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Automotive_chassis_design 底盘设计讲解 英文.pdf

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AUTOMOTIVE_CHASSIS_DESIGN 底盘 设计 讲解 英文
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Automotivedesign Automotive design Chassis* design *pronounced: chas‐e – singular chas‐e‐z – pluralIntroduction Introduction ? Loadsduetonormalrunningconditions: Loads due to normal running conditions: – Vehicle transverse on uneven ground. – Manoeuver performed by driver. p y ? Five basic load cases: – Bendingcase Bending case – Torsion case – Combined bending and torsion g – Lateral loading – Fore and aft loading gBending Payload Bending ? Due to loading in Engine Occupants Fuel tank Payload vertical (X‐Z) plane. ? Due to weight of componentsalong Engine components along the vehicle frame. ? Static condition vehicle Wheels/ bk i Suspension structure can be treated as 2‐D beam. – Vehicle is approximately symmetric in x‐y plane. braking pp y y yp ? Unsprungmass – Components lie below chassis – Donotimposeloadsinstaticcondition Do not impose loads in static condition.Bending moment/ Shear force diagram of a typical hi l passenger vehicleBending Bending ? Dynamic loading: – Inertia of the structure contributes in total loading – Always higher than static loading – Road vehicles: 2.5 to 3 times static loads – Off road vehicles: 4 times static loads El ? Example: – Static loads ? Vehicle at rest . ? Movingataconstant velocity onaevenroad. m g Moving at a constant velocity on a even road. ? Can be solved using static equilibrium balance. ? Results in set of algebraic equations. – Dynamic loads Vhil i b d t ttli t F ? Vehicle moving on a bumpy road even at constant velocity. ? Can be solved using dynamic equilibrium balance. ? Generally results in differential equations. m g m a FTorsion Torsion ? When vehicle traverse on an d l uneven road. ? Front and rear axles experiences a moment. ? Puresimpletorsion: Front axle Rear axle ? Pure simple torsion: – Torque is applied to one axle and reacted by other axle. – Front axle: anti clockwise torque (front view) – Rear axle: balances with clockwise torque – Resultsinatorsionmoment Results in a torsion moment about x‐axis. ? In reality torsion is always accompanied by bending due t it to gravity.Torsion Torsion l Front axle Rear axleCombinedbendingandtorsion Combined bending and torsion ? Bending and torsional loads are super imposed. imposed. – Loadings are assumed to be linear ? One wheel of the lightly loaded axle is raised on a bump result in the other wheel go off ground. ? Allloadsoflighter axleisapplied toone Bending Torsion ? All loads of lighter axle is applied to one wheel. ? Due to nature of resulting loads, loading symmetry wrtx‐z plane is lost. ? R’ R can be determined from moment balance g balance. ? R’ R stabilizes the structure by increasing the reaction force on the side where the wheel is off ground . ? The marked – – Side is off ground – Side takes all load of front axle – Side’s reaction force increases – Side’s reaction force decreases to balance the moment. Combined bending and torsionLateralloading Lateral loadingLateralloading Lateral loading ? For a modern car t = 1.45 m and h 051m = 0.51 m. ? Critical lateral acceleration = 1.42 g ? In reality side forces limit lateral y acceleration is limited within 0.75 g. ? Kerbbumping causes high loads andresults inrollover and results in rollover. ? Width of car and reinforcements provides sufficient bending stiffness to withstand lateral forces forces. ? Lateral shock loads assumed to be twice the static vertical loads on wheels.Longitudinalloading Longitudinal loading ? When vehicle accelerates and dltit i f decelerates inertia forces were generated. ? Acceleration –Weight transferredfromfrontto back transferred from front to back. – Reaction force on front wheel is given by (taking moment abtR R ) ? Deceleration –Weight transferred from back to front. Reaction force onfrontwheelis – Reaction force on front wheel is given byLongitudinalloading Longitudinal loading ? Limiting tractive and g braking forces are decided by coefficient offrictionb/wtiresand of friction b/w tires and road surfaces ? Tractive andbraking Tractive and braking forces adds bending through suspension. ? Inertia forces adds additional bending.Asymmetricloading Asymmetric loading ? Results when one wheel strikes a raised objects or drops into a pit. ? Resolved as vertical and horizontal loads. ? Magnitudeofforce dependson ? Magnitude of force depends on – Speed of vehicle – Suspension stiffness Wheelmass Raised object` – Wheel mass – Body mass ? Applied load is a shock wave hh h l d – Which has very less time duration – Hence there is no change in vehicle speed – Acts through the center of the wheel.Asymmetricloading Asymmetric loading ? Resolved vertical force causes: – Additional axle load – Vertical inertia load through CG – Torsion moment tomaintaindynamicequilibrium to maintain dynamic equilibrium. ? Resolved horizontal force causes: – Bendinginx‐zplane Bending in x z plane – Horizontal inertia load through CG – Moment about z axis to maintain dynamic equilibrium. ? Total loading is the superposition of all four loads.Allowablestress Allowable stress ? Vehicle structure is not fully rigid y g ? Internal resistance or stress is induced to balance external forces ? Stress should be kept to acceptable limits ? Stress due to static load X dynamic factor ≤ yield stress – Shouldnotexceed67%ofyieldstress Should not exceed 67% of yield stress. ? Safety factor against yield is 1.5 ? Fatigue analysis is needed g y – At places of stress concentration – Eg. Suspension mounting points, seat mounting points points.Bendingstiffness Bending stiffness ? Importantinstructural stiffness Important in structural stiffness ? Sometimes stiffness is more important than strength strength ? Determined by acceptable limits of deflection ofthesideframedoormechanisms. of the side frame door mechanisms. – Excessive deflection will not shut door properly ? Localstiffness offloorisimportant Local stiffness of floor is important – Stiffened by swages pressed into panels – Secondmomentofareashouldbeincreased Second moment of area should be increasedBendingstiffness Bending stiffness ? Thinpanelsseparatedbyhoneycomb Thin panels separated by honeycomb structure reduced vibration ? Localstiffness hastobeincreasedat: ? Local stiffness has to be increased at: – Door B – Bonnet – Suspension attach points – Seating mounting points – Achieved by reinforcement plates and brackets.Torsional stiffness Torsional stiffness ? Allowable torsion for a medium sized car: 8000 to 10000 N‐ / m/deg ? Measured over the wheel base ? Whentorsionstiffnessislow: When torsion stiffness is low: – Structure moveup and downand/or whip – When parked on uneven ground doors fail to close – Doorsfail toclosewhilejacking ifjackpointsareatacorner Doors fail to close while jacking if jack points are at a corner ? Torsion stiffness is influenced by windscreens ? TS reduces by 40% when windscreens removed O h il iff ? Open top cars have poor torsional stiffness ? Handling becomes very difficult when torsional stiffness is low.Chassistypes‐Ladderframes Chassis types Ladder frames ? Used by early motor cars ? Early car’s body frame did not contribute much for vehicle structure. Cross beam – Mostly made of woodwhich has low stiffness ? Carried all load (bending and torsion) torsion) ? Advantages: – Can accommodate large variety of bodyshapesandtypes body shapes and types – Used in flat platforms, box vans, tankers and detachable containers ? Still used in light commercial Side rails vehicles like pick up.Chassistypes‐Ladderframes Chassis types Ladder frames ? Side rails frequently have open channel section ? Open or closed section cross beams ? Good bending strength and stiffness ? Flanges contribute large area moment of inertia. ? Flanges carry high stress levels O i f fi i ? Open section : easy access for fixing brackets and components ? Shear center is offset from the web ? Localtwistingofsideframeis ? Local twisting of side frame is avoided ? Load from vehicle is applied on web – Avoidsholesinhighlystressesflanges Avoids holes in highly stresses flanges ? Very low torsional stiffness.Chassistypes‐Ladderframes Chassis types Ladder frames ? Torsion in cross member is dbbd f d Clockwise side frame bending reacted by bending of side frames ? Bending in cross frames are td bti f id reacted by torsion of side frames ? All members are loaded in torsion torsion ? Open sections are replaced by closed sections to improvetorsionalstiffness Anti‐clockwise cross frame torsion improve torsional stiffness – Strength of joints becomes critical – Max bending occurs at joints – Attachment of brackets becomes l more complexChassistypes‐cruciformframes Chassis types cruciform frames ? Can carry torsional loads , no elements of the frame is subjected to torsional moment. ? Made of two straight beams ? Haveonlybendingloads Have only bending loads ? Has good torsional stiffness when joint in center is satisfactorily designed designed ? Max bending moment occurs in joint. ? Combiningladderandcruciform ? Combining ladder and cruciform frame provides good bending and good torsional stiffness ? Crossbeamsatfrontandbackat ? Cross beams at front and back at suspension points are used to carry lateral loadsChassis types‐Torque tube back bone f frame ? Main back bone is a closed Back bone box section ? Splayed beams at front and rear extent tosuspension rear extent to suspension mounting points ? Transverse beams resist lateralloads Transverse lateral loads ? Back bone frame: bending and torsion beam ? Splayed beams: bending ? Transverse beams: tension orcompression or compression Splayed beamsChassistypes‐Spaceframes Chassis types Space frames ? In all frames till now length in one dimension is very less compared h h di i to the other two dimensions ? Increasing depth increases bending strength ? Used in race cars ? All planes are fully triangulated ? Beam elements carry either y tension or compressive loads. ? Ring frames depends on bending of elements – Windscreen, back light – Engine compartment, doors – Lower shear stiffness ? In diagonal braced frame s stiffness provided by diagonal elementChassistypes‐Integralstructures Chassis types Integral structures ? Modern cars are mass produced Sh ttl i d t ld ? Sheet steel pressings and spot welds used to form an integral structure ? Components have structural and other functions Sid f dh f i d ? Side frames + depth + roof gives good bending and torsional stiffness ? Geometrically very complicated ? Stress distribution by FEM only ? Stress distribution is function of applied loads and relative stiffness between components ? Advantages: – Stiffer in bending and torsion – Lower weight – Less cost Quietoperation – Quiet operationStructural analysis by Simple Structural f ()hd Surfaces (SSS) method ? Many methods to determine loads and stresses ? Elementary method is beam method, FEM is advanced method and SSS is intermediate ? Developed by Pawlowski in 1964 ? Determines loads in main structural elements ? Elements are assumed to be rigid in its plane p ? Can carry loads in its plane – Tension, compression, shear and bending ? Loads normal to plane and bending out of plane is invalid and not allowedSSS method –Analysis of simple van ( ) (torsion case)SSS method –Analysis of simple van ( ) (torsion case) ? Ten structural components are considered ? Ifgeometryisknownand ? If geometry is known and axle loads are known, edge loads (Q s) can be dd determined. ? For a fully laden van front axleloadislighter. axle load is lighter. ? By moment balance R’ r can be determined. '' ** 22 rf rf RR tt =SSS method –Analysis of simple van ( ) (torsion case) ? The equilibrium of SSS‐2 and SSS‐ 3areobtainedbytaking 3 are obtained by taking moments as R f and R’ r are known. ? SSS‐2 (front cross beam) ? SS 3(Rearcrossbeam) 2 *0 2 f f R Pw t ?= ? SS‐3 (Rear cross beam) 3 ' *0 2 r r R Pw t ?= ? P2 and P3 will be equal in magnitude as they act at the width of the vehicle and the torqueatthefrontandrear must torque at the front and rear must be equal.SSS method –Analysis of simple van ( ) (torsion case) ? Considering SSS‐6 g ? Q 1 to Q 5 will occur around periphery ? Applies opposite moment to P 2 and P 3 ? Taking moment at A 31 2 3 31 2 3 4 41 2 22 21 ()()() 0 Pll l Qll ll Qhh Qh Pl + +?+ + +????= ? Consider SSS‐4 (front panel) 62 1 0 Qh Qw ?= ? Consider SSS‐5 (rear door frame) 61 3 0 Qh Qw ?=SSS method –Analysis of simple van ( ) (torsion case) ? Consider SSS‐8 (floor panel) ? Consider SSS‐9 (windscreen frame) 61234 2 ()0 Qllll Qw +++? = ? ConsiderSSS‐10(Roof) 612 5 () 0 sin Qhh Qw α ? ?= Consider SSS 10 (Roof) ? Six unknowns Q 1 to Q 6 65 4 0 Ql Qw ?= ? Substitute Q 2 , Q 3 and Q 4 in the eqnof SSS‐6 Q b bt i d d ? Q 6 can be obtained and hence rest of the unknowns can be derivedSimple Structural Surfaces representingasalooncar in representing a saloon car in bending Material from J.H. Smith, 2002Passengercar Passenger car ? Morecomplex thanboxtypevan More complex than box type van ? Detailed model vary according to mechanical components components – Front suspensions loads applied to front wing as for strut suspension for strut suspension – Rear suspension (trailing arm or twist beam) loads toinnerlongitudinalmemberunderthebootfloor to inner longitudinal member under the boot floor – SSSs varies with body types VehiclestructuresrepresentedbySSS Vehicle structures represented by SSS Busorboxtypevehicle Van P SSSand Not SSS Bus or box type vehicle Van Passenger car Structuresthatarestructuralsurfaces Structures that are structural surfaces Image from J.C.Brown,2002Structures that are NOT simple l f structural surfaces Image from J.C.Brown,2002Halfsaloonmodel Half saloon model ? Limitedto5Loads Limited to 5 Loads – F1z= (radiator, bumper, battery)/2 F2 = (engine)/2 – F2z= (engine)/2 – F3z= one front passenger and seat F t dhl ffl tk – F4z = one rear passenger, seat, and half fuel tank – F5z = (luggage)/2 (b h) ? 1 UDL (body weight)Process Process ? Calculate reactions at front and rear axles Cacuate eacto sat otadea ae s ( taking moments and vertical force equilibrium) – R zf /2 zf / – R rz /2 ? Calculate forces in each of the SSS ? 11 equations with 11 unknowns ( K 1, .. K 10 , M) can be evaluated from SSS1 to SSS8 ? Equilibrium of right frame to be verified with forces and momentsHalfSalooncarmodel ‐Bending Half Saloon car model BendingSSS1 ? TransverseSSS Figure SSS 1 ? Transverse SSS representing the strut tower tower ? Resolving Forces K 1 +K 2 –R f /2=0 K 1 + K 2 R fz / 2 0 ? Moments K 1 = R fz *w 1 /(2*(w 1 +w 2 )) 1 fz 1 /( ( 1 2 ))SSS2 ? Upperfrontlongitudinal Figure SSS2 ? Upper front longitudinal ? Resolving Forces K K (l +l ) 0 K 1 –K 3 –u (l 1 +l 3 ) = 0 ? Moments K l u*((l +l ) 2 /2) M=0 K 1 l 3 –u*((l 1 +l 3 ) 2 /2)‐M =0SSS3 Figure SSS3 ? Lowerfrontlongitudinal Lower front longitudinal ? Resolving Forces F 1z +F 2z +K 5 ‐K 2
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