260 ksi yield strength needed

[Jboy333]

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

 

[Jboy333]

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.

 

[israelkk]

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].

 

[Jboy333]

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.

 

[israelkk]

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.