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Robbie
Robbie MegaDork
3/10/20 1:25 p.m.

So I recently posed this question in my f-dat build thread. I'm at the point where I am getting ready to weld in front supports for my roll hoop, and I will go ahead and make the whole thing into a truss while I am at it. Here are my two design ideas (ignore the snowpants drawn by my son and the incorrect math drawn by me): The dotted lines are 1 inch tube and the solid lines are 1.75 tube. The floor and the front and rear uprights are already done so therefore are the same in both designs and not changing unless there is serious reason.

I got some suggestions that the design on the left is stronger (from some pretty darn smart and experienced people I may add) so I was considering that direction. But then, disaster! I didn't have enough tube. I would need to go get another 55ish inch 1.75 tube. Which is annoying, and does hit the challenge budget for $20-30 (1-1.5%!)

So I wondered if I could find out exactly how much stronger the design on the left actually is. I remember from engineering school that you actually can figure this out, but unfortunately I did not remember HOW. Enter http://skyciv.com - a totally cool free tool on the web for doing the math for you in just such situations. So after a bit of futzing figuring out a new system, I had a couple designs, and I loaded them up with 4500lbs of force downward and 2250lbs of force squeezing in on each of the uprights at the top. First thing I notice is they are similar in weight, but the right one is 1/2lb heavier and the CoG is higher and further back which is slightly less optimal.

Next I hit calculate and overlaid the displacements and the axial forces inside the bars. wow! Note that the displacements are exaggerated in the animation and not to scale.

Here are my overall findings so far. Does this pass the sniff test? Note that I'm not super interested in absolutes here, only the relative comparisons, since my applied loads are sort of arbitrary anyway.

10.95 kg total

11.18 kg total (.5 lbs heavier, 1lb total for both sides)

With ~4500lbs applied evenly to floor, and ~2250lbs applied inward at front and rear top point:

3.1mm max deflection

0.7 mm max deflection

Member #1 is overloaded (stress above 250 MPa, exceeds yield of 260 MPa)

None of the members are overloaded

Front and rear uprights are in tension

Front and rear uprights are in compression

Large horizontal displacement at front upright

Large horizontal displacement at rear upright

CoG lower and further forward

CoG higher and further back

 

TLDR, looks like both designs are strong, but according to this the right one may actually be stronger, which is surprising to me. What say you?

Mr_Asa
Mr_Asa HalfDork
3/10/20 1:43 p.m.

At first blush I'd agree with the right being stronger from loads along the y-axis.  The most resistant to deflection in that direction would be a column, the weakest would be a diving board.  I'd have to refresh myself on some of  it and take a closer look to fully be comfortable in that, though.

Sidenote: As a leftover from my days at FSAE, I'd try to triangulate those beams and get them tied in at the same node.

alfadriver
alfadriver MegaDork
3/10/20 1:50 p.m.

Based on how you loaded it, the numbers make sense to me.  The question is- does that loading represent something real or important or a goal of your design?

Seems to me, the key load is the squeezing horizontal load at the top of each frame- where the right design has a direct load path from one load to the other.

BTW, does that code do 3 dimensional structures?  Add that to the open source aero model, and you could do a lot do design a car.

And is there open source suspension modeling?

AngryCorvair
AngryCorvair MegaDork
3/10/20 1:57 p.m.

Not sure if feasible, but i'd like to see a new layout, with tubes from top left to bottom right and bottom left to top right, tied together where they cross. TL to BR should be made of the larger diameter material.

AngryCorvair
AngryCorvair MegaDork
3/10/20 1:59 p.m.

I was drawing while those guys were typing.  Mr_Asa and I are on the same page re. reacting into nodes rather than into middle of lengths.

And my diameter suggestions are loosely based on your available material.

McDesign
McDesign New Reader
3/10/20 2:13 p.m.

In reply to Robbie :

In general - never end a member into any part of another member - at joints only. 

That means if you have to go into the middle of a second tube, then bend or segment that second tube slightly, and hit it at that point.

Think about a tight string that you push on the middle of.  No matter how tight, you can always deflect it.

That's because the resisting force to a perpendicular load would have to be infinite to resist it.

Look at Georg Plasa's cage transverse dash bar, with its two diagonals from the center of it to the front strut towers - I'm confident that's what allowed his cage to collapse and kill him - they buckled the dash bar backwards, which pulled in the ends, and consequently the front legs of the cage at the A-pillars.

That one crash photo is why I designed this cage and a swing "up" door V-bar -

Forrest in Atlanta

Robbie
Robbie MegaDork
3/10/20 2:46 p.m.

In reply to alfadriver :

The loading I gave it was just a guess at a significant bump event. I dunno how accurate that is.

The tool does do 3d frames and trusses. I have no idea about suspension modeling however.

Robbie
Robbie MegaDork
3/10/20 2:47 p.m.

In reply to AngryCorvair :

I can try that

jharry3
jharry3 HalfDork
3/10/20 3:08 p.m.

You could go to a website called RISA 3d.  You can download a free sample of their 3D structural analysis package.  It has a limit on nodes but what you have won't take more than 10 or 12.     It takes a little time to figure out.   Input joints as x,y,z coordinates in what ever units you choose.  Input members as joint to joint while choosing your section size, od and wall thickness, yield limit,  35 ksi pipe in your case.   Then add loads as joint loads the load case section and combine load cases in the load combination section.   You have to set boundary conditions to restrain it.  It takes a second to run and you can get a 3d analysis of stresses and deflections.   

Stampie
Stampie UltimaDork
3/10/20 3:11 p.m.
AngryCorvair said:

Not sure if feasible, but i'd like to see a new layout, with tubes from top left to bottom right and bottom left to top right, tied together where they cross. TL to BR should be made of the larger diameter material.

I actually agree with you. I'd also like to see a top bar if possible. 

Robbie
Robbie MegaDork
3/10/20 3:17 p.m.
AngryCorvair said:

Not sure if feasible, but i'd like to see a new layout, with tubes from top left to bottom right and bottom left to top right, tied together where they cross. TL to BR should be made of the larger diameter material.

This design (according to the software) is only very marginally stronger than landing the bar in the middle, for this specific load setup.

max deflection 3.09 mm and almost a full kg heavier than the original 'left' design. The software is still saying that member 1 is overloaded (member 1 is the bottom one on the left half). INTERESTING!

Robbie
Robbie MegaDork
3/10/20 3:19 p.m.
jharry3 said:

You could go to a website called RISA 3d.  You can download a free sample of their 3D structural analysis package.  It has a limit on nodes but what you have won't take more than 10 or 12.     It takes a little time to figure out.   Input joints as x,y,z coordinates in what ever units you choose.  Input members as joint to joint while choosing your section size, od and wall thickness, yield limit,  35 ksi pipe in your case.   Then add loads as joint loads the load case section and combine load cases in the load combination section.   You have to set boundary conditions to restrain it.  It takes a second to run and you can get a 3d analysis of stresses and deflections.   

Cool! yeah that is almost exactly what this skyciv is doing. and yes they do take a few minutes to figure out!

Robbie
Robbie MegaDork
3/10/20 3:25 p.m.

even adding a top bar to the x design doesn't reduce the deflection in the bottom noticeably. I think I'm learning the triangle design really benefits from two things:

a - having something holding up the middle of the bottom bar

b - being a little bit 'taller' overall than an x design (the top bar of the triangle is higher than the middle point of the x

Robbie
Robbie MegaDork
3/10/20 3:27 p.m.

Now, we all have to keep in mind that if my guessed loading isn't really valid then none of this is proving anything.

Stampie
Stampie UltimaDork
3/10/20 3:28 p.m.

In reply to Robbie :

How about a bar straight down from that x in the middle?

Robbie
Robbie MegaDork
3/10/20 3:36 p.m.
Stampie said:

In reply to Robbie :

How about a bar straight down from that x in the middle?

Heaviest yet (by 1.5kg), and it does reduce the deflection in the bottom beam to 1.15 mm.

Stampie
Stampie UltimaDork
3/10/20 4:42 p.m.

In reply to Robbie :

Are you going to have a dash bar?

Robbie
Robbie MegaDork
3/10/20 9:03 p.m.

In reply to Stampie :

Sort of. The front upright is a hoop that also mounts the steering wheel. So that is sort of a dash bar.

AngryCorvair
AngryCorvair MegaDork
3/10/20 10:49 p.m.

In reply to Robbie :

Cool, thanks for running these!  I draw free body diagrams a lot.  I use f=ma a lot, and Statics is just the subset of f=ma where a=0.

sleepyhead the buffalo
sleepyhead the buffalo Mod Squad
3/11/20 3:21 a.m.
Robbie said:

Now, we all have to keep in mind that if my guessed loading isn't really valid then none of this is proving anything.

a distributed 4500# load seems a bit high for a single side to take?  That's equivalent to the whole car taking a 9000# vertical load?  which, if you're shooting for ~1500#s, would mean 6gs down?  I mean, that's valid for FAR part23 utility catagory.  It seems a bit high, unless you're expecting to have 1500#s of downforce; then it kind of represents a loading of 3g.  I'm not sure how I feel about 2500# loads coming in from each suspension at right angles to that point.

it does make sense to me that the "right" design is strongest, in this case.  Since the bottom beam has the largest load applied, and it's being forced into bending along its length.  So you're trying to find ways to reduce that bending, which means trying to 'shorten' the length of the bottom beam.

the right design does this by supporting that beam 'in the middle' in a way that the load is spread in tension to the two verticals and the upper piece (it's all a bunch of triangles). and the heaviest piece is helping take the main load up and covering the most 'open' section of the design.  You could make the 'right' even stronger by placing a vertical from the bottom node up to the middle of the upper bar, but doing that probably falls outside the realm of diminishing returns (based on this loading... going 3D it might have some gains).

In comparison, the 'left' design is supporting the bottom beam mainly by passing loading to the right; but the left side of the structure is largely unsupported, and is represented by it being a funky rhombus thing.  AngryCorvair's design makes it a bunch of triangles again, but doesn't have the benefit of reducing the length of the bottom beam's loading... something Stampie's design 'cures'.

standard caveats:
it's been a long time since I've done statics, I could be remembering these things wrong, ymmv, etc.

alfadriver
alfadriver MegaDork
3/11/20 6:21 a.m.
Stampie said:

In reply to Robbie :

How about a bar straight down from that x in the middle?

That's a hard temptation to not do- but doing that, the bar you suggest is a two force member.  In the old fashioned ideal world, they were useless, but in this case, since the members actually deflect and the loads are more distributed- it can be used.  Still- the better methods of joints is at a node instead of the middle of a member- the load path is a lot more effective.

Goes back a long way for me, Angry- I taught statics at Michigan, but have barely used them in my professional career

Robbie- does the tool tell you which members have the highest loads going through them?  If so, perhaps you can take some weight out of the lower loaded ones.  

Robbie
Robbie MegaDork
3/11/20 10:22 a.m.

In reply to alfadriver :

Yeah, the tool does show tons of different things (shear, moment, axial, torsion, displacement, stress, buckling, dynamic frequency, etc). Many I don't really understand. 

But I do see the axial load, and by far the largest is in the top bar (on the triangle). However I don't think this design is limited by these axial loads. Instead there is a large stress at the bottom corners, and that seems to be the bigger concern. 

If I could optimize for weight, I wouldn't use 1.75*.120 tube, but then I wouldn't meet the nhra roll bar rules.

Robbie
Robbie MegaDork
3/11/20 11:01 a.m.

In reply to sleepyhead the buffalo :

I just tried to pick loading that would represent something higher than I would expect to see, but not two orders of magnitude higher haha.

Also I have no idea what sort of g's I'll hit when hitting a bump. Especially with just the tire sidewall and a small rubber puck for suspension. The impulse will be large. How large? Hard for me to know currently.

sleepyhead the buffalo
sleepyhead the buffalo Mod Squad
3/11/20 2:27 p.m.

In reply to Robbie :

I understand.  I was putting that together so you'd get an idea of 'how big a ballpark' you guessing were having you play in.

I also did some digging around on Palatov's development website... because I remember he'd done FEA analysis with SolidWorks, and I thought I remember mention of using 'hitting an apex curb while cornering at 2g' as a design parameter.  Did exactly find that, but I did find a number of references of using '4g' when checking the strength of uprights and bellcranks... which you should probably scale for weight distribtion per tire.  So, let's say 50/50 front rear, and a 1650# target weight, and 500#s of downforce (hey, I can be optimistic!  laugh).  So you'll have a 'doward load on the frame of ~2150#s... so 4g = 8600#s.  So, I guess your distributed load is in the ballpark.

But, then, that means you'll have an either tranverse or 'negative longitudinal' force through the suspension of 2150# via cornering or braking.

so, it looks like your 'out of the air' forces are correct.  I'm not a 100% sure about the sign on the 2500# force trying to squeeze the frame... but that's the kind of thing I always had trouble working through on FBD's.  I tend to get wrapped around the axle of forces and reactive forces and mass resiting this, and that and the other thing.  So, I'll leave that up to someone else to give feedback on.

Robbie
Robbie MegaDork
3/11/20 3:18 p.m.

In reply to sleepyhead the buffalo :

Wow, that's useful - thanks!

As for forces pushing in at the tops, keep I'm mind this is just the middle section of my frame. The front suspension is forward of this and the rear behind. So any force up on the wheels would push both up and in on this section of frame.

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