper Control Arm( )

    42.263.499

    58.1206

    63.499

    1020.1723

    10525.939.90

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Front Lower Control Arm

    ( )

    42.263.499

    58.1206

    63.499

    1020.1723

    10525.939.90

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Upper Control Arm

    ( )(

    42.263.499

    58.1206

    63.499

    1020.1723

    10525.939.90

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    Rear Lower Control Arm

    ( )(

    42.263.499

    58.1206

    63.499

    1020.1723

    10525.939.90

    12

    3

    ==

    =

    ×

    ×

    =−

    SF 

     MPaS 

     

    SPINDLES

    FATIGUE ANALYSIS

    Maximum Vertical Force = 2891.2 NMinimum Vertical Force = 0 N

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  • 8/9/2019 Design of Formula SAE Suspension Components

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    Maximum Bending Moment = (2891.2)(0.082)= 237.08 N-m

    Minimum Bending Moment = (0)(0.082)=0 N-m 

    ( )   4844 1022.7014.0019.04

    m I   −

    ×=−Π

    =  

    Maximum Stress 

    ( )( ) MPa

     I 

     MyS    39.62

    1022.7

    019.008.237

    8max  =

    ×

    ==−

     

    Minimum Stress

    ( )( ) MPa

     I 

     MyS    0

    1022.7

    019.00

    8min  =

    ×

    ==−

     

    Mean Stress

     MPaS S 

    S m

      20.312

    039.62

    2

    minmax=

    +=

    +

    =  

     Average Stress

     MPaS S 

    S a

      20.312

    039.62

    2

    minmax=

    =

    =  

     MPaS 

     MPaS 

     MPaS 

     MPaS 

    e

    YT 

    UT 

    5.412

    5.742

    25.701

    825

    1000

    =

    =

    =

    =

     

    Soderberg Approach

    ( )   be

    b

    c

    e

    e

    e

    a

    YT 

    m

    S  N 

     MPaS 

    /110

    65.32

    120.31

    25.701

    20.31

    1

     

      

      −

    =

    =

    =+

    =+

     

    ( )   ( )

    126.35.412

    5.742loglog

    085.0

    5.412

    5.742log

    3

    1log

    3

    1

    22

    1000

    1000

    =

    ==

    −=

     

      

     −=

     

      

     −=

    c

    S c

    b

    S b

    e

    e

     

    ( )   085.0/1085.0126.3

    65.3210  −

     

      

     

    = N   

    N=1.39X1019 cycles 

    UPRIGHTS

    FRONT UPRIGHT CORNERING FORCE

    1445.7 N

    FUBJ 

    FLBJ 

     y

    z

    ∑=+==+

    −−==+

    0)239.0()142.0)(7.1445(0

    7.14450

    UBJ 

     LBJ UBJ  y

     F  M 

     F  F  F 

    r

     

     N  F 

     N  F 

     LBJ 

    UBJ 

    65.2304

    95.859

    =

    −=

     

    REAR UPRIGHT CORNERING FORCE  

    1445.7 N

    FUBJ 

    FLBJ

     y

    z

    ∑=+==+

    −−==+

    0)272.0()147.0)(7.1445(0

    7.14450

    UBJ 

     LBJ UBJ  y

     F  M 

     F  F  F 

    r

     

     N  F 

     N  F 

     LBJ 

    UBJ 

    02.2227

    32.781

    =

    −=

     

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    HUBS

    FRONT HUB CORNERING FORCE

    1445.7 N

    FUL 

    FLL 

     y

    z

    ∑+==+

    −−==+

    )0.98()0.205)(7.1445(0

    7.14450

    UL

    UL LL y

     F  M 

     F  F  F 

    r

     

     N  F 

     N  F 

     LL

    UL

    87.4469

    17.3024

    =

    −=

     

    REAR HUB CORNERING FORCE

    1445.7 N

    R UL

    R LL 

     y

    z

    ∑∑ +==+

    −−==+

    )0.98()0.205)(7.1445(0

    7.14450

    UL

    UL LL y

     R M 

     R R F 

    r

     

     N  R

     N  R

     LL

    UL

    87.4469

    17.3024

    =

    −=

     

    PULLRODS

    TENSILE STRESS

     A

     P=σ    

    ( )   2522 10432.20077.00095.04

    m A  −

    ×=−Π

    =  

    ( )

    61.289.118

    26.310

    89.118

    10432.2

    7.14452

    5

    ==

    =

    ×

    =−

    SF 

     MPaσ  

     

    Downloaded from SAE International by Brunel University, Tuesday, October 21, 2014