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?

 

[rb1957]

(again) I think there's a significant difference between driving an impactor into a body using axial force and screwing a pointy shaft into a body. Surely you can see that how the metal will flow is completely different. No, I don't have a simple equation for you and would suggest testing; and I not willing to invest much more thought into the problem. If you don't like my replies, you could try to "red flag" me and complain to the mgmt.

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

 

[shaneelliss]

Yikes, dhengr, I guess I have come across as a jerk. I apologize for that. It wasn't my intention. I enjoy the discussion, but I feel like a lot of times the details distract people from the question, so I was trying to get back to that. I have read your responses to other people's questions over the years and have come to respect your experience and knowledge, so I am a little bummed that you deleted a page of thoughts and comments that I am sure I would have liked to read.

 

[moon161]

I expect that the force to indent as described would depend highly on the square of the depth of the indentation. I'm just going by the area of the anvil interacting. Maybe in the form of (Depth ^2) * some constant * SY.

As an experiment, you might try using a cutting tip, like the end of a big drill, instead of your 45 degree anvil. It might be a lot easier to drive it. You might have to experiment with the pitch of the driving thread. As a SWAG, maybe use a 3/4"-16 thread? Maybe use a ball screw to minimize driving force as well.

Maybe use a 3/4-16 rod with an adaptor, threaded at one and and a socket for a 5/8 or 3/4 stubby drill at the other. Drive the same, back it out, and snug a bolt into the new hole.

 

[IRstuff]

shaneelliss

I would caution you against making assumptions about what is, or isn't, a critical detail. Often, there are trivial constraints that could have monumental effects, had someone been made aware of them.

To wit, the fact that the shaft is turning raises issues not only about friction, but also the possibility of galling. While magnesium is supposedly "resistant" to galling, applying that much pressure to it could change the dynamic.

TTFN
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[CoryPad]

Btrueblood linked to an article dealing with conical indentation related to hardness testing (Rockwell method with Brale indenter). The equations in this and similar articles provide high accuracy for small indents, but I am not sure they scale to your sizes. I think you will find my simple equation is a good starting point and physical testing would be needed to refine the value.

 

[3DDave]

I would try an energy method - look at the problem as one of breaking the crystals in the metal. In that view it is displacing at least the volume of penetration, from which one could determine the amount of energy to nearly liquefy a similar volume of metal.

Sensibly, if the material is already near the melting point, the force required lessens.

At each increment in depth there will be a volume of material intersected by the cone, so divide the energy required for that incremental amount by the distance required to get the force required for that increment. Since the volume incrementally displaced is the cube of the depth, the forces should be increasing at least as the cube of the depth with some factor for increasing friction; simple enough to test with a small setup and more limited depths.

The inaccuracy to this is that it doesn't account for residual strength in the displace material, and it doesn't include material outside the cone volume that is also displaced. It also avoids the contribution of elastic deformation outside the plastic zone and the friction between the cone and the material being displaced. Accounting for torque applied and thread friction is a separate issue.

I don't imagine the amounts from these contributors are equal to the initial volumetric displacement, so doubling the values from melting should be representative.

Altogether I would not expect a simple formula.

I would probably pre-drill the indentation, although since similar amounts of material are displaced, one could look at the power absorbed during the process to also make an estimate similar to melting the material.

 

[racookpe1978]

I very specifically do not think you can scale-up from a single pin-tip (rockwell hardness type imprint) to a deep conic imprint into a flat plate. Even a shear energy comparison is wrong (or at least inaccurate) because a shear force cuts a ring through a constant distance around the shear in the limited area of the base metal between the shear and the die. It really is only that metal in the ring that is displaced.

In this case, the volume of the cone is pushed INTO the base metal around the cone (no metal is removed at all) and so the plasticized volume is 2x the conic volume. First you have to displace the base metal from the cone, then you have to push it through the base metal that surrounds the final cone.

 

[CoryPad]

3DDave, there is no need to refer to melting, the process would be solid-state, so plasticity and strain energy/work would be used.

racookpe1978, for indentation, the metal would flow in the opposite direction of the cone, up the sides of the indenter and create a "pile up" on the surface around the indenter.

 

[shaneelliss]

I was misinformed slightly about the procedure they were using to drive the screw. They don't use an impact wrench, they just turn the bolt with a ratchet and a multiplier. There are two pointed tips. One is stationary on the opposite face of the material from the turning screw and then the pointed tip of the turning screw itself. So turning the screw with a ratchet applies compression and both tips sink into the material on opposite sides.

I went out and we tested the torque it would take to drive the point in and found that the turning tip embeds slightly further than the fixed tip on the opposite side. A 5/8" embedment of the turning tip (which embedded the fixed tip about 9/16") required 925 ft-lbs of torque. We got the same results with four tests. We didn't have any way to test for a smaller indentation because the torque wrench we had had a minimum setting of 50 ft-lbs and our torque multiplier was 18.5:1. So we basically just torqued it until we hit the minimum 50 lbs of torque on the torque wrench (which using the multiplier was 925 ft-lbs) and that gave us the 5/8" embedment. This was quite a bit less than was reported to me originally but they said they were going off memory from doing this several years ago, so they probably just remembered wrong.

I wanted to keep playing with it for science, but the plant operations guys said they had better things to do so the experiment was ended. I can't get any of the mathematical models I tried to match the experimental results. The closest I got was about 4x more than the experimental results with a model based on delta = PL/AE. But I didn't feel like that was the model I would have expected to be the most accurate. Oh well, time to move on to something more productive I guess.

 

[CoryPad]

What yield strength were you using for the magnesium? A log of that size likely is as-cast, with low strength, probably lower than 100 MPa (15 ksi).

 

[shaneelliss]

The log is coming directly off a casting machine when I pick it up. I used 18 ksi in my calcs.

 

[rb1957]

I notice that the conversation seems to have returned to the impactor type of story (like a Charpy test) we started with. But the OP has stated that the cone tip is torqued, rotated, into the log. This is surely a completely different type of story, no? and may explain why one bit beds in more than the other (one's turning and the other isn't, no?).

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

 

[Compositepro]

I agree that rotation of the tip is a completely different situation than simple penetration. Rotation will cause plastic shear flow of the material in contact with the tip. Once it is flowing due to shear it will flow in the direct of any pressure gradient with little or no additional energy input. The pressure gradient in this case is from the tip of the penetrator to the surface of the ingot, where pressure is zero (15 psia).

 

[gruntguru]

The OP has told us that only one of the two indenters is rotated. He therefore has a direct indication of how much difference the rotation makes. One would suspect that the rotated indenter will penetrate slightly more, however the non-rotated indenter can be be used to verify whatever theory is used to model this somewhat simpler situation.

je suis charlie

 

[GregLocock]

A lower bound estimate of the work to remove the conical plug is the yield stress times the area of the cone times say a quarter of the diameter of the cone. How does that stack up with your other estimates? Also bear in mind you are creating two conical holes, not one, with just one screw.

I am amazed at the small difference turning vs plunge.

Cheers

Greg Locock


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[CoryPad]

This topic is similar to bolted joints. It is desirable to avoid indenting clamped parts with screws and nuts. The allowable pressure is higher for a stationary fastener (which provides compression only) compared with a rotating fastener (compression plus shear).

 

[tbuelna]

After reading back thru the posts I did not see any mention of what the diameter of the penetrator is. It was noted that the penetrator cone has a 45deg included angle and a "sharp" tip. But in theory, if the steel penetrator had a very sharp point and a very small diameter it would not require much force to penetrate 5/8" into the magnesium surface. As long as the steel penetrator has sufficient diameter to prevent buckling from the compressive load and unsupported length.

 

[rb1957]

from back on the 17th ...
from the OP ... "The shaft diameter is 2.25" and I am only trying to push the tip in about 5/8", ..."

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