Design trailing arm suspension 2

[Spanish Wells]

Hi to all; I wonder if someone could help me please ?
I'm designing / building a trailer using a trailing arm suspension system. My problem is to design the arm to accommodate the torsional stress involved.
The wheel / tire is 32" diameter, 10" wide.
Static load of 4,000 lbs per tire.
The stub axle(inc brakes) will be attached to an arm approx 20" long from axle center to pivot point.
This arm will be a "bell crank" transferring suspension loads to an air bellows suspension.
My problem is to specify the section of the arm to resist the torsion loads.
I want the arm to be as thin as practical to minimize the overall width of the system.
My thoughts are to flame cut the arm/bell crank from A36 structural plate of the appropriate thickness.
Could somebody help with the appropriate formulae so I can look at various sections of the arm,
for example: what width would I need for 1" plate, or 1.5" plate, etc.
I'm thinking I could build a table using Microsoft excel to include all the variables such as: load, g force, twisting moment, length of arm, material thickness, material width, mechanical properties of the steel, etc and play with the section of the arm to give me a good solution.
Appreciate any help.
Bob

 

[Spanish Wells]

LittleInch,
You may be correct.
80 ft, include drawbar of 10" x 10" x 3/8 tube = 3,800 (correction, doesn't need to be full length of boat).
00 lbs.
20 ft 4' x 4" x 3/8" cross-members = 340 lbs
6 trailing arm assemblies @ 60 lbs each = 360 lbs
3 complete Dexter 10,000 lb axles @ 435 lbs ea = 1,300 lbs.
So far that's 5,800 lbs......
I might get it to under 8,000 lbs.
Tires have a capacity of 6,000 lbs each, and there are stronger axles available.
I need to check how much tongue weight the tow truck can handle. It has to be a very substantial number for stability. My Ram 2500 truck calls for 10% of trailer weight as tongue weight.
As with any design process, it's one step at a time. I had the basic concept in mind, but the issue that was really bothering me initially was the trailing arm design, so that's where I started.
Thanks
Bob

 

[SwinnyGG]

Is my logic sound, please ?

I think you are moving in the right direction- but not there yet.

Before I dive in, let me say - I admire your effort and I hope it doesn't appear I'm attacking you. Merely trying to help.

1.5 degrees is a LOT of deflection. Remember that when the trailing arm absorbs load, because the primary loads are eccentric, it will deflect in multiple planes. Effectively the individual tires are going to be 'steered' in different directions by their loads. You need to control this in order for the trailer to be safe and predictable behind the truck. If I were in your shoes I would probably be targeting stub axle deflections somewhere around an order of magnitude smaller than what you're seeing, under the worst case loading condition (all loads combined simultaneously). This is why stiffness is controlling your design, not strength.

Also: 48,000 psi is a high stress level. If you were to build from A500 tube, which in the highest available grade has a yield stress of about 50,000 psi, you basically have a safety factor of 1. That's bad.

Keep in mind that all the loads we're talking about so far (even that potential 8G+ pothole) are NOT factored. Those are true loads that are easily possible in the real world.

Your approach for sizing these member seems, so far, to be this:

1) Guess what the loads are (no slight against you- you have no choice)
2) Increment up member size until what you've selected just barely meets the requirement

I would challenge you to take a 180 degree turn in your approach. If I were you, my approach for this would be:

1) Determine all functional requirements of the design: you've done this already.. you need it to lower, you need clear span on the inside, you need to mate to the stub axle arrangement you've already selected, etc
2) Determine stress limit: choose a commonly available material i.e. A500 tube, divide the yield stress by a reasonable safety factor, use that value as your yield stress limit. Important clarification: you must use yield strength for this, not ultimate tensile strength. If you exceed yield, things bend. Do it a lot, and they continue to bend until they break. Do not use UTS.
3) Package everything (decide if you're putting the frame over or under the hub centerline, etc). Determine what space you have for that trailing arm to live in
4) Select the largest sections possible based on the envelope for each component
5) Calculate stresses for the individual members. You'll need to account for all loads at once. This will be hard. Start with the thinnest wall tube available in your selected sizes.
6) If your stress calcs check out, move on. If not, repeat step 5 with the next thickest wall available in your selected size.
7) Once you meet your member stress requirements, start checking connections. If connections don't work, you may have to go up in wall thickness so that your welds are strong enough. Again, repeat until you pass.
8) As a final check, look over the design making sure it can actually be built in the real world. Engage your fabricator if they are willing (some are, some aren't)
9) Submit for review by your licensed engineer

When some things are designed, you design the physical dimensions of the part and then pick a material that works.

Steel tube is available in very limited alloys and sizes at reasonable cost, which is why you want to approach from the other direction. This approach also gives you the stiffest configuration available in the packaging space you have. And stiffness is going to control your design more than ultimate strength is. Suspension members need to be very stiff; make them stiff enough, and generally they will be strong enough.

 

[Spanish Wells]

Thanks Swinny for once again taking the time to contribute.
I value your opinion and am certainly considering your input.
For discussion:
- In my post of 28 July, 13:30, I "reversed engineered" a Dexter 10,000 lb trailer axle to see if I could figure out the loads they are using. If you consider it as a beam, with 2 supports (the leaf springs) and loads at each end, using 4g gives a stress of 66,000 psi. So it seems they are considering less than 4g. Do you think my logic is sound ?
- The effect of a shock load is mitigated by the tires, and suspension components, I believe. Solid tires, and no suspension puts every bit of that load into the structure.
When I was considering 1026 DOM round tube for the trailing arm, I found a yield stress value of 60,000 psi. I didn't think to change the value in my equation for A500 steel - good catch !
- It is interesting to consider one of your earlier comments when you suggested I use square rather than round tube. Your statement was that square tube is stiffer than round tube per unit of volume. However, 1026 tube has a yield stress of 60,000 psi, whereas A500 is 50,000 psi, highest grade. So using 6g, a safety factor of 2, 60,000 in-lb torque, I could use:
5' x 5' x 5/16" sq with a SF of 1.9, or (18.5 lbs / ft). I don't know how to calc twist.
5" dia x 3/8" DOM with give an SF of 2.0. (19.1 lbs/ft). Apparently 0.9 degrees twist.
It seems that the differences in yield strength overcome some of the benefits of sq section, although I don't know whether there a significant benefit in twist.
This is very big material, and all the connections and pivots will also need to be similarly sized !
Your comment on twist or deflection is interesting, once again my only reference is Dexter, and their system appears to be liable to flexing also.
Regarding your comments on process, I fully agree.
With my overall width limitation of 102", the stub axle length, tire widths, etc and a "minimum" trailing arm thickness, the clearance between the frame members "will be what it is". I don't have a particular dimension in mind. Right now it's looking like about 40"; that's OK. I am trying to define the section of the trailing arm, perhaps the most critical component in the whole design concept.
I've just realized that increasing the section of the trailing arm actually increases the torque loads, as they are based on the centerline of the trailing arm member to the center of tire !!
Greg, I imagine you are still following this discussion.
One item not yet resolved is the g force to be considered. If I am correct in my logic, it seems Dexter is using <4g. What are your thoughts based on the measurements you have seen, please ?
Secondly, for discussion once again, the tires and suspension mitigate shock loads. The tires will mitigate the shock load into the trailing arm, and the suspension will further mitigate it into the frame. Would it be reasonable to say that the tires see the full g load, the trailing arm sees less, due to the mitigation of the tires, and the frame sees even less ?
Thank you
Bob

 

[r13]

tires and suspension mitigate shock loads. The tires will mitigate the shock load into the trailing arm, and the suspension will further mitigate it into the frame. Would it be reasonable to say that the tires see the full g load, the trailing arm sees less, due to the mitigation of the tires, and the frame sees even less ?

IMO, the tires, suspension and hydraulic cylinder will soften the below/shock, but will not reduce the force. Simply as a loaded spring, it must have a rigid base to provide the reaction that equal to the applied force.

 

[Spanish Wells]

To continue my argument further, if we add a spring suspension seat into the model, clearly the driver does not feel much of the effect of a large shock load; clearly that is the point of suspension. Whereas with solid wheels, no suspension and sitting directly on the frame the driver would feel every bit of the g force directly.
Secondly, I just realized that in my comparison with the Dexter axle above, they appear to have used <4g, but also 0 safety factor, in their calculations, if my logic is sound.

 

[r13]

I think it is due to effect of damping, some energy is lost. Maybe you should use conservation of energy concept to evaluate the impact, with the input from the shock/suspension supplier.

 

[SwinnyGG]

- In my post of 28 July, 13:30, I "reversed engineered" a Dexter 10,000 lb trailer axle to see if I could figure out the loads they are using. If you consider it as a beam, with 2 supports (the leaf springs) and loads at each end, using 4g gives a stress of 66,000 psi. So it seems they are considering less than 4g. Do you think my logic is sound ?

The tube of a solid axle is fundamentally different than a trailing arm - it's basically a simply supported beam with two point loads, and all of the loading is in bending; torsional load is very close to zero.

I'm not sure how you calculated the stress value in your analysis of that axle; I get a much, much lower stress.

 

[r13]

Agree with SwinnyGG.

 

[Spanish Wells]

Swinney,
Maybe I'm wrong, it's been 45 years since I did structural stuff in college.
My point is to try to define what g and safety factor Dexter has used in their design.
In the beam axle, 5' OD, 1/4" wall,10,000 lbs load rating.
70" long, 42" between spring hangers,
Moment of Inertia for beam calculates at 10.55. (pie (D^4 - d^4))/64.
Moment of 5,000 lbs x 14" = 70,000
Stress = My/I = 70,000 x 2.5 / 10.55 = 16,600 psi
Yield stress of 1026 DOM = 60,000 psi.
So they must have used a combined g factor / safety factor of <4 ??
This is my point.
So it seems that the biggest trailer axle manufacturer is using a combined g and safety factor of <4 .
Does my logic make sense ?
Thanks
Bob

 

[SwinnyGG]

Moment of 5,000 lbs x 14" = 70,000

This is where your error is.

The section of a solid axle between the flexible support (the spring) and the end (where load is applied) is NOT a simple cantilever with a fixed end.

They are fundamentally different. Analysis of a simply supported beam is more complicated.

 

[Spanish Wells]

Swinney
OK. I went to Engineer's edge and found:
"Beam Stress and Deflection equations and calculation for beam supported at both ends and two equal loads"
They show the equation: Stress = Wx/Z or Wxz/I
for my example, 5" tube, 1/4" wall, 70" long, 42" between supports, equal loads of 5,000 lbs, MofI = 10.55
I get: Stress at support (spring hangar) = 5,000 x 14 x 2.5 / 10.55 = 16,500 psi.
Again my point is that with a yield stress of 60,000 psi, g + safety factor is <4.
If my calculation method is wrong, would you please give me your thoughts on the method I should use ?
Thank you
Bob

 

[r13]

Here is a case of damaged suspension system for your info use, and to remind you that beware of the effect of lateral force on the connections. Link

 

[SwinnyGG]

I'm not saying the calculation method is wrong - I'm saying your suspension configuration is fundamentally different than a solid axle connected to a chassis with leaf springs. It is compliant against impact in ways your system is not and cannot be; compliance means lower loads during impacts. Using that system as a basis of comparison regarding the safety factor of your components is dangerous.

At the end of the day you have to decide. If you're comfortable using 4g loads, use 4g loads. We can't stop you.

 

[Spanish Wells]

Swinney
Appreciate the input once again.
I believe I understand the notion of compliance. However I'm not sure that I follow how the two systems are so fundamentally different.
As I see it, the "g" force is an acceleration; "compliance" in the system slows that acceleration down so that the forces are lower. Compliance is in the suspension springs themselves, including the tires, also in the components of the mechanism.
I did not intend to use a 4g factor because it seemed too low, and also it is quite easy to include a much larger g force in the calculations, using a 4" square section.
My difficulty was bridging between an apparent g + SF of <4 according to the calculations for the beam axle, and your assertion that I should consider 8-10g + SF.
My calculations, based on the input received here, for:
- static load 5,000 lb
- stub axle length 10"
- trailing arm length 14.5"
- g force of 6g
- material section 4" x 4" x 3/8" square tube
Stress = 30,000 psi.
This gives me a safety factor of 50/30 = 1.67. or an overall factor g + SF of 10.
Do you think this is sufficient to overcome the fundamental differences in the two systems ?
Thanks
Bob
ps, if I used a rectanglar section 4" wide, 5" tall, 3/8" wall, this factor would be nearer 13.

 

[SwinnyGG]

The fundamental difference is that in your 'baseline' (the commercially available solid axle) the member absorbing stress is 70" long. Your cantilever is 10" long. This means that the impulse that member is required to absorb is significantly higher. Imagine two springs- one 7x longer than the other. Even if they absorb the same load, the longer one does so with less strain.

4" is also very small as far as connections. Remember that your stub axle must transmit a big moment (over 4,000 ft-lb) per your math above) into that section.

And that 4,000 lb is just a single load case.

You have to consider the worst case loading condition when you size your members- that's max braking plus maximum longitudinal impact plus max transverse impact multiplied by your chosen safety factor. All the analysis you're doing so far is valuable as a though exercise, but you won't get close to knowing your 'real' sizes until you calculate that maximum load case.

I would encourage you not to size everything based on this one load case. You'll wind up having to start over once you consider the aggregate case.

 

[Spanish Wells]

Thanks Swinney