Lift Assist Torsion Bar 1

[JacobP12345]

What would be an ideal material for a torsion bar assisting lifting a 200 pound lid with the rod being subjected to a displacement of 90 degrees? Hardened 4140 has been attempted but it yielded and is not ideal. Looking elsewhere has suggested higher carbon steels such as 1074, but before trying this I want a better idea of what would work. What kind of spring steel if any would work best? The bar will be 36 inches long and of 1/2 inch diameter. From my calculations using the material properties of 4140 I am getting a maximum surface stress of 240 ksi. Any Suggestions?

 

[MintJulep]

Assuming that your calculations are correct find a solution where the maximum stress is below yield and below endurance limit.

When you find that impossible, remove the arbitrary constraints on geometry.

 

[FEA way]

I guess that you used the linear elastic mechanics of materials approach: T = (φ*G*I)/L ---> τ_max = 2T/πr³ Right ? Then your results are in general correct. I checked it and my result is 1762 MPa (255,6 ksi). However the twist angle is large and if you want to obtain correct values I suggest using nonlinear Finite Element Analysis. This way you will be able to find material with the desired response to this prescribed rotation.

 

[JacobP12345]

I will give that approach a try. Given that yield strength is roughly accurate, is there any material you know of or could recommend that could either be heat treated to have that effect yield strength on the surface or otherwise? My trouble so far isn't in the kinematics but the material science. I'm not well versed in metallurgy or material science beyond an introductory course and need insight into materials that can meet that stress if anyone could give any tips it'd be greatly appreciated

 

[FEA way]

I forgot to add that the small difference between our results is probably caused by the fact that we assumed different shear modulus. Anyway, most metals (including titanium) will yield at such extremally high stress. Even special high yield strength steels have a limit of up to 1000 MPa (145 ksi). The only thing that comes to my mind is maraging steel. But first I would advise you to consider changing the geometry of this bar or reducing twist angle. This stress seems exceptionally high. And of course perform nonlinear FEA to make sure that this value is actually meaningful.

 

[JacobP12345]

So I came to realize I made a pretty substantial error in my calculations, and used the diameter in place of the radius.
Considering a = deflection in radians, and our deflection to be 90 degrees, that gives us 1.57 radians.
Applying this to the equation, 1.57 = length (36 inches) * (shear) / (radius (.5 inches) * shear modulus(assumed to be roughly 11000 ksi)) I got a rough expected stress of 120 ksi which is far more reasonable. Going under that assumption, could I utilize hardened 1074 or 1095? Or would I still need to go with something like hardened 300m (which can be hardened to a yield strength of rougly 280 ksi) or 4340?

 

[EdStainless]

Too much twist in 36". 30deg/ft is a lot, and I am guessing that 36" is your total length and not your free length.
There are a couple of ways around this.
Use a stiffer bar, twist it less, and use a mechanical linkage to get 90deg of output rotation.
you might want to use info from some of the aftermarket auto guys;

https://swayaway.com/tech-room/torsion-bar-wheel-rate-calculator/

Often these bars are hollow (to allow larger stiffer bars without making them too heavy) since the ID does little to increase stiffness.
While 200lb isn't much load the short bar and huge rotation result in very high stress.
Even going to a higher strength material will have issues:
1. Materials don't usually like to be loaded for long times at stress that high
2. Fatigue life is likely to be very short

If you really need that much twist you should look at lower modulus alloys.
While 4140 is common for steels many OEM bars in autos are actually 1065 or something like that.
For lower modulus you could look at Al and Ti alloys.
A Ti 6/4ELI bar might not be a bad option.

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P.E. Metallurgy, Plymouth Tube

 

[JacobP12345]

Thanks for the input, the issue we're having is on a few we've seen in the field they operate off a 1/2'' hex torsion bar and are simply mounted by hex sleeves on both sides, overall length of 36'', and rotate the full 90 degrees and act as a lift assist for the lid. It's a simple straight bar and I'm completely lost as to what kind of metal it is and have been trying to figure it out for a couple weeks now. Would 1065 fill that role? The only metal I found that had a yield strength high enough was 4340 and 4340m but again I'm not a metallurgist so I don't put much weight behind my voice on the subject. We're considering using an alternative method to reduce the twist angle while trying to get the same amount of spring action. Any suggestions? We expect it to have very few cycles of use, maybe a couple times a year.

 

[EdStainless]

Are you sure that the bars are steel?
Can you buy them from the guys that supplied the ones that you have seen?
Or can you at least get one to test?

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P.E. Metallurgy, Plymouth Tube

 

[JacobP12345]

We have one to test but it's welded in place to the frame of the door we are improving on. It has rust on the surface from end to end so I'm pretty sure its steel. That's why I've been pounding my head against a wall for a solid few weeks trying to figure out what they're using. I assume its similar to the torsion bars on lift gates and the easy lift systems on truck bed doors. So far we have tried hardened 4140 and it has failed to duplicate the performance. I was considering giving 4340 a shot.

 

[FEA way]

Indeed the transformed formula is τ=(φ*G*r)/L. But with your data it still gives 239,85 ksi ...

 

[JacobP12345]

From my math, not doubting just wanting to make sure, I used .25'' for r as my diameter is .5'' so my distance to stressed surface (r) should be half that, or .25''. I used 11*10^6 psi for G, 1.57 for alpha, and 36'' for L. This yielded t= 1.57*11*10^6*.25/36 which gives me approx 120 ksi.

 

[FEA way]

Oh, my mistake. I was still thinking that 0.5 inch is your radius. That’s because in one of your previous posts you gave the formula with „radius (.5 inches)”. I assumed that this is the corrected value (as you began this post saying that you took the diameter in place of radius). I should have checked the first post to make sure.

Anyway, you’re right that the stress is about 120 ksi (830 MPa). Still large but high yield strength steels may handle it. However nonlinear FEA is necessary to confirm this. I’ll give it a try tomorrow.

 

[JacobP12345]

Thank you both for all your help by the way. I have narrowed down my search to hardened 4340 and 6150. Both seem to have yield strengths (all be it very close) to our calculated 120 ksi. I imagine if I had them heat treated it could increase their fatigue resistance and strength modestly. 300m is also being considered but it is substantially more expensive than 4340. Using just some linear interpolation to help confirm my observations from

https://swayaway.com/tech-room/torsion-bar-maximum-angle-twist/,

it seems that a bar of .5'' diameter could potentially handle the angle of twist I'm expecting. I look forward to seeing what you find tomorrow, again thank you for your help!

 

[EdStainless]

4140 should work fine at 0.5" diam, it will through harden and you shouldn't have any trouble getting 180-200ksi UTS with an 850F temper treatment.
So if the working stress is 120ksi then this should be fine.
4340 will not give you any higher strength.
You would need a grade like H-11 or D6AC in order to get 220-240ksi UTS, but it doesn't sound like you need it.
The use of hex bar is clever. But you have to be careful of the 'corners'. If they get nicked it could be an issue with fatigue. They would need to be rounded off.

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P.E. Metallurgy, Plymouth Tube

 

[JacobP12345]

Thank you for your help, we have tried 4140 round from McMaster which claimed to be hardened, but the yield strength was around 100 ksi. Their hardened 4340 is around 122 ksi. We really were looking at a buy-and-try kind of idea to test out, but if you're suggesting simply hardening 4140 instead of going to 4340 we'd save a lot on price per pound there so we'll give it a try. Is there any sort of reference sheet I can go off of in terms of the different tempering's and their relative UTS and yield strengths for different metals? The only reference material I can find are for the materials untreated properties which has made this search very frustrating. Is there any disadvantage to moving to 4340 as a side note?

 

[FEA way]

I performed nonlinear FEA with 3 different levels of model simplification. And the maximum shear stress is 242 MPa (35.1 ksi).

 

[EdStainless]

This is clipped from an aerospace material spec for medium strength steels.
Yes 4140 can be through hardened in this size to much higher strengths.

[URL unfurl="true"]https://res.cloudinary.com/engineering-com/raw/upload/v1565271009/tips/steel_HT_table_fe8zj4.docx[/url]

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P.E. Metallurgy, Plymouth Tube

 

[JacobP12345]

Thank you. This should be my last question for a little while till after I do some testing. What is the relationship between UTS and yield strength? Is their a relatively consistent means of estimating maximum shear/yield strength from the UTS? I am considering testing a larger diameter rod just to see what benefits there might be (9/16'') but the added stress might be an issue.

 

[FEA way]

According to the "SINTAP - Structural Integrity Assessment Procedure for European Industry" project report the following approximate relationship exists between UTS and yield strength for most steels:

UTS=σ_y*(1+2*((150/σ_y)^2.5))