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260 ksi yield strength needed

260 ksi yield strength needed

260 ksi yield strength needed

(OP)
We're working on a project with a 1/2" square torsion rod that is being used to act as a variable spring by moving a slider with a square hole in it up and down the length of the bar to change the effective length.

We're needing a minimum yield strength of 260 ksi for our application.  Can anyone please recommmend our most cost effective option for a part like this?  

The part is about 10 inches long with about a 5 inch leg at a right angle where the load is applied.

We were originally trying to use a heat treated 4140 steel, but have had a hard time getting consistent results (and are probably asking 4140 to do too much for us).  In a previous post some people had suggested 4140H, but nobody seems to carry it, and because of that, those that do carry it charge an arm and a leg.  We're looking to be as cost-effective as possible.

Thanks in advance for your advice!

-Jesse

RE: 260 ksi yield strength needed

You will need to upgrade to either 4340 or 300M.

RE: 260 ksi yield strength needed

(OP)
The part will be going through about 7.5 degrees of torsional deformation.  Will 4340 and 300M perform well under those conditions.

RE: 260 ksi yield strength needed

AnoOther option would be maraging steel.

RE: 260 ksi yield strength needed

How did the 4140 perform? Did it yield? Did it fracture? The torsional stiffness of 4340 or 300M will be the same as 4140, but you will have increased yield strength.

RE: 260 ksi yield strength needed

(OP)
The 4140 seems to be performing well.

According to our FEA analysis we're exceeding the yield strength, but we aren't seeing any visible plastic deformation.  I'm sure there's some work hardening going on there though.

In a previous post, I was trying to get advice on why our material was cracking during heat treating.  That's been our major obstacle in using the 4140 - trying to get consistent results, or even any usable parts because they were cracking during the heat treating process.

Do any of you have any experience with Cryogenically treating parts?  We're looking at that as a possible option as well.

RE: 260 ksi yield strength needed

What issue do you hope to adress with cryogenic processing?

RE: 260 ksi yield strength needed

(OP)
We were hoping to increase our yield strength.....but unfortunately it would come at the expense of losing ductility.  

RE: 260 ksi yield strength needed

Cryogenic treatment does not seem to be a good choice for your part.

If you think 4140H is expensive, don't bother with 4340, 300M or maraging steels.  That being said, they all are better than 4140 if you intend to use them at 260 ksi yield strength.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

RE: 260 ksi yield strength needed

(OP)
We were quoted $1190 for a 48"x12"x.5" plate of 4140H

Any ideas what that might run in a 4340 or 300M?

RE: 260 ksi yield strength needed

On commodity-type products, 4340 is approximately twice the cost of 4140.  

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

RE: 260 ksi yield strength needed

(OP)
Cory,

Thanks.  That helps.  That 4140H cost was about 5 times the cost of a comparable piece of regular 4140.  4340 might be an option then. Thanks.

RE: 260 ksi yield strength needed

Is this bar loaded in both directions or only to one direction?

Is it for static or dynamic (cyclic) use?

If it for static use is it continuously under deflection? If it is you may have severe relaxation with time.

Since you only deflect the bar 7.5 degrees how do you account for the gaps between the slider square hole to the bar? This gap may affect the accuracy for of deflected angle.

http://israelkk.googlepages.com/home

RE: 260 ksi yield strength needed

According to MIL-H-6875H, Heat treatment of steel, process for, the 4140 should not be heat treated above 200-220ksi ultimate strength. On the otherhand 4340 can be treated to 260-280ksi ultimate tensile strength but not the yield strength.

300M can be heat treated to 280-300ksi tensile strength

RE: 260 ksi yield strength needed

I'd go for precipitation-hardening steels. In addition, I would have a check at the stress using strain gages..

RE: 260 ksi yield strength needed

Square section torsion bar has closed form stress and deflection formulas. I do not see the need for FEA analysis or even strain gages testing. This is a simple part to analyse.

I took the liberty to calculate the torsion stress when the 10" bar is deflected 7.5 degrees and the stress is 97.6 ksi. The problem will be when you use only part on the bar length. I think you should reevaluate your design. To my opinion the concept that the 1/2" square SOLID torsion rod is being used to act as a variable spring by moving a slider with a square hole in it is wrong for this size bar and for the 7.5 degree deflection.

RE: 260 ksi yield strength needed

From your post it seems that you are cutting the rods from plate. IMO this is not the best approach to making a torsion rod. I would use the best available bar stock in what ever alloy you intend to use.
The use of plate can give you up to 20% variability in physical properties depending on the material and your cut in respect to rolling direction.

RE: 260 ksi yield strength needed

(OP)
israelkk -

We're you calculating the bar in bending or in torsion - I'm not sure that I've explained the concept clearly enough and that you might be misinterpreting how the device is functioning.  The design is functioning perfectly fine, we're just trying to get the right steel for the job so that we have a good safety factor at an acceptable cost.

unclesyd -
We're only temporarily laser cutting the parts for prototypes- just haven't made the jump yet to the tooling for the forged parts.

RE: 260 ksi yield strength needed

I analysed it for torsion. What is the shorter length that you deflect when you move the slider?

RE: 260 ksi yield strength needed

(OP)
It's about 1.167" before it begins to go into the .5" radius bend of the short leg where the load is being applied.

RE: 260 ksi yield strength needed

Are you deflecting the 1.167" long portion of the bar 7.5 degrees?

Where do you measure the 7.5 degrees at the end of the 5" leg?

If is the case then you are actually measuring the bending of the 5" leg combined with the torsion of the bar 1.167" long (or any other length depends where the slider is) and the complicated area of the 1/2" radius bend which has a combination of torsion and bending and stress concentration (stress raiser) due to the bend radius. The torsion bar between the leg and the slider see bending load too because the bar is supported only at the slider and not near the 1/2" bend radius (if I am not mistaken).

The result is that your torsion-bending bar is completely non-linear and unnecessarily over stresses.

Commonly a torsion bar is designed in such a way that it is supported at both ends to avoid bending completely. When the torsion bar is 10" I assume this is your worst case due to the added bending moment = [Force] X [load at the end of 5" leg].

RE: 260 ksi yield strength needed

(OP)
Actually, the bar is supported on either end by a pair of brass bushings that are free to rotate.  The slider acts inbetween the bushings.  Yes, you are correct as well about the bending of the 5" leg as well, that is seeing some bending deformation while applying the torsion to the longer portion of the bar.

RE: 260 ksi yield strength needed

If this the case then the bending of the 5" leg is constant no matter how long is the torsion bar (assuming constant load at the end of the 5" leg). Therefore, the most stressed area is the 1/2" bend. If you could use a straight bar without the bend and use a stiffer leg with a square hole as the slider to apply the load then your bar will be stressed only under shear. However, the stress will be as I calculated to achieve the 7.5 degrees deflection.

In your case the torsion bar never deflect 7.5" degrees because the leg bends. The deflection of the leg is a function of the leg length by power of 3. The deflection of the torsion bar is a function of the torsion bar length (10" max) but by power of 1. Therefore, I suspect that you actually have a bending bar instead of torsion bar because most of the deflection is the result of leg bending.

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