Minimum thickness required for contact area

[bf109g]

Hi all. I have a triangular load distribution due to contact between a plate and a hinged fitting. I know load magnitude and contact area.

How do I go about determining minimum plate thickness required to prevent this thing being a can opener? It seems that if I consider the load magnitude as the volume of a wedge, I get an astronomically high running load along the edge I'm concerned about. Am I missing something here?

 

[GregLocock]

Well, you will get a stress raiser at the edge. It's actually a big deal when you are joining aluminium hardpoint plates into composite structures, but even your scenario can be a big deal. So, are you allowing for the flexibility in both the can opener and the plate?

Are you FEAing this or is it a hand calc?

Cheers

Greg Locock

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

 

[Dan320]

I once managed to avoid a similar problem by introducing a radius at the edge of the stiffer fitting.

Dan

 

[bf109g]

The materials are 7075 aluminum. Hand calculations are being performed and I haven't taken flexibility into account.

The hinged fitting is radiused along all contact edges.

 

[Dan320]

Show us a sketch, maybe it's easier to visualize.

 

[rb1957]

what if you applied the resultant load at the centroid of the loading ... maybe you have a running load, consider an individual pitch ... but like dan posts above, a pic is worth 1,000 words

 

[bf109g]

Here's a sketch of what we have. Load and dims are for a previous part. Don't worry bout the right side not being radiused - the actual part will be.

This part will react against a piece of machined plate. I'd like to have an idea on how to determine the minimum plate thickness so that we don't end up punching thru the plate with the part.

 

[SWComposites]

Looks like a can opener ...

Need a drawing of the structure that the part reacts against: plate and substructure. And need dimensions in the width direction (into the paper on your first sketch)

 

[rb1957]

well, you can solve for R1 (by statics) ... hint, about 12,000 lbs

a triangular bearing load (as shown) is a reasonable assumption, so you know the peak load per unit thickness.
hint, about 24,000 lbs/in

divid by thickness to be less than the bearing allowable (or either side of the interface). hint, about 0.25" (assuming Al)

lick of paint ...

 

[Timelord]

If the pin and hole locations and sizes are not "real" exact, you will get point loading at the end. If you are doing this analysis for safety, I would assume that the loading is concentrated at the end of the part.

Timelord

 

[rb1957]

i'd be more worried about bending of the link, at the pin ...

the moment in the link is about 10,000 in.lbs,
at 0.25in thk the bending stress is something like 60,000/(1^2*0.25) = 240ksi ... better be steel

 

[bf109g]

All -

Thank you so much for your responses. I've been trying to clean up another inherited mess (another idiot who thought shearing stress due to bending is V/A).

rb -

I think we're on the same wavelength here. I guess my brain is vapor locked on this:

divid by thickness to be less than the bearing allowable (or either side of the interface). hint, about 0.25" (assuming Al)

Could you elaborate a little on this?

BTW - the link has two legs approx .6 wide each.

 

[Dan320]

What if you check the link for bearing stress at the edge and the plate it bears upon as if the link was, indeed, a can opener. That is, check for shear in the plate along the edge of the link. (shear area = link width + 2x distance to (say) R2 x plate thickness.) If the link is radiused, this calculation will probably be very conservative and you will never have to worry.

 

[rb1957]

"divide by thickness to be less than the bearing allowable (or either side of the interface). hint, about 0.25" (assuming Al)"

triangular distributed load, peak = A, span = B
resultant = A*B/2 = R1
R1 = 12,000lbs
B = 1in (near enough)
A (peak contact load) = 12,000*2/1 = 24,000 lbs/in

bearing stress = peak load/thk = 24,000/0.25 = 96ksi

but look at bending through the pivot M = 10,000 in.lbs
stress, for 0.25" thk, = 6*M/(0.25*1^2) = 240ksi

of course if there are two links then this'd be the combined thickness. how are the links joined together ? how is the load applied to the combination (to has link ? to the middle of the spacer ??)

 

[bf109g]

rb -

So what you did is figure out the thickness of the link required to bring it under Fbry, right?

My concern is the thickness of the material under the link, i.e. the machined base, not the link itself. I have the link pretty well figured out.

The link is loaded by contact along the vertical face. I've attached an old solid model of the part to give you an idea.

 

[rb1957]

well i figured that 0.25" would be sufficient width for Al for contact, but this is not sufficient for bending about the pivot.

but it seems you're asking not about the link, but about the structure that link contacts. but that's a whole different thing. this it one thick flange ? does it have stiffeners ? aligned to the link axis or normal (so that only a small area is fully effective) ? on one level, how much area would have to shear for the links to break the supporting structure ? how does the supporting structure react to the link loads (bending ?) ??

 

[bf109g]

Bingo! I'm more concerned about the shearing of material under the links. The base sits atop standard honeycomb floor panel. For simplicity consider the base as a plate of uniform thickness.

 

[rb1957]

tabanac !

so it's either ...
1) the panel is essentially rigid, what pressure does it take to crush the honeycomb; or
2) the panel is flexible (being "rigidly" supported somewhere else) and bends like a beam.