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Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela , Orissa

Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela , Orissa. Visit us at : www.saenitr.in. Steering & suspension. Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela , Orissa. Marketing. Transmission. Brakes. Materials.

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Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela , Orissa

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  1. Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela, Orissa Visit us at : www.saenitr.in

  2. Steering & suspension Mission :BAJA SAE ASIA 2010 Team : BLACK MAMBA Institution: N.I.T.Rourkela, Orissa Marketing Transmission Brakes Materials Vehicle Engineering Electronics system Visit us at : www.saenitr.in

  3. MATERIAL OF CHASSIS • Tube Size: • Outer diameter=1 inch. , Inner diameter=0.74 inch. • Material selected: • 1018 carbon steel because of following reasons: • High tensile strength. • Good weldability. • Low cost. • Availability Visit us at : www.saenitr.in

  4. Material Composition: Visit us at : www.saenitr.in

  5. Mechanical Properties: Visit us at : www.saenitr.in

  6. WELDABILITY A measure of whether a steel can be easily welded is to determine the carbon equivalent (Cequiv) of the steel. For 1018 Carbon Steel : Cequiv=0.2+(0.9/6)=0.35 (Easily weldable) Visit us at : www.saenitr.in

  7. ROLLCAGE WITH DIMENSION Visit us at : www.saenitr.in

  8. WELDING PROCESS Tungsten Inert Gas (TIG) welding is preferred for welding the parts of rollcage. • The TIG (Tungsten Inert Gas) welding process generates heat from an electric arc maintained between a non consumable tungsten electrode and the part being welded. • Gas is fed through the torch to shield the electrode and molten weld pool to prevent oxidization. • Filler rod , same as that of base metal is added to the weld pool separately. Visit us at : www.saenitr.in

  9. TIG Welding Benefits • Superior quality welds •  Welds can be made with or without filler metal •  Precise control of welding variables (heat) •  Free of spatter •  Low distortion Visit us at : www.saenitr.in

  10. ANSYS REPORT The vehicle collides with a rigid wall at 60kmph (sixty kilometers per hour) and comes to a standstill in 0.2 seconds. We analyze the impact from four directions. This value of deceleration coupled with the full loaded mass of the vehicle which is approximately 350 kg (three hundred fifty kilograms) amounts to forces of approximately 29 kN (twenty nine kilo Newton). The calculations are as follows : acceleration in meters per second squared=(60)*(5/18)*(1/0.2) =83.33 Force is mass times acceleration. So force in Newton =350X83.33 =29166N ~29kN Visit us at : www.saenitr.in

  11. FRONT IMPACT • The load is applied on the two horizontal front extremity members while keeping the rear plane fixed .Each force is having a magnitude of 14.9kN and directed inwards along the X-axis. The massive force no doubt bends the bars substantially, but it doesn’t deform the driver’s compartment much. Max Stress= 4.56 x 108 pa. Min Stress= 0 Visit us at : www.saenitr.in

  12. SIDE IMPACT • A sideways impact is on that poses maximum threat to the driver’s safety. This has the potential tendency of crushing the compartment space. We have applied a load of 29 kN at the side bracing upper member while keeping the other side fixed. Once again no doubt the deformation is present but the driver has a good chance of staying unhurt. Max Stress= 2.64 x 109 pa. Min Stress= 0 Visit us at : www.saenitr.in

  13. REAR IMPACT • Similar to the front impact case, here also we have two 14.5 kN forces doing the damage. The forces are applied on the horizontal members inwards along the X-axis while the front extremity plane is kept fixed at its position. This does much less deformation because of the extensive frame work at the drivers back resting plane. Max Stress= 9.21 x 108 pa. Min stress= 0 Visit us at : www.saenitr.in

  14. ROLL OVER • The applied load for analysis is 3500N. From the Von-Mises diagrams we see that the long and mostly unsupported member has bent quite some amount but not enough to render the driver hurt. So all the analysis confirm the safety of the rollcage. The material employed for analysis is structural steel but the one to be employed for actual fabrication is the much tougher and lighter alloy steel which thus guarantees better results. Max Stress= 2.36 x 108 pa. Min stress= 0 Visit us at : www.saenitr.in

  15. Basic Suspension System The primary purpose of a suspension system is to support the weight of the vehicle and give a smooth ride. It is desirable that it should also: • allow rapid cornering without body roll when the car leans to one side. • keep tires in firm contact with the road at all times and conditions. • prevent body squat (tilting down at rear) when accelerating. • prevent body dive (tilting down at front) when braking. • allow front wheels to turn for steering. • keep the wheels vertical and in correct alignment at all times. Visit us at : www.saenitr.in

  16. Double wishbone suspension • Double wishbone suspension is an independent suspension design using two parallel wishbone-shaped arms to locate the wheel. • Each wishbone (or arm) has two mounting positions and attached to the knuckle via ball joints. • The shock absorber and coil spring mount to the wishbones to control vertical movement. • Double-wishbone designs allow to control the motion of the wheel throughout suspension travel, controlling such parameters as camber angle, caster angle, toe pattern, roll center height, scrub radius, scuff and many more. Visit us at : www.saenitr.in

  17. Why Double Wishbone ? • It is fairly easy to work out the effect of moving each joint, so one can tune the kinematics of the suspension easily and optimize wheel motion. • It decreases the cambering effect. • It checks the body vibrations. Material to be used: MS tube Outer Diameter : 1 inch Wall Thickness : 12 SWG Visit us at : www.saenitr.in

  18. Ackermann steering geometry • The steering arm makes 74° with the axis parallel to wheel axis . It is calculated by the formula • tanθ=wheel base/ (track width/2) Or,θ=74° Wheel Parameters • Diameter :21 inch • Rim Diameter :10 inch • Tread width :8 inch Visit us at : www.saenitr.in

  19. Roll center The roll center is the point about which the vehicle rolls when cornering and is different for front and rear part. The front roll centre is usually lower than the rear, thereby transferring weight to the rear during cornering. In our design we have parallel wishbone system which results in the roll center appearing very near to the ground which is beneficial for stability during cornering. Visit us at : www.saenitr.in

  20. Track rod length=62-(2*14.61)-(2*8)=16.78 inch • Track rod offset from center of wheels=3*sin74° ,where 74° is the Ackerman angle. • Tie-rod length=14.32-(3*cos74°)=13.49 inch{Taking length of steering arm=3 inch making 74° with axle} Calculation of scrub radius: Caster trail=10.5*tan10°=1.85 inch Due to KPI of 10°, shift = 10.5*sec 10°*tan 10° =1.88 inch Zoom view Caster trail Visit us at : www.saenitr.in

  21. It consists of steering wheel , steering shaft , rack and pinion gear box with gear ratio 18:1. • Selected over recirculating ball steering system for its lesser complexity of design. Visit us at : www.saenitr.in

  22. Design calculation FOR THE DESIGN OF AXLES AND TIE ROD WE WILL USE MILD STEEL AND THE ENTIRE CALCULATION IS BASED ON THE FOLLOWING DATA: Material Property Magnitude Modulus of Elasticity 200GPa Tensile Strength 455MPa Yield Strength (tension) 250MPa Poisson's Ratio 0.29 The maximum weight of the vehicle as per the specification is: 230Kg. Taking factor of safety into account while design let the load be: 250Kg. The aim is for the car to weigh approximately 250 kilograms and it is expected to generate about one times the force of gravity in cornering. As stated previously these forces were increased to give a suitable factor of safety. It was determined that a cornering force of two times the force of gravity and a weight force of 250 kilograms would be suitable. After applying the weight force in the centre of the tyre and the cornering force across the tread surface of the tyre, and working out the moments around the wheel bearing centres, it was possible to work out the forces on the top and bottom suspension mounting points to be approximately 2500N and 7500N …..(2500+2500*2) respectively. STEERING RATIO: It is defined as the angle made by the steering wheel for each 1deg movement of wheel.It has been found that car requires 580deg of steer wheel for 35deg movement of wheel .Thus the steering ratio is: 580/35 or 16.5:1. It should be within 12 to 20 .So the design is safe. Minimum Turn Radius( as pe Track (T)= 48 inch WHEEL BASE (W)=87 inch θ=35deg. R1= T/2 + W/sin θ =175.68 inch=446.23 cm=4.5m(approx) Visit us at : www.saenitr.in

  23. Determining the rack load R=steering wheel radius: 300mm r=pinion pitch circle radius: 5 mm T= number of pinion teeth:5 P=linear or circular pitch: 2*pi*r/t:6.28 mm E=input steering effort= 2*20= 40N W= output rack load: If the pinion makes one revolution; input steering wheel movement Xi = 2*pi*R  output rack movement Xo = 2*pi*r movement ratio(MR): 2*pi*R/(2*pi*r)= R/r=60. Assuming no friction: MR= W/E. W=E*MR= 40*60=2400. Visit us at : www.saenitr.in

  24. Analytically finding the shear stress through the bolt for the bottom spherical rod end. Using the following values: F = 7500N (as found previously for the bottom of the front upright); d = 10mm (diameter); Finding the area of the bolt: A= pi*d2/4= 78.5mm2. Finding the shear stress in the bolt: σ=F/A= 7500/78.5= 95.54MPa. This is much lower than the maximum permissible shear stress(455MPa) for high tensile steel. Analytically finding the stress in the tie-rods.( data taken from maruti alto) Using the following values: W = 2400N (as found previously for the rack output load); d = 15mm (inside diameter of tie-rod); D = 18mm (outside diameter of tie-rod); L = 335mm (length); FOS = 2 (factor of safety); and Visit us at : www.saenitr.in

  25. E = 200GPa (modulus of elasticity). From Euler's formula: W =π2 *EI/L2 I= FOS*W*L2/( π2 * E) =2.73*10-4 m4 Also I=pi*( D4-d4)/ 64 = 2.73*10-4 m4 Putting d=15mm we get Dmin= 4√50.62 = 15.000004. But we have D= 18 mm .Thus our design is safe. Finding stress at the base of front upright F = 7500N (as found previously for the bottom of the front upright); L = 80mm (length); D = 40mm (outer diameter); d = 30mm (inner diameter); and y = 20mm (centroid). Find the moment about the base of the upright: M = F ¤ L = 7500 ¤ 80 = 600000Nmm Find the second moment of inertia: I= π( D4- d4)/64 =85904.9 mm4. Find the stress at the base of the upright: σ=M*y/I = 600000*20/85904.9= 139.6 MPa. This value is lower than the 250MPa . So the design is safe. Visit us at : www.saenitr.in

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