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4.1 primary structure
4.1 primary structure
Primary structure starts with suspension mounting point, cockpit opening and regulation in order to detemine size and position of the main hoop ,front hoop and front bulkhead.
suspension point
cockpit opening
regulation
After getting the size and position of the main hoop and front hoop, we create trusses following the sketch 2 with enough space for other component.
First model problems
complained about height of isolate front from Team.
geometry has been redesigned for driving performance.
battery design team need more space.
steering rack touch with the primary structure
Improve the primary structure.
create member that can connect with new geometry.
change member for not touching between frame member and steering rack.
increase space for battery pack, result is longer isolate front and position changing of main hoop and front bulkhead.
fig.1 recheck space for battery pack, cockpit opening and drivetrain.
Replace pipe with the material is AISI 1020 size diameter 25.4 mm thickness 2 mm form STEP 1 1.2 material selection.
In this step can receive weight of primary structure that equal to 43 kg. less than the 45 kg. (target), therefore, design criteria weight topic is passed.
4.1.1 Torsional stiffness
Given the load applied 1000 N at each node point. [2][3][4] because this load is maximum that can apply on the primary structure without failure.
test in two cases
fix rear/ load on front suspension mounting point.
fix front/ load on rear suspension mounting point.
fig. 2 torsional stiffness condition.
neglect horizontal deformation, use vertical deformation to calculate chassis rotation [5].
fig. 3 maximum vatical deform point of fix rear/ load on front case.
result: fix rear/ load on front case has chassis rotation = 1.05 degree
likewise, fix front/ load on rear has chassis rotation = 0.76 degree
Torsional rigidity is defined in unit Nm/deg. and can defined multiple ways, the design team chose the method that appeared most frequently during benchmarking [6]. F, the applied moment of the center line of the chassis is calculated. this moment is the divide by the angular rotation of the chassis.
For the obtained torsional stiffness, the design team use lower torsional stiffness to determine the torsional stiffness of primary structure. Thus, torsional stiffness of primary structure = 1513.75 Nm/deg. > 1200 Nm/deg. (Target)
4.1.2 Failure criteria
The force due to inertia in racing on track causes from three situations.
acceleration
braking
cornering
Between acceleration and braking case there is a similar force acting on the primary structure. We will simulate only the braking case because it is 1.5g more than in the acceleration case. cornering case is considered at latteral acceleration equal to 2g.
In addition, design team need critical torsions to verify failure. critical torsion can occur in two case is cornering until rollover and driving on uneven road.
the track is a flat load. So, driving on uneven road is not necessary but still calculate asymmetry case (statics) during transportation. therefore, design team chose critical torsion form maximum force produced by rollover or asymmetry to determine failure.
DETERMINE mounting point.
Drivetrain
accumulator container for battery pack
driver
impact attenuator
4.1.3 racing case
4.1.3.1 Braking case
maximum deceleration = 1.5g.
Use the Ansys remote force function to apply force on mounting point. And the location of the force is CG of that component. fix support at both rear and front suspension mounting point.
result : maxsimum combined stress = 21.0 Mpa, minimum combined stress = 10.7 Mpa.
*mark*: Use the functions in Ansys to evaluate the shape drawn as a Beam with the desired cross-sectional dimensions. which this function can use to calculate stress only max. and min. conbined stress.
max stress 21 Mpa. < yield stress 294 Mpa. safety factor = 14
4.1.3.2. Cornering case
Centripetal Acceleration = 2g.
Use the Ansys remote force function to apply force on mounting point. And the location of the force is CG of that component. fix support at both rear and front suspension mounting point.
result : maximum combined stress = 77.6 Mpa, minimum combined stress = 7.0 Mpa.
max stress 77.6 Mpa. < yield stress, tensile 294 Mpa. safety factor = 3.78
4.1.4 Critical torsion
4.1.4.1 Select maximum torque
Rollover case
Rollover is cornering at heigh accelerate that the formula rollover. Frist, to find the rollover acceleration value. the next stage, to calculate total roll moment, that is 1765.80 Nm, balancing by vertical load Fr and Ff at the front and rear suspension mounting [7].
But design team don't know relationship between Frt and Fft, and only one equation. A ration must be assumed between these two unknowns. Therefore, it will be assumed that front suspension provides nM of the total roll moment, when n = 0.5 - 0.7 [7]. we select n = 0.7 because it creates high roll moment.
maximum force from rollover case = 1030 N
Asymmetry case
From suspension team, maximum force = 1500 N
Select maximum force (from asymmetry case) = 1500 N to check failure
4.1.4.2 Failure Check
condition:
fix rear/ applied moment on front suspension mounting.
fix front/applied moment on rear suspension mounting.
front lock and rear lock both conditions are no failure.
fix support at front condition creates maximum stress = 110 Mpa with safety factor = 2.6
all Failure criteria = pass
4.2 aerodynamics
4.2 aerodynamics
Because of time, team decided to use old model of aerodynamic devices (front wing, rear wing, sidepod, undertray) and will design angle of nosecone (seta) and choose the design that will occur less coefficient of drag.
4.2.1 Nosecone
we will adjust seta only and calculate cd, drag force and lift force and also consider area of impact attenuator.
Parameter set
velocity of car = 36 m/s (car max speed)
area = 0.21 m^2
air density = 1.225 kg/m^3
temperature = 288.16 K
Mesh Setting
First model (62.1 degree)
Second model (55.9 degree)
Third model (72.9 degree)
1 st model: 62.1 degree
drag force = 36.80 N
cd = 0.22
lift force = -8.57 N
cl = -0.05
Velocity contour
Pressure contour
2 nd model: 55.9 degree
drag force = 37.01 N
cd = 0.22
lift force = -12.92 N
cl = -0.08
Velocity contour
Pressure contour
3 rd model: 72.9 degree
drag force = 39.21 N
cd = 0.24
lift force = -5.84 N
cl = -0.04
Velocity contour
Pressure contour
For nosecone, we want less drag force and cd, although 2 nd model has drag force and cd close to 1 st model but it generates more down which we do not want. In model 3 rd, it generates less down force but most drag force. So we choose 1 st model, 62.1 degree, for this car.
4.2.2 Rear wing joints
Rear wing is the main part which generate down force so the joints and weld points must have enough durability. Because we did not calculate cd of cd so we will apply down force and mass of rear wing to the mounting rack.
from
Parameter set
Load = 540 N. (S.F. = 1.9)
Wing = 9.2 kg. = 91 N.
Downforce = 200 N.
Distributed load
Joint
Material: AISI 1020
Thickness = 5 mm.
Deformation
Max deformation occurs at attachment point between rack and wing which equals to 13.97 mm.(max)
Von-Mises Stress
At welding points has Von-mises stress equal to 112.7 MPa which is not exceed Yield stress (294 MPa).
Max = 120.4 MPa
S.F. = 2.6
Shear stress
At welding points has Shear stress equal to 33.5 MPa which is not exceed Yield stress (294 MPa).
Max = 46.8 MPa
S.F. = 8.7
4.2.3 Weight
Aero dynamic parts have total weight equal to 32.6 which are not exceeding the target (32.7 kg.)
Material: carbon fiber
Density = 1.15 g/cc
4.2.4 Conclusion
Assembly
4.3 Seat and belt joints
4.3 Seat and belt joints
4.3.1 belt joints
4.3.1 belt joints
Belt joints will be attach to 6 point as shown in the picture and we design the joint to be ring shape in order to rotate to the best angle for attaching belt. The ring will be welded to attach to primary structure
point A-A : 20cm each from the center
point B-B : 10 cm each from the center
point C-C : each ends of the tube
FBD
After inspection the detail of belt direction while impact, the approximated direction had been founded as shown.
Then apply force 40g at driver (75kg) and apply force to the calculation to get the final belt tension result.
equation
1.) Fx=0 2(0.785A+0.383B+0.163C) = 30,000
2.) Fz=0 0.453A-0.663B-0.925C = 0
3.) Mcenter=0 A-B-C = 0
Result:
belt A = 13170.5N
belt B = 11428.8N
belt C = 1741.69N
We can assured that the regulation recommendation is correct(15000N)
Design
The design will be tested in ANSYS simulation and maximum equivalent stress will be use to consider the failure.
Simulation setup : Fix support at attached area , 15,000 N compression force at inner ring in forward direction
The first design is a ring that attach to primary structure which has 1-1.5 cm radius and made from steel.After apply 15,000N force, the simulation show the failure occur at the middle part of the ring (Max stress is near the cross section that the ring attach to the primary structure but can not use that as referent because it is close to fixed support which is not accurate).
In the final design , the ring has more cross section area by becoming half ring with 2 part attach to the structure and has 100 mm^2 cross section area and have better force distribution. The simulation after apply 15,000 N (from regulation) show that the ring can withstand the force with S.F.=1.27.
4.3.2 seat joints
4.3.2 seat joints
Seat will be attach to 4 point as shown in the picture and we design the seat joint to be fixed by steel bolt screwed through seat and attach to primary structure. A and B bolt are attach 9 cm from center.
FBD
After inspection the joints. the impact force (40g of 75kg from driver and 10 kg from seat )will be distributed to each joint calculated by moment equation(assume all the driver impact force send to seat because the shape of seat obstruct force path)
Fs = 10kg x 40g ≈ 4,000N
Fp = 75kg x 40g ≈ 30,000 N
MboltB = 0
0 = 2AL - 2/3(4,000N) - 2/4(30,000N)
Bolt A = 12,583.33 N (compression at side of the bolt)
F = 0
2B + 2A - 30,000N - 4,000N = 0
Bolt B = 4,416.67 N (tension at top of the bolt)
Design
The design will be tested in ANSYS simulation and maximum equivalent stress will be use to consider the failure.
Simulation setup : Fix support at floor , 12,583.33 N compression force at side of the bolt and 4,416.67 N tension force at top of the bolt)
The first design is 1 cm radius and 2 cm length steel bolt attach to 2 mm steel floor. The simulation show the failure occur at thin floor in bolt A case (compression force case) due to exceeded force had been distributed to small area.
In the final design. steel bar(4 x 24 x 1 cm )with bolt hole has to be attach to the floor by welded to create more area for distribute force along the floor. The simulation after apply each force show the result that the joint can handle each force with S.F. = 2
4.4 impact attenuator
4.4 impact attenuator
IA must withstand a collision force at a speed of 7 m/s, and the car has a mass of 300 kilograms. IA must absorb the energy of 7350 J, and the AIP plate must not deform more than 25 millimeters.
Mutiple model
In the first step, we designed the shape to be simple, which is a rectangular shape, cylindrical shape, and a honeycomb shape to visualize the energy absorption calculation before performing it in a more complex shape. The rectangular shape has dimensions of 140mm x 140mm with a thickness of 2 millimeters, while the cylindrical shape has a diameter of 145 millimeters (which gives a radius of 72.5 millimeters) and a thickness of 2 millimeters by using an aluminum alloy as the material choice is beneficial due to its lightweight nature and relatively low cost compared to other materials. The AIP will be in the form of a square plate with a side length of 400 millimeters and a thickness of 5 millimeters. To adhere to the given specifications, the AIP will be fabricated using steel material in accordance with the designated regulations.
rectangular shape
cylindrical shape
According to the regulation, the energy absorption must be at least 7,350 J under the conditions of a car with a weight of 300 kg and a collision speed of 7 m/s, and Decelerates the vehicle at a rate not exceeding 20 g average and 40 g peak.
In reality, the car would need to collide with a wall, but in the simplified model, an Impact Attenuator is used, and a rigid body with a mass of 300 kilograms is used to represent the car for simplicity.
Analytical Estimation of Impact Values
Given that, a car of m = 300 kg travels at vimpact = 7 m.s−1 to hit a rigid wall. Design that can fully stop the car by absorbing at least Ke = 7350 J of the kinetic energy that the car exerts. The decelerations of such an event are not higher than aaverage = 20 g on average and apeak = 40 g at peak, with g = 9.81 m.s−2.
Approximation of impact time = vimpact/aavg
= (7 m/s)/(20 x 9.81 m.s−2)
= 0.0356 s
There are two versions of IA presented. According to the analytical result, the simulation time was preset to 0.04 s
In the ANSYS software, the explicit dynamic module is used to simulate the situation. This is because the simulated scenario involves a dynamic problem, which is easier to create using the explicit dynamic module. In this module, the collision speed is set to 7 m/s, and the simulation time is set at 0.04 seconds. The longer simulation time is set to accommodate the completion of the collision process before concluding the simulation. It allows sufficient time for the objects to interact and reach a stable state after the collision. The conditions set may be similar in terms of specifying the speed and time, but the shape of the Impact Attenuator (IA) may vary. The specific design and geometry of the IA can change depending on the requirements and constraints of the simulation as shown in Figure 1 and Figure 2 . Fixed support as shown in Figure 3.
The results obtained from the designed shape are as follows. By examining the total deformation of the AIP (Impact Attenuator Plate) and the energy probe of the IA (Impact Attenuator)
To determine how much energy the IA absorbs, you cannot directly observe it. Instead, you need to compare the total energy with the kinetic energy generated within the IA. By comparing these values, you can assess the energy absorption capabilities of the IA.
According to the program, it appears that the rectangular shape, when subjected to a collision, results in the IA reporting an "energy error too large." This is due to the object deforming to a point where it cannot be further analyzed. However, in the case of the cylindrical shape, it is possible to create a complete simulation. This indicates that the cylindrical shape of the IA can withstand collisions better than the rectangular shape
The deformation results of the AIP indicate that it has a deformation value less than the specified limit of 25 millimeters. Hence, the model remains unchanged until deformation reaches or exceeds 25 millimeters.
Therefore, in order to meet the specified criteria, an additional design modification was made by reducing the size of the IA cells and increasing the number of cells. The aluminum thickness remained the same, but the diameter was reduced to 65 millimeters, and the number of cells was increased to four. And it maintains the same length of 200 millimeters as before. shown in Figure below.
The updated model is then tested in the ANSYS software with the same specified parameters as before to evaluate the energy absorption capabilities. This allows us to examine the energy absorption values of the modified IA design.
From the ANSYS software, it is observed that the updated design of the IA absorbs significantly more energy than before. The IA is now capable of absorbing 35,628 J of energy, indicating a significant improvement in its energy absorption capabilities.
Based on the deformation analysis, the AIP exhibits deformation values below 25 millimeters. Therefore, it can be concluded that this model is suitable for the current design.
To summarize, the AIP has a weight of 6.54 kilograms, while the IA weighs 0.55 kilograms. When combined, the total weight is 7.09 kilograms. Additionally, the IA is capable of absorbing energy up to 35,628 Joules.
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