International Journal of Mechanical and Production Engineering, ISSN(p): 2320-2092, ISSN(e): 2321-2071 Volume- 6, Issue-6, Jun.-2018, http://iraj.in PERFORMANCE ENHANCEMENT AND ANALYSIS BY REDUCING WEIGHT OF UNSPRUNG MASS OF FORMULA STUDENT RACING CAR 1ARUN NAIR, 2SHRUTI KAWADE, 3BRYAN DIAS, 4IVAN GEORGE, 5DHIRAJ K. BHANDARKAR 1,2St. John College of Engineering and Management, Palghar (East), Maharashtra 5Assist. Prof., Faculty of Mechanical Engineering, SJCEM E-mail: [email protected], [email protected], [email protected], [email protected] Abstract - This paper is to redesign the unsprung mass of SAE SUPRA vehicle with a primary focus on reducing the weight of overall assembly. The design of the wheel assembly is critical due to the forces acting on the wheel assembly during accelerating, braking, cornering and tilting. Furthermore, the Wheel Assembly is an important part of an automobile and its failure is hazardous endangering human life. Therefore it is required to design the Wheel Assembly and its components considering all the factors leading to the failure by developing a safe design. It must also be noted, the components should be designed in such a way that they have a minimum weight at the same time care must be taken that they do not cross a certain limit of stress value. In this study, the design of the Wheel Assembly for R10 Rims has been presented. The weight of the Vehicle is considered to be 300 kg along with the driver. The forces acting on the components, the failure criteria is illustrated. The study deals with finding out the dimensions of the individual components and also detecting the probable regions of stress concentration. The main objective is to redesign and reduce the unsprung weight by 30% compared to 2017 model by the team. The design procedure follows all the rules laid down by FSAE Rule Book for Formula Type Cars. Keywords - Unsprung Mass, Reducing Weight, Wheel Assembly, SAE SUPRA, FSAE I. INTRODUCTION II. COMPONENTS OF UNSPRUNG MASS SAE SUPRA is a Student project that involves a 1. Wheels complete design and fabrication of an open wheel 2. Hub formula-style race car. Being a race-car, the primary 3. Upright goal is to achieve the best performance to weight 4. Brake disc ratio. The reduction of weight in any area will allow 5. Brake caliper for better vehicle performance overall. An automobile is said to function appropriately only when all its III. DESIGN CONSIDERATION systems are working as they are required to work. Most of the vehicle’s weight is supported by a 1) Getting the prerequisite parameters from the suspension system. It suspends the body and suspension and steering geometry associated parts so that they are insulated from the Considering the Front Wheel Assembly the road shocks and vibrations that would otherwise be Parameters are as follows transmitted to the passengers and the vehicle itself. King Pine Inclination: 6.1° However, other parts of the vehicle are not supported Caster Angle: 0° by the suspension system such as wheel axles, wheel Track Width: 1200 mm bearing, hub, tires, brakes, steering and suspension Wheel Diameter: 406.4 mm parts are not supported by the springs. This mass is called unsprung mass. Light weight construction is 2) Brake Calculation important for the racing. Thus generally unsprung Mass of vehicle (m) = 300 kg weight should be kept as low as possible. At 40 kmph The unsprung mass must be lower than the sprung Velocity (v) = 11.11 m/s Distance (s) = 8 m ² mass and also should be as least as possible to Braking force = = 2314.35 N provide proper drive stability and load balancing of . Force on one wheel= = 578.59 N the vehicle. There are a lot of forces acting on the wheels in the static and especially in the dynamic Braking torque on one wheel condition. Thus utmost care must be taken while = 528.59 x 9.5 x 25.4 x 10-3 designing the Wheel Assembly. The objective of = 139.614 Nm Optimization is always to find the best possible and Similarly, At 100 kmph suitable dimension. v = 27.78 m/s Here, R10 rims have been used along with 16 x 6 x s = 8 m 10 tires. The Weight of the vehicle is considered to be Braking torque = 872.897 Nm 300 kg along with the driver. All the forces have been found out on the above basis and according to the Wilwood PS-1, above mentioned Wheels. The design procedure Piston diameter (d2) = 1.12 inch = 28.448 mm follows all the rules laid down by SAE Rule Book for Area (A2) = 0.99 inch2 = 638.7 mm2 Formula Type Cars. Using Pulsar rear Performance Enhancement and Analysis by Reducing Weight of Unsprung Mass of Formula Student Racing Car 38 International Journal of Mechanical and Production Engineering, ISSN(p): 2320-2092, ISSN(e): 2321-2071 Volume- 6, Issue-6, Jun.-2018, http://iraj.in Diameter of bore (d1) = 14mm Therefore selecting SKF 6008 as the bearing suitable Area (A1) = 153.94 mm2 for front wheel assembly Force of 2000 N Pedal ratio = 4:1 Therefore C=13200 Force on master (F1) = 2000 x 4 = 8000 N Therefore P actual=3583.03N As this is greater than the previously assumed load . F2 = = = 33192.153 N Therefore, it is safe for our design . b) Rear Bearing For 2nd piston, Similarly, Considering 300 kg vehicle the load on Total Clamping Force = 2 x 33192.1= 66384.306 N rear will be 200 kg Assuming entire load acts on a single bearing Frictional Force = 66384.306 x 0.4 = 26553.7224 N Therefore, radial force is considered 1.5G Therefore Fr = 1.5*200*9.81 =2943 N Case 1: Considering 180 mm disc C = 16263.15 N Braking torque = 26553.7724 x 0.09 =2389.84 Nm Approximately C = 16500 N Therefore selecting SKF 510001 as the bearing Case 2: Considering 120 mm disc suitable for rear wheel assembly Braking Torque = 1593.223 Nm 4) Design of Hub Case 3: Considering 150 mm disc a) Determining the forces acting on the Hub: Braking torque = 1991.53 Nm Torque on the Brake Disk Petal: A torque of 105Nm is acting on the Brake Disk Petal. 3) Bearing Calculation The Force acting on each hole a) Front Bearing ÷ = Let us consider 50:50 weight distribution of load Therefore considering 300 kg vehicle the load on ÷/ front will be 150 kg = = 846.77 N Considering that this entire load acts on a single Torque on the Wheel Petal: - bearing The Force acting on each hole = 486.11 N Therefore, radial force is considered 1.5G Therefore F = 1.5 × 150 × 9.81 =2207.25N r Force due to Side Impact: - Here the side Impact force is taken to be 2G Axial force is considered 1G = 2 × g × vehicle mass Therefore F = 1 × 150 × 9.81=1471.5N a Impact force = 2 × 9.81 × 300 = 5886 N Now let us consider P = 1.5Fr…standard practice Impact force on 1 petal = 5886/4 = 1471.5 N P = 1.5 × 2207.25 = 3310.875N Now, = (′10) Loads on Bearing: - For ball bearing k=3 The load on 1 bearing is 750 N Therefore let us consider l’10 =50mr The load on 2nd bearing is 750 N Therefore C =3310.875 × 50 The axial load on the bearing is 1450 N = 12197.368N Approximately, C =12500N b) Design of Wheel Petal Shear Failure of Petal Therefore from PSG page no 4.12, 4.13 Allowable stress in the Hub in shear = τ = 56.875 N/mm2 D D B C C0 N Force acting on Petal as obtained is 846.77 N Bearing (mm) (mm) (mm) (N) (N) (rpm) Now, τ = SKF6007 35 62 14 12500 8800 13000 . 56.675 = SKF6008 × ( × ) 40 68 15 13200 9800 10000 t x b = 7.43 =7.5 6008-RZ 40 68 15 17800 11600 22000 Where, t= thickness of Wheel Petal SKF6206 30 62 16 15300 10000 13000 b= distance between the hole and the end of petal SKF6207 If t = 10mm b= 0.743 mm 35 72 17 20000 13700 10000 If t = 15mm b= 0.495 mm SKF6208 40 80 18 22800 16000 10000 This bending is due to the force of 846.77 N. Table 1 The radius of effective bending is 54 mm Bearing selection Mb = 846.77 × 54 Performance Enhancement and Analysis by Reducing Weight of Unsprung Mass of Formula Student Racing Car 39 International Journal of Mechanical and Production Engineering, ISSN(p): 2320-2092, ISSN(e): 2321-2071 Volume- 6, Issue-6, Jun.-2018, http://iraj.in Mb = 45738 N-mm 2) Geometric modelling of Brake disc. Now, By Flexural Equation, σ = σb= 113.75 N/mm2 y= 2b=d I = x t x d3 . = / t=10mm, d=16mm Figure 2: Brake disc Geometric model Total thickness is = width + diameter of hole 3) Geometric modelling of Upright =16+10 = 26 mm c) Design of Brake Disk Petal Allowable stress in the Hub in shear =τ =56.875 N/mm2 σ = σb= 113.75 N/mm2 I = x b x t3 t= thickness of petal b= width of petal =26 mm as found from above calculation Force acting on Petal as obtained is 846.77 N Now, . = / t = 9.62 The thickness of petal is taken as 10 mm. 56.675 = × (×) t x b = 7.44 If t = 4mm, b = 1.86 mm Figure 3: Upright Geometric model Thus the width of the petal is taken to be 5mm. V. DESIGN ANALYSIS IV. GEOMETRIC MODELLING 1) Analysis of Hub 1) Geometric modelling of Hub.