Automotive design

1. Automotivedesign Chassis*design *pronounced: chas‐e –singular chas‐e‐z –plural

2. Introduction • Loads due to normal runningconditions: • Vehicle transverse on unevenground. • Manoeuver performed bydriver. • Five basic loadcases: • Bending case • Torsioncase • Combined bending andtorsion • Lateralloading • Fore and aft loading

3. Bending Payload Occupants Fueltank • Due to loadingin Engine • vertical (X‐Z)plane. • Due to weight of componentsalong the vehicleframe. • Static condition vehicle structure can be treated as 2‐Dbeam. Suspension Wheels/ braking • – Vehicle is approximately symmetric in x‐yplane. • Unsprungmass • Components lie belowchassis • Do not impose loads in staticcondition.

4. Bending moment/ Shear force diagram of a typical passenger vehicle

5. Bending • Dynamicloading: • Inertia of the structure contributes in totalloading • Always higher than staticloading • Road vehicles: 2.5 to 3 times staticloads • Off road vehicles: 4 times staticloads • Example: • – Staticloads m g • Vehicle at rest. • Moving at a constant velocity on a evenroad. • Can be solved using static equilibriumbalance. • Results in set of algebraicequations. F – Dynamicloads • Vehicle moving on a bumpy road even at constantvelocity. • Can be solved using dynamic equilibrium balance. • Generally results in differential equations. m a m g F

6. Torsion • When vehicle traverse onan unevenroad. Rearaxle • Front and rear axles experiences amoment. • Pure simpletorsion: Frontaxle • Torque is applied to oneaxle and reacted by otheraxle. • Front axle: anti clockwise torque (frontview) • Rear axle: balanceswith clockwisetorque • Results in a torsionmoment about x‐axis. • In reality torsion is always accompanied by bendingdue togravity.

7. Torsion Rearaxle Frontaxle

8. Combined bending andtorsion • Bending and torsional loads are super imposed. • – Loadings are assumed to belinear • One wheel of the lightly loaded axle is rais on a bump result in the other wheel go off ground. • All loads of lighter axle is applied to one wheel. • Due to nature of resulting loads, loading symmetry wrt x‐z plane islost. • R’R can be determined from moment balance. ed Bending Torsion • R’R stabilizes the structure by increasing the reaction force on the side where the wheel is off ground . • The marked– • Side is offground • Side takes all load of frontaxle • Side’s reaction forceincreases • Side’s reaction forcedecreases • to balance the moment. • Combined bending andtorsion

9. Lateralloading

10. Lateralloading • For a modern car t = 1.45 m andh • = 0.51m. • Critical lateral acceleration = 1.42 g • In reality side forces limit lateral acceleration is limited within 0.75 g. • Kerb bumping causes high loads and results inrollover. • Width of car and reinforcements provides sufficient bending stiffness to withstand lateral forces. • Lateral shock loads assumed to be twice the static vertical loads onwheels.

11. Longitudinalloading • When vehicle accelerates and decelerates inertia forces were generated. • Acceleration – Weight transferred from front toback. • Reaction force on front wheel is given by (taking moment abt RR) • Deceleration – Weight transferred from back tofront. • Reaction force on front wheel is givenby

12. Longitudinalloading • Limiting tractive and braking forces are decided by coefficient of friction b/w tires and road surfaces • Tractive and braking forces adds bending throughsuspension. • Inertia forces adds additionalbending.

13. Asymmetricloading • Results when one wheel strikes a raised objects or drops into apit. • Resolved as vertical and horizontalloads. • Magnitude of force depends on Raised object` • Speed ofvehicle • Suspensionstiffness • Wheelmass • – Bodymass • Applied load is a shockwave • Which has very less timeduration • Hence there is no change in vehiclespeed • Acts through the center of thewheel.

14. Asymmetricloading • Resolved vertical forcecauses: • Additional axleload • Vertical inertia load throughCG • Torsion moment • to maintain dynamicequilibrium. • Resolved horizontal force causes: • Bending in x‐zplane • Horizontal inertia loadthrough CG • Moment about zaxis • to maintain dynamicequilibrium. • Total loading is the superposition of all fourloads.

15. Allowablestress • Vehicle structure is not fullyrigid • Internal resistance or stress is induced to balance externalforces • Stress should be kept to acceptablelimits • Stress due to static load X dynamic factor ≤ yieldstress • Should not exceed 67% of yieldstress. • Safety factor against yield is1.5 • Fatigue analysis isneeded • At places of stressconcentration • Eg. Suspension mounting points, seat mounting points.

16. Bendingstiffness • Important in structuralstiffness • Sometimes stiffness is more important than strength • Determined by acceptable limits of deflection of the side frame doormechanisms. • Excessive deflection will not shut doorproperly • Local stiffness of floor isimportant • Stiffened by swages pressed intopanels • Second moment of area should beincreased

17. Bendingstiffness • Thin panels separated by honeycomb structure reducedvibration • Local stiffness has to be increasedat: • Door • Bonnet • Suspension attachpoints • Seating mounting points • Achieved by reinforcement plates andbrackets.

18. Torsionalstiffness • Allowable torsion for a medium sized car: 8000 to 10000 N‐ m/deg • Measured over the wheelbase • When torsion stiffness islow: • Structure move up and down and/orwhip • When parked on uneven ground doors fail toclose • Doors fail to close while jacking if jack points are at acorner • Torsion stiffness is influenced bywindscreens • TS reduces by 40% when windscreensremoved • Open top cars have poor torsionalstiffness • Handling becomes very difficult when torsional stiffness is low.

19. Chassis types‐ Ladderframes • Used by early motorcars • Early car’s body frame did not contribute much for vehicle structure. • Mostly made of wood which has low stiffness • Carried all load (bending and torsion) • Advantages: • Can accommodate large variety of body shapes andtypes • Used in flat platforms, box vans, tankers and detachablecontainers • Still used in lightcommercial Cross beam Siderails vehicles like pickup.

20. Chassis types‐ Ladderframes • Side rails frequently have open channel section • Open or closed section cross beams • Good bending strength andstiffness • Flanges contribute large area moment ofinertia. • Flanges carry high stresslevels • Open section : easy access for fixing brackets andcomponents • Shear center is offset from theweb • Local twisting of side frame is avoided • Load from vehicle is applied onweb • – Avoids holes in highly stressesflanges • Very low torsionalstiffness.

21. Chassis types‐ Ladderframes Clockwise side framebending • Torsion in cross memberis reacted by bending ofside frames • Bending in cross framesare reacted by torsion of side frames • All members are loadedin torsion • Open sections are replaced by closed sections to improve torsionalstiffness Anti‐clockwise cross frametorsion • Strength of joints becomescritical • Max bending occurs atjoints • Attachment of brackets becomes morecomplex

22. Chassis types‐ cruciformframes • Can carry torsional loads , no elements of the frame is subjected to torsionalmoment. • Made of two straightbeams • Have only bendingloads • Has good torsional stiffness when joint in center is satisfactorily designed • Max bending moment occurs in joint. • Combining ladder and cruciform frame provides good bending and good torsionalstiffness • Cross beams at front and back at suspension points are used to carry lateralloads

23. Chassis types‐ Torque tube backbone • frame • Main back bone is aclosed Backbone • boxsection • Splayed beams at front and rear extent to suspension mountingpoints • Transverse beams resist lateralloads • Back bone frame:bending and torsion Transverse beam • Splayed beams:bending • Transverse beams: tension orcompression Splayedbeams

24. Chassis types‐ Spaceframes • In all frames till now length in one dimension is very less compared to the other two dimensions • Increasing depthincreases bendingstrength • Used in racecars • All planes are fullytriangulated • Beam elements carry either tension or compressiveloads. • Ring frames depends on bending ofelements • Windscreen, backlight • Engine compartment, doors • Lower shearstiffness • In diagonal braced frame s stiffness provided by diagonal element

25. Chassis types‐ Integralstructures • Modern cars are massproduced • Sheet steel pressings and spot welds used to form an integralstructure • Components have structural and otherfunctions • Side frames + depth + roof gives good bending and torsionalstiffness • Geometrically very complicated • Stress distribution by FEMonly • Stress distribution is function of applied loads and relative stiffness betweencomponents • Advantages: • Stiffer in bending andtorsion • Lowerweight • Lesscost • Quietoperation

26. Structural analysis by Simple Structural Surfaces (SSS)method • Many methods to determine loads andstresses • Elementary method is beam method, FEM is advanced method and SSS isintermediate • Developed by Pawlowski in1964 • Determines loads in main structuralelements • Elements are assumed to be rigid in itsplane • Can carry loads in itsplane • – Tension, compression, shear and bending • Loads normal to plane and bending out of plane is invalid and not allowed

27. SSS method – Analysis of simplevan (torsioncase)

28. SSS method – Analysis of simplevan (torsioncase) • Ten structural componentsare considered • If geometry is known and axle loads are known, edge loads (Q s) can be determined. • For a fully laden van front axle load islighter. • By moment balance R’r can be determined. • R ' r *tr Rf*tf • 2 2

29. SSS method – Analysis of simplevan (torsioncase) • The equilibrium of SSS‐2 and SSS‐ 3 are obtained by taking moments as Rf and R’r areknown. • SSS‐2 (front crossbeam) P2w Rf *tf 0 • 2 • SS‐3 (Rear cross beam) • P3wR ' r *tr  0 • 2 • P2 and P3 will be equal in magnitude as they act at the width of the vehicle and the torque at the front and rear must be equal.

30. SSS method – Analysis of simplevan (torsioncase) • ConsideringSSS‐6 • Q1 to Q5 will occur around periphery • Applies opposite moment to P2 andP3 • Taking moment atA • P3(l1l2l3)Q3(l1l2l3l4)Q4(h1h2)Q2h2 P2l10 • Consider SSS‐4 (front panel) • Q6h2 Q1w  0 • Consider SSS‐5 (rear doorframe) • Q6h1Q3w0

31. SSS method – Analysis of simplevan (torsioncase) • Consider SSS‐8 (floor panel) • Q6(l1l2l3l4)Q2w0 • Consider SSS‐9 (windscreenframe) • Q6(h1h2)Q5w0 sin • Consider SSS‐10(Roof) • Q6l5Q4w0 • Six unknowns Q1 toQ6 • Substitute Q2, Q3 and Q4 in the eqn ofSSS‐6 • Q6 can be obtained and hence rest of the unknowns can bederived

32. Simple Structural Surfaces representing a saloon carin bending Material from J.H. Smith,2002

33. Passengercar • More complex than box typevan • Detailed model vary according to mechanical components • Front suspensions loads applied to front wing as for strutsuspension • Rear suspension (trailing arm or twist beam) loads to inner longitudinal member under the bootfloor • SSSs varies with body types

34. Vehicle structures represented by SSS Bus or box typevehicle Van Passengercar SSSand NotSSS

35. Structures that are structuralsurfaces Image fromJ.C.Brown,2002

36. Structures that are NOT simple structuralsurfaces Image fromJ.C.Brown,2002

37. Half saloonmodel • Limited to 5Loads • F1z= (radiator, bumper,battery)/2 • F2z=(engine)/2 • F3z= one front passenger andseat • F4z = one rear passenger, seat, and half fueltank • F5z =(luggage)/2 • 1 UDL (bodyweight)

38. Process • Calculate reactions at front and rearaxles • ( taking moments and vertical forceequilibrium) • Rzf/2 • Rrz/2 • Calculate forces in each of theSSS • 11equations with 11 unknowns ( K1, .. K10, M) can be evaluated from SSS1 toSSS8 • Equilibrium of right frame to be verifiedwith forces andmoments

39. Half Saloon car model ‐Bending

40. Figure SSS1 • TransverseSSS • representing the strut tower • ResolvingForces • K1 + K2 – Rfz / 2 =0 • Moments • K1 =Rfz*w1/(2*(w1+w2))

41. Figure • UpperfrontlongitudinaSlSS2 • ResolvingForces • K1 –K3 – u (l1 +l3) =0 • Moments • K1l3 – u*((l1+l3)2/2)‐M=0

42. Figure SSS3 • Lower frontlongitudinal • ResolvingForces • F1z + F2z + K5‐ K2‐ K4 =0

43. Figure SSS4 • engine firewall • Resolving Forces and by symmetry • K5 ‐ K6 =0

44. Figure SSS5 • Floor Cross beam(Front) • Resolving forces and bysymmet • K7‐K4‐F3z=0 ry

45. Figure SSS6 • Longitudinal underboot • Resolvingforces K9+ K8‐ Rrz /2 + F5z=0 • Moments: K9 = (Rrz*l6/2 – F5l10) /(l5+l6)

46. Figure SSS7 • Floor cross beam(rear) • Resolving forces and bysymmet • K9‐K11‐F4z=0 ry

47. Figure SSS8 • RearPanel • Resolving forces and by symmetry • K10‐K8 =0

48. Figure SSS9 • Right‐hand side‐frame • Resolvingforces • K6 – K7 + K11 + K10 – u*(L + l6 – l3) =0 • Moments aboutA • K10*(L + l6 – l3)+K11*(L – l3 – l5)–K7*(l4 – l3)–u*(L + l6 – l3)2/2 =0

49. Conclusion • SSSs 1 to 9 are subject toloads • The rear boot top frame, rear screen, roof, windscreen, floor panel and boot floor have no loads applied tothem • The side‐frame carries the major loads and is the main structural memberfor determining the bending stiffnessand strengthof thecar.

50. SSS representation of a saloon carin torsion • Front axle is assumed to be lighter thanrear. • Maximum torque that can be applied is: • Rfz *tf R ' rz*tr • 2 2 • Rfz and R’rz arereaction • loads at suspension mountingpoints • R’rz can beobtained.