Force required to dimple metal with sharp tip 3

[shaneelliss]

What formula or method would you use to determine the force required to push the sharpened tip of a round shaft a certain distance into a very thick piece of metal (magnesium in my case, but I assume the formula would be the same using the proper E or Fy or something for whatever material)? To be specific, I want to push a sharpened steel shaft (45° point) 1/2" into a piece of magnesium that is 30" thick and 18" x 18" square. How do I calculate the axial force required to do that?

 

[CoryPad]

Your load case is essentially plane strain indentation. For a flat punch onto a flat surface, the applied pressure would need to be 3 times the material yield strength. For your case with a pointed indenter, the applied pressure would be lower, but probably not that much lower. How accurate do you need to be?

 

[IRstuff]

What diameter is the shaft, i.e., are you pushing any part of the full diameter of the shaft into the magnesium?

TTFN
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faq731-376 forum1529

 

[shaneelliss]

The shaft diameter is 2.25" and I am only trying to push the tip in about 5/8", so it is still on the angled portion of the sharpened tip, not on the full diameter of the shaft. As far as accuracy, I would like it to be accurate to within a couple thousand pounds of axial force. I just assumed there was some theoretical equation that I could use, but maybe not?

I am a structural guy so for steel there are bearing calculations that are essentially 1.8 x Fy x Bearing Area to theoretical failure (which is quite a bit less than the 3xFy stated above by CoryPad), but it doesn't seem quite right to call this a bearing surface area failure.

 

[IRstuff]

I would have guessed that it's more like a shear strength problem

TTFN
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homework forum: //

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[rb1957]

is there a reason why you don't want to cut a CSK/dimple ? Plastically forming the dimple would require yielding a lot more material than just the cross-section of the shaft.

another day in paradise, or is paradise one day closer ?

 

[shaneelliss]

The pointed shaft is part of a lifting rig where the shaft presses into a large 10,000 lb magnesium log and lifts it. So a crane drops the lifting rig over the magnesium log and the shaft is pressed into the side of the log on each side and then the crane lifts the log. I appreciate the help, but try not to get distracted by trying to solve a problem with a number of other solutions. I have a solution that works, but am trying to figure out the mechanics behind it with engineering principles. The shaft is threaded and is turned through a fixed nut, which presses the tip into the magnesium. I want to figure out if I can use a smaller shaft, but I don't know how to calculate the force required to make the dimple. I know the torque required to turn the shaft and using some assumptions on friction I can back calculate the axial force, but I don't know if my assumptions on friction are accurate enough to base a design on. So I thought I would start from scratch and see of I could calculate what the force needs to be. But it turns out that I don't know how to do that so I thought I would ask the good mechanical engineers on this site. If you have any insight on how I could do that, I would appreciate it. Thanks.

 

[rb1957]

you want to rotate the shaft ? why not use something like a Really big Phillip's head screw driver ? force to embed into the "log" would be much lower.

another day in paradise, or is paradise one day closer ?

 

[MintJulep]

https://en.wikipedia.org/wiki/Conta...id_conical_indenter_and_an_elastic_half-space

 

[shaneelliss]

MintJulep, that was exactly what I was looking for. Thank you!

 

[shaneelliss]

Hmm, after working through the math, something does not seem right. We get a depth of penetration of about 5/8" with a force (back-calculated from torque) of about 113,000 lbs. Per the equations on the wikipedia page and another reference I found once I knew what to look for, I calculate that for 5/8" penetration, I would need a force of 1,700,000 lbs. My force number back-calculated from torque might be off by a factor of as much as 2, but not more than 10, which these equations suggest. My shaft couldn't handle that much force to begin with. Is it known that these theoretical equations don't match real life by a factor of 10?

 

[rb1957]

"a force (back-calculated from torque)" ... ?

another day in paradise, or is paradise one day closer ?

 

[racookpe1978]

How are you driving the shaft now, and how accurately do you think you are determining the torque you think you are applying? Hydraulic force twisting a threaded rod?

You are driving a cone into the large diameter "log" so the initial penetration starts with a near-zero-area point touching a near-zero area tangent wall on the side of the log. Then, as this point goes in, the contact area increases as two things happen: Unlike a shear punch through a flat metal, the contact area increases proportional as all three of the depth of the cone AND the angle of the cone AND the depth of the side of the log increase, while the shear force of the flat metal remains the same all of the time as the diameter of the punch.

So I expect your initial pressure and initial torque to be very low, and initial penetration depth starts very low then increases to some "stall point".

 

[gruntguru]

Reading mintjulep's link, it appears the formula provided will only apply up to the elastic limit of the Mg. Beyond that, the effective modulus will drop significantly. That probably accounts for the difference since in your case, most of the penetration will be in the plastic region.

je suis charlie

 

[racookpe1978]

It has got to be into the plastic region - actually, well past the local plastic yield since both of the tips of the probes go into the "log" a "long way" on both sides.

 

[shaneelliss]

OK, so to clarify some things. The log of magnesium is rectangular, not round, so the point goes into a flat face. And the shaft is more like a bolt. It is threaded with a nut welded on the end. The shaft/bolt is driven by an air impact wrench with a multiplier that allows more torque to be applied than the air gun itself can give. A guy is standing there operating the impact wrench, which is turning the shaft/bolt through a threaded fixed plate into the magnesium log. There is a simple equation that can be used to convert torque applied to a bolt into the axial force on the bolt. I don't consider the equation very accurate because it relies on a coefficient of friction between the bolt and the nut, which has been shown to vary, but in ideal conditions, it has been tested and shown to be accurate. The formula is T = c x F x d, where T=torque, c = the friction coefficient, F = the axial force, and d = the diameter of the bolt (

https://www.engineersedge.com/torque.htm).

In the end, I still think that all of that information (except maybe the flat face bit) is not required to answer the question, if there is an answer. It doesn't matter why I want to do it this way or how the shaft is driven, I just want to know if there is an equation that describes the normal force required to drive a shaft into a material. MintJulep gave a good resource, but as others have clarified, it only works in the elastic range. Does anyone have a reference for an equation in the plastic region?

 

[rb1957]

I think there's a significant difference between driving an impactor into a body and screwing a pointy shaft into a body.

another day in paradise, or is paradise one day closer ?

 

[shaneelliss]

Ok rb1957, in what significant way will it change the equation? In the end isn't it still just a normal force pushing a rod into a material? In my case the rod might be rotating while it is being pushed in, but I don't see how that changes things significantly. But I am willing to be taught if you have something meaningful to say.

 

[dhengr]

Shaneelliss:
NO... How’s that for an answer without any questions or discussion?

If you don’t want to discuss your problem, we don’t really give a damn what you do, or why, or how. There is a lot of engineering knowledge and experience here on E-Tips, and many people more than willing to waste their time on you problem and to try to be of help. But, with your attitude, who cares. I just deleted the page of thoughts and comments I had started for you to consider, since I don’t have a quick formula.

 

[btrueblood]

I don't believe there is a simple formula that will get you there, though you may get close using your experimental data and the equations shown in the opening paragraphs of this:

http://www.sciencedirect.com/science/article/pii/S0020768306004239