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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Topic: Floor failure!!!Posted: 10 Aug 2019 at 12:31pm |
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I do wish you lived near.. I owe you more than a beer of choice for all this effort!
Your 10,000# number is stated as breaking. Do you consider Breaking and fracturing to be the same? A past colleague of mine is also a Mechanical Engineer, yet you have so much more knowledge of the trailer than even what I could pass along to him. He defines break and fracture differently.. His break number alines with yours, fracture is lower.. I'm going to come out of all this smarter than when I started. I really liked not knowing all this (not yet ready to laugh about that)... |
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offgrid
Senior Member
Joined: 23 Jul 2018 Online Status: Offline Posts: 5290 |
Posted: 10 Aug 2019 at 9:08am |
The frame bending stress calc is a bit more complicated unfortunately (you probably guessed that). ![]() First you need to select the beam loading case you want to use. I am assuming a uniformly loaded single ended cantilever beam. (case 18 in the attached link). This isn't quite right of course but short of creating a detailed finite element model of the trailer its probably about as good as we can easily do. R1 in this model is the hitch and R2 is the axle. This gives a maximum moment at the axle of wa^2/2 where w is the load per unit length and a is the overhang. My 179 frame rail overhang is 80 inches and assuming a total weight of 3800 lbs (the 179 gross weight rating) and an overall length of 240 inches I get a moment of 25333 in-lbs for each frame rail. Bending stress is moment divided by section modulus so that would be 20.76 ksi. Mild steel allowable design yield is 36 ksi so that gives a safety factor of 1.73. The allowable yield point is the point at which plastic deformation of the steel will begin to occur, meaning that the frame rail will not return to its original shape after being loaded. But that's not the whole story. We need to take into account the axle mounting brackets and the effect of the torsion axle on the frame rails. That is caused by the moment applied to the frame by the torsion axle trying to twist the frame up in back and down in front of the attachment point. Check out this link to see what I mean. This is a weakness of a torsion axle design compared to a leaf spring axle. This load will be additive to the bending load of the cantilever beam we calculated above (bad), but the peak load will now be at the back end of the axle mounting bracket so the overhang will be reduced from what I calculated above (good). Short of having an FEA model of the trailer about all I know how to do is calculate the moment created by the torsion arm and weight on the wheel and then apply half this moment at the back end of the axle attachment bracket (the other half I assume will be taken at the front end of the bracket). With a torque arm length of 7 inches, a weight of 1750 lbs on the wheel, (half the axle gross weight rating) I get a moment of 6125 in-lbs and a bending stress of 5020 psi from the torsion axle. Recalculating the bending stress for the trailer beam load for the 6 inch shorter overhang behind the end of the axle bracket I get 17.8 ksi. Adding these two together I get a total bending stress of 22.8 ksi for a safety factor of 1.58. Or looked at another way 6000 lbs force on the whole trailer to reach allowable design load. Using the ultimate yield point (when the steel would actually break) for mild steel of around 61 ksi corresponds to about 2.67 g's or about 10000 lbs total load on the trailer. Not too good considering that from what I've read vehicle frames should be designed to allow for a minimum of 2.75-3 G bump loads if not higher. There will be additional stiffness added by the trailer structure itself which will reduce the loads on the frame rails, but still, its pretty clear to me that the rPod frame and axle are under designed at least for the heavier rPods and we need to be careful in how we load and handle rough roads and potholes. Keeping as much load forward as we can without exceeding the tow vehicle tongue weight rating is the way to go to minimize frame and axle bending as well as trailer sway risk. Ok, that's enough engineering for one day for me ![]() |
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1994 Chinook Concourse
1995 RV6A Experimental Aircraft 2015 Rpod 179 - sold |
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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Posted: 10 Aug 2019 at 8:11am |
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Here's a screen shot of the spreadsheet I built (Mac) and the results at .10 steel.
I did run the examples to test the formula (shown on the left).
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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Posted: 09 Aug 2019 at 6:05am |
I've really confused myself this morning, nothing new.. Using the 1.22 modulus, with 192" of length, how much force you come up with? Here's what I have: 2x4x192x1.22 = 1,873.92 pounds of force to bend. Update: I ha a friend come over with his calipers and re-check. He got a reading of .09865.. I did it old-school and see that's much closer to what my cheapie calipers read.. The .1 Off-grid speaks of is what I will go with now.
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offgrid
Senior Member
Joined: 23 Jul 2018 Online Status: Offline Posts: 5290 |
Posted: 08 Aug 2019 at 5:18pm |
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I don't think a bent frame would effect whether you had too much negative camber or not, it would just make the trailer sit tilted on the axle. Might create an illusion though. What I get from most folks with reported axle failure is that the negative camber is pretty noticeable just by standing back and looking at the wheels.
I don't believe that the axles are custom set by FR either. Several folks on the forum have gotten new axles and no one has mentioned any custom adjustments. I have a spreadsheet that calculates changes in tongue and axle weight when you move load around in the trailer. If you want that PM me. I don't recall creating a spreadsheet for the axle and frame loads when I did those calcs but I'll take a look. One minor point on the section modulus calc, it is the moment of inertia divided by the distance from the neutral (center) axis to the most extreme fiber (2 inches in this case). Moment of inertia for a hollow rectangular shape is (bd^3-hk^3)/12 so Z= (bd^3-hk^3)/6d. This comes out slightly larger than what you calculated. But I measured my tube wall thickness at 0.1 inch rather than 0.12 so ended up almost the same as you for Z at 1.22 in^3.
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1994 Chinook Concourse
1995 RV6A Experimental Aircraft 2015 Rpod 179 - sold |
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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Posted: 08 Aug 2019 at 12:29pm |
I do appreciate your comments and efforts here! Safe travels! Not sure how to check camber with a bent frame.. Another person mentioned that not all axles are set to a certain crown. The statement was the axle is set according to the unit.. I'd love to here from another 182g owner that is willing to check their axle to compare. I'd like to 'think' FR went to the extent of setting according to each unit they sell. Yet for some weird reason... I doubt it. What I've read so far is you developed a spreadsheet to calculate the load. Is this something you would share? |
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offgrid
Senior Member
Joined: 23 Jul 2018 Online Status: Offline Posts: 5290 |
Posted: 08 Aug 2019 at 12:04pm |
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I’m on the road right now, I should be able to look it up later. IIRC, I think the axle has about a 2x safety factor at the 3500 lb max loading. That’s not a great safety margin for vehicle structural design. So about 7000 lbs would likely fail it. If you are down to 1/8 inch crown it’s probably been bent a little. It’s going to have to come off to fix the frame anyway so you might want to get it recambered while it’s off. Is your wheel camber ok?
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1994 Chinook Concourse
1995 RV6A Experimental Aircraft 2015 Rpod 179 - sold |
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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Posted: 08 Aug 2019 at 10:22am |
I did a bit of searching and don't see the Load calculations, found the thread I believe you are speaking of.. Do you recall the actual name of the thread with the calculations? I thought so on the inches cubed. My numbers went to 8400# with even load and center axle (which we don't have). My engineering friend called me and we went through several areas he required to even estimate the force required to fracture the beam. He came up with 6300#. I see you folks have discussed bent axles before, so with my new numbers of 6300#, won't my axle have bent? |
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offgrid
Senior Member
Joined: 23 Jul 2018 Online Status: Offline Posts: 5290 |
Posted: 08 Aug 2019 at 9:08am |
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I did the frame load calc and posted it a few months ago, you probably can search for it. I assumed the frame rails were uniformly loaded beams cantilevered aft of the axle. But there is in addition the torque from the torsion axle that has to be taken by the beams at the same location, so it’s worse than you get just by assuming simple beam loads.
The US units of section modulus are inches cubed. |
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1994 Chinook Concourse
1995 RV6A Experimental Aircraft 2015 Rpod 179 - sold |
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Olddawgsrule
Senior Member
Joined: 20 Sep 2017 Location: New Hampshire Online Status: Offline Posts: 1014 |
Posted: 08 Aug 2019 at 7:28am |
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Okay, I found a site with an example of our Tube steel strength with the formula and result. Now if only I was smart enough to understand the result.. Hopefully one of you can!
Quoted: The strength of a beam in bending is proportional to the beam's section modulus. This value is calculated from the geometry of a beam's cross section. A google search will turn up tables of section moduli for common shapes like I-beams, circles, and rectangles. In your case I will assume that the beam is being bent "the hard way." The section modulus for a rectangle is (b*h^2)/6, where b is the width of the rectangle (the base) and h is the height. A tube is actually composed of two nested rectangles, one defining the outer wall and one defining the inner wall. To calculate the modulus of the tube shape, we must subtract the inner rectangle's section modulus from the outer rectangle's section modulus. For the 2"x4"x0.12" tube: ((2*4^2)/6)-(((2-.24)*(4-.24)^2)/6) = 1.18 [in^3] 1.18 pounds of force per inch to the 3rd power? 1.643 pounds of force per inch? or: 192" to the 3 power x 1.18. This actually makes more sense to me.. It would be so nice to be smart!!! |
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