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Being apart of a bigger project while taking on the responsibility of an aspect of it.

And the potentially more nuanced goal of this sprint is to design a functional trailing arm! Note that we’re going to be designing a trailing arm that will go on THE LEFT SIDE of the car. It isn’t important in the actual design, but makes the loading conditions easier to describe later on.

Topics Covered

  • Design Concepting

  • Computer-Aided Design

  • Design for Manufacturing (DFM)

  • *Design for Assembly (DFA)

  • Finite Element Analysis (FEA)

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In the top view show above the system becomes a bit more complex. We can see an additional branch/arm the connects to the pivot axis. This is to help distribute forces which you’ll see soon enough.

At this point, start thinking about what can be changed in the concept. For instance:

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I’ll quickly note here that only one of the rear wheels will be powered (ie. have the motor in it). I want to create two trailing arms which are mirrors of one another, and for the side that will not have the motor we will create a spacer to make up the distance and mounting face to the

At this point, start thinking about what can be changed in the concept. For instance:

  • How far does the wheel need to be from the pivot point?

  • How far does the shock mount need to be from the pivot point?

  • Do all the points need to lie in the same line? (Can the shock be moved vertically?)

  • Does trailing arm need to be horizontal? (At what angle is it optimized?)

If you’re getting a bit overwhelmed, that’s normal! This is likely the first time you’re approaching a design problem in this way, so it takes some time to get used to, and also feel free to ask questions! Don’t think you’re bothering us, when you ask questions, you show you’re interested and we want to help build that interest!

Now, back to the technical stuff. To answer a lot of these questions we need to set some design constraintconstraints.

Design Constraints

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Anchor
Design-Constraints
Design-Constraints

You’ll probably learn some different terminology on this, but I’m using it as the values we know that we should design around. A big part of designing parts is finding number that you need to work around, but it’s very time consuming, so I’ve done the tedious stuff so you don’t! So, I’ll list them out here, and I’ll add in sections that show how I determined these values. You don’t need to know it, but I want to feed people’s curiosity where I can, and it stands to show a good example of documenting your design decisions.

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Maximum Angular Displacement for the Trailing Arm: 10 degrees from the rest position

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titleReasoning

From McMasterCarr McMaster Carr (a good supplier, but expensive) we can see that swivel bearings usually are reated rated for around 20 degrees of swivel. We don’t have a specific bearing specified, we’ll need to judge our movement and loading requirements before we can do that.

So for now, we’ll limit ourselves to +/- 10 degrees with the thought that reducing it is better

Angle between the Pivot and the Wheel Mount: Greater than Zero at Max Extension

Expand
titleReasoning

When the car brakes, it applies more force to the front wheels, thus less force to the back wheels. As the coilovers and shocks are under compression in the resting position, they will extend to meet the ground.

In braking

NEEDS A DECISION

Loading From the Wheel

Expand
titleReasoning

The regulations define the loading condition in terms of acceleration; 1G Steering, 1G Braking, 2G Bump. We can see how load transfer effects the loading conditions on the wheels. Since the trailing arms are used on the rear of the car we’ll specifically look at those values.

Image Removed

The above screenshot is from a spreadsheet I developed that will calculate the the load distribution under a 1G brake and 1G steer. In a braking scenario there is more weight loaded to the front however. So I made some adjustments to the calculator and we can see that the rear should expect around 111 kg of mass.

Image Removed

So by the 2G bump case, we should expect an upward of force of around 2177.82 N

How the steering and braking cases impact our loading conditions are through the friction between the tire and the ground. If the car was turning with a centripetal acceleration of 1g, it would require a force equal to it’s weight. This force would be supplied by the friction from the tire, which is calculated by the coefficient of static friction and the normal force. Based on generally accepted values, the coefficient has a value less than 1, which means it cannot produce a force equal to the the weight of the car.

So now we need to make a decision. The car needs to be safe, but we don’t want unrealistic loading conditions either.

NEEDS A DECISIONthere is a backwards force applied to the wheel mounting point. This will create a clockwise moment around the pivot point, pushing the wheel down. This will contribute to an increase in downward force on the wheel. This will ideally distribute the load transfer under braking better than if it was below the horizontal.

Realistically, the moment generated will be negligible (assuming small angles) to the rest of the load transfer seen by the loading condition, and therefore we will not adjust the mass proportion present on the back.

Loading Cases From the Wheel (must pass all four loading cases)

Case 1

Case 2

Case 3

Case 4

4355.64 N Upward

4355.64 N Upward

4355.64 N Upward

4355.64 N Upward

2177.82 N Forward

2177.82 N Forward

1569.6 N Backward

1569.6 N Backward

2177.82 N Inward

2177.82 N Outward

2177.82 N Inward

2177.82 N Outward

Expand
titleReasoning

The regulations define the loading condition in terms of acceleration; 1G Steering, 1G Braking, 2G Bump (https://www.americansolarchallenge.org/ASC/wp-content/uploads/2021/12/ASC2024-Regs-EXTERNAL-RELEASE-A.pdf, Appendix F, F.2). We can see how load transfer effects the loading conditions on the wheels. Since the trailing arms are used on the rear of the car we’ll specifically look at those values.

Image Added

The above screenshot is from a spreadsheet I developed that will calculate the the load distribution under a 1G brake and 1G steer. In a braking scenario there is more weight loaded to the front however. So I made some adjustments to the calculator and we can see that the rear should expect around 111 kg of mass.

Image Added

So by the 2G bump case, we should expect an upward of force of around 2177.82 N

How the steering and braking cases impact our loading conditions are through the friction between the tire and the ground. If the car was turning with a centripetal acceleration of 1G, it would require a force equal to it’s weight. This force would be supplied by the friction from the tire, which is calculated by the coefficient of static friction and the normal force. Based on generally accepted values, the coefficient has a value less than 1, which means it cannot produce a force equal to the the weight of the car.

So now we need to make a decision. The car needs to be safe, but we don’t want unrealistic loading conditions either. Based on our priority to have a race-worthy car rather then a high performing car, we’ll assume the higher loading which means that the expected loading is:

Loading Condition

Load

Notes

2G Bump

2177.82 N Upward

Will be applied in all simulations

1G Acceleration

1088.91 N Forward

Will create higher compressive stresses in the part

1G Brake

784.8 N Backward

Will create tensile stresses in the part

1G Steer Left

1088.91 N Inward

1G Steer Right

1088.91 N Outward

There two additional to consider on top of this; the application of a safety factor, and the location of the loading.

At the time of writing this, the highest safety factor that needed to be applied in the previous car was 2 and therefore will be applied to this sprint. However, if this changes I’ll add it to the end of this section. So the new loading conditions are:

Loading Condition

Load

2G Bump

4355.64 N Upward

1G Acceleration

2177.82 N Forward

1G Brake

1569.6 N Backward

1G Steer Left

2177.82 N Inward

1G Steer Right

2177.82 N Outward

By the regulations, the loading condition is applied to the contact path of the tire (where the tire makes contact with the road). Because of this separation between the application point and the wheel mounting point, there will also be moments generated around the wheel loading point. The unloaded diameter of the wheel is 557 mm, so, there will be a moment from the acceleration/brake condition and the steering conditions. The moment arm for the 2G bump case comes from the geometry of the wheel assembly WHICH NEEDS TBD. There is also pneumatic trail which would create more moments around the mounting point, which should not be neglected. Based on the tire specifications under Sources and some comparison to other rolling resistance coefficients, the rolling resistance coefficient given is likely not a unitless coefficient, but a measurement of the pneumatic trail as rolling resistance is the force required to overcome the moment created due to a non-uniform loading at the tire. Units are not provided, and thus will be assumed as millimeters. Meaning the pneumatic trail is 3.02 mm at it’s worst (larger moment arm, larger moment).

NEEDS A DECISION

At max compression the shock should be tangent to a circle concentric to the pivot point.

Expand
titleReasoning

Based on M = F*r*sin(theta) where theta is the angle between the moment arm, we see that if the angle is anything other than 90 degrees the moment generated is not maximized. At max compression, when the spring will need to apply the greatest force, losing any of it to a bad angle seems suboptimal.

GET THE MOUNTING PATTERN FOR THE MOTOR

Timeline

Week 1 - Concepting

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The idea’s pretty simple, come up with ideas! In concepting we want to come up with as many possible solutions as possible that can solve the problem we have. In our case, the problem is how do we support the wheel given all the information in INSERT SECTION HERE Design Constraints.

What also might be useful to consider is how to make the part. If you take the best design possible that can’t be manufactured to a machinist, they still can’t make it. We won’t have the training/info session on manufacturing methods until later in the sprint, so only consider it at this stage if you have the time.

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An extension to this part of the design sprint would be to make your design parametric. A parametric design means that the dimensions of the part are defined “externally” from the feature file so it can automatically update if something needs to be changed. In essence, it makes it easier to tweak your design by not needing to dig through your feature tree to change a dimension.

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I hope at this stage you are looking at your design to make changes to make it easier to manufacture. We can can’t purely rely on “if there’s a will, there’s a way” when we want to make something, and this means something need to give, either our bank account, or our design. We rather our design whenever possible.

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If you have a preference on who you’re paired up with for your group be sure to let us know! We’ll probably create a form or a Lettuce Meet link to schedule the sessions. TBA.

Time Slot

Group A

Group B

2:45pm - 3:30pm

3:30pm - 4:15pm

4:15pm - 5:00pm

5:00pm - 5:45pm

5:45pm - 6:30pm

Week

Jens Dekker

Shangheethan Prabaharan

Week 1 and Week 3

Group A

Group B

Week 2 and Week 4

Group B

Group A

Time Slot

Group A

Group B

2:45pm - 3:30pm

3:30pm - 4:15pm

4:15pm - 5:00pm

5:00pm - 5:45pm

5:45pm - 6:30pm

Week

Jens Dekker

Shangheethan Prabaharan

Week 1 (Sep 17)

and Week 3 (Oct 1)

Group A

Group B

Week 2 (Sep 24) and Week 4* (Oct 22)

Group B

Group A

*On Oct 22nd I’ll be at a wedding for most of the day, but as it is a final review, Shangheethan can lead the discussion on the different designs so you can see what everyone created.

Sources
Anchor
Sources
Sources

View file
name2020 BRIDGESTONE SOLAR CAR TIRE DATA.pdf

Credits

This design sprint was heavily inspired by the one created by Aidan Lehal, Min Qian Lu, Kevin Bui, and Emily Guo! Big shoutout to them for the hard work they put in!

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