Any engineers or engineering students here?

[Dave R]

It doesn't hurt to use the axle's max torque load but if the engine/drivetrain is not capable of generating that load, then your over engineering.

The force applied to each of the tubes depends on how far away from the centerline you place them, not how far apart they are from each other.

Also, if it makes a difference, the tubes do not both see compression loads at the same time. And generally speaking only the lower will see max load, unless your planning on using full torque in reverse and even then it'll be a much lower load as reverse gear is rarely ever as steep as first. That and for articulation is why 3 link (wishbone center) set-ups are popular.

 

[Bray D]

I've been following you guys. I'm a mech. engineering student. I'll transfer next year to U of I. I've completed most of the math so far (through diff eq.) and am currently enrolled in my second physics class (electricity and magnitism) and statics. haven't taken dynamics or strength of materials yet so I'm a little behind. sorry I don't have anything useful to add to the discussion but I enjoy lurking haha

 

[Jim Oaks]

Still sticking with the 1.75 OD x .250 Wall 1.25 ID tubing and:

I = .3399
Z = .3885
A = 1.1781
k = .5372

(hopefully A & K came out right)

E = 29,000,000
Yield Strength = 70,000

48-inch length link
1,250 pounds on center of link (5,000lb / 4)

I had calculated a stress yield of 38,610psi

I used Buckling = (pi^2*E*I) / (L^2) for a result of 42,229psi

Thus far if these calculations are correct, I'm good because the 38,610psi is less than the tubes yield of 70,000 and below the buckling rate of 42,229psi. Sound right so far?

The torque is a little confusing..........

Krug, you indicated Fa = T/(a+b). I'm calculating my links at the axle will be 10.5-inches apart. Unsure if the lower link will be at the axle center or below it. What Dave states about the force on the tubes being affected by their distance from the axles centerline makes sense.

If (a+b) isn't the distance between the links, is it then more of a + & - calculation? Example, -2.5" below axle centerline + 8" = 5.5?

For T (torque), am I suppose to use inch or foot pounds?

Just playing with the formula right now I used:

Fupperlink = Fa/cos(Angle a)

T = 66,000psi (Multiplied the 5,500lbs X 12)
a+b = 10.5 (I used distance between a & b)
Angle a = 5 (degrees)

Force = 22,159psi

 

[krugford]

It doesn't hurt to use the axle's max torque load but if the engine/drivetrain is not capable of generating that load, then your over engineering.

Dave is right there. If your engine can't produce enough torque with your crawl ratio, then the axle will never see it.
I had a nice long chat with a friend of mine about this thread tonight. We were in the same classes in college. He mentioned that I should have included a ground force and it was at that point I realized that I had assumed a very worst case scenario. That image that I had attached did not include a ground force, because I had assumed that the tire was going to be in a clamped, or trapped, position. This means that ALL of the force goes into the links. I do realize that this could very well happen while rock crawling, but I don't think that you're going to get stuck and just mash on the gas till something gives. Because it won't be the links!

I didn't check the math because I left my calculator at work. So I'll assume your numbers are correct. I would say you are good on the 1250 lb force applied to the center of the link. You'll have a factor of safety of around 2. You don't compare this result to the buckling equation however. The buckling equation is only used for axial loading. This would be the forces applied to the links through the axle torque, the truck standing up on end, etc.

As for the load applied because of the axle torque, the forces applied to the links alone are going to be equal and opposite because of the torque moment applied through the axle. If you sum the forces in the x direction (horizontal) for the image I posted, the only two forces are the forces on the links. These are your "resolved" forces. Meaning the actual force in the links are broken down into their individual components. The actual forces in the links are going to depend on their angles (front to back and side to side) but they will be equal when broken down into their front to back horizontal force components.

The distances of a and b are the distances of the links from the centerline. It doesn't need to be broken up like this, since both equations are the same and both use (a+b), all you really need is the length, or distance between the links. One other thing to keep in mind, if your axle mounts are going to be swept forward from the axle (in the shape of a "C"), this is going to change (increase) the forces on the links because it will change the angle from tangential (angle "a") on the drawing.

The torque should be 66,000 in*lbs. (a+b) should be 10.5 inches. And the cosine 5 deg should be really close to 1. Your force on the links due to axle torque alone should be just over 6000 lbs. This is also assuming that the 5500 ft'lbs of max torque is what can be delivered to one side of the vehicle. Remember, you have two upper and two lower links. Your forces due to axle torque are small compared to the forces from landing on a rock in the middle of the span.

Which leads to my next question. Has anyone on here actually broken/bent a link from anything other than an impact or bad welds?

Bray D, would that be U of Iowa or U of Illinois?

 

[Bray D]

U of Illinois. My bro graduates this year with an AgE degree so next year I get to slide right into his apartment

 

[Jim Oaks]

How do you calculate a factor of safety in?

Hmmm.....I got a 42,229psi buckling result and a force rating from torque of 22,159psi. I'm assuming 42,229 is what the column can withstand and 22,159 is what I could be applying from the max axle torque. If that's the case, the axle is going to give before the links which is what I would want.

How do we recalculate this with ground force? This make me think of drag factor/coefficient of friction in my technical crash formulas.

 

[Jim Oaks]

Ok, how about this:

If my 1.75"x.250" tubing will buckle at 42,229psi and the buckle formula was designed for columns, then we would be essentially discussing the weight of the vehicle being applied at the top of the link end and how much could be applied before buckling?

I'm a little confused here with the buckle rating and weight of the vehicle. The vehicle weighs 5,000lbs. If my link end was 1.75" surface, would you divide the 5,000lbs by the 1.75 to get PSI? That wouldn't be very high.

Now that I have my buckling yield, what do I do with it?

 

[Jim Oaks]

Hey Krug, you still here?

 

[Will]

This is why rule-of-thumb is used so often. Unless you can computer model all of the possibilities, it's a waste of time. When you have an axle hopping and slamming up and down on a climb, what is the stress then? Who knows. depends on how much energy a deformed tire with certain lugs or a spring or anything can store and release. If it were a stationary structure like a bridge you could do all of this calculation and throw in a safety factor of 4 and it would be valid.

Maybe I've already said this but some old guy told my wife one time "Good enough for governmnet work--you know what government work is don't you?

Measure it with a micrometer.

Mark it with chaulk.

Cut it with an axe."

 

[krugford]

Yeah, I'm still here. I've been swamped at work lately and haven't had time to really sit down and think. I get a more well thought out post later tonight.

 

[Dave R]

I'd pretty much have to agree with Will. Your attempting to design something by using a very narrow set of parameters. So far you seem to have concentrated on static loads, unfortunately your creation is going to have to venture into the real world. The first rock you encounter that puts a slight deformation in one of the tubes will toss all of your calculations out the window. I haven't looked for this sort of info out on the WWW but there are plenty of people who have gone before you. Though I'm sure you've already thought of this but, I'd suggest finding some of their work and improving upon it. Personally, I look for the guys who's designs have failed. That way you know what to either avoid or what to.

I've not done much machine design since just after university and even then the work was of an extremely controled nature. We could safely predict 99.9% of the forces that the machine would encounter in it's environment. Unfortuately, you can't even begin reaching that level. So for me, that leaves you with SWAG and a nice big factor of safety. Hell, build it, break it, learn why it failed and build it again.

 

[IMenriched]

WOW
looks like you on the right path....I'm no engineer, by any means, but spent my fair share of time in a desk.

Just curious...I noticed the that your OD of material changed as well as the wall thickness from one material option to another...why not run thicker wall, .125 compared to .25 still using 1.5OD??? OR didn't gain that much tensil strentgh for you?

I realize that your trying equate the worse case ....But I think I'm getting the senerio of balancing the weight of the truck basically on 1 tire (like in an rocking crawling)...maybe I missing it but what about the forward motion (good ole momentum), of this rolling mass.
I get that your not going to running 40mph rock crawling, 1-5mph still would project a whip affect ..wouldn't it???. hence the center of weight would shift the the impact stresses??

it all depends on what you want to break 1st ...your suspension or your axles?? If you plan on your link suspension with a bit of give that may end up saving your axels or is that the whole point?

I'll shut up and sit down...next

 

[JohnnyU]

A suspension link (with a ball-type pivot on both ends) is basically a two force member. Fix one degree of freedom and things get more complicated.

As a degreed ME and a Design Engineer, I'm going to have to agree with Will on this one. As long as the dynamics work out, you're going to be set. The assumptions youd have to make would simplify this system enough to require a large safety factor (8x maybe). However if you don't make those assumptions, and instead try to account for and quantify those parameters, your problem quickly becomes over complicated.

I see no reason to over-engineer this process, which is exactly the route you are heading. Half of the "hardcore" rigs on Pirate weren't designed using FEA software or the like. Many were built just using good old-fashioned SOTP Engineering. Use your intuition, if it looks too small, it likely is, of course the oppostie is also true.

 

[Jim Oaks]

There is actually a really neat Excel program set up by Dan Barcroft and Greg Blanchette at pirate for computing 3 and 4 link suspensions. I'm not trying to duplicate or improve their work, just get an understanding of some of the basics for myself.

There are too many factors to calculate. I'm just looking for some rough base lines to start with. Obviously if the vehicle was moving and landing it's going to have different variables. There's a lot that can happen and probably a formula for everyone, but I'm sure in the hell not going to try and figure them out.

I don't like to take peoples numbers and use them if I don't understand them. There's no guarantee their even right.

I see no reason to over-engineer this process, which is exactly the route you are heading. Half of the "hardcore" rigs on Pirate weren't designed using FEA software or the like.

You'd probably be surprised at how many people have actually used programs such as Dan Barcroft's Excel program to design their buggy suspensions.

From what I see, what links that are damaged are done so by perpindicular force applied to the link such as bending it on a rock. Mostly though links fail at the joint.

Just curious...I noticed the that your OD of material changed as well as the wall thickness from one material option to another...why not run thicker wall, .125 compared to .25 still using 1.5OD??? OR didn't gain that much tensil strentgh for you?

I'm not following you. .25 would be thicker wall than .125. Using 1.50 OD tubing with .125 wall would have a stress rating of 87,594psi. and a .250 wall would be 56,515psi. Going to a thicker .50 wall would have a stress rating of 45,918psi. Just going to 1.75 OD tubing with a .250 wall would lower the stress rating to 38,610psi.

 

[IMenriched]

There are too many factors to calculate. I'm just looking for some rough base lines to start with. Obviously if the vehicle was moving and landing it's going to have different variables. There's a lot that can happen and probably a formula for everyone, but I'm sure in the hell not going to try and figure them out.




I'm not following you. .25 would be thicker wall than .125. Using 1.50 OD tubing with .125 wall would have a stress rating of 87,594psi. and a .250 wall would be 56,515psi. Going to a thicker .50 wall would have a stress rating of 45,918psi. Just going to 1.75 OD tubing with a .250 wall would lower the stress rating to 38,610psi.

Sorry i jumped in on topic and didn't realize there was more(2nd page).
I find it intersesting, it's been a while since i used this stuff.

Agreed there SOOO many variables to try to take into a account. at some point build it and test.

BUT.....Your kidding....running thicker wall of material actually lowered the tensil strength???? lodgically that doesn't track....common sense would seem to indicate that the thicker the wall the stronger that it is (yeild strength). if your after bend and warp...mod of elastisity would be a major concideration. which I'm sure that your trying to do.

Also you may concider Stainless Steel (SS) as material choice..what you have selected is low carbon....yes a higher carbon content steels will increase brittleness, but yeild strenth goes up as well (1020 vs 1045).....somewhere I remember that some types of SS (ie 416) offering high tensil strength and High mod of elastisity,(bending more and returing to oringinal shape/position)....maybe others would help confirm or debunk this. yes