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542 Views, 4 Replies

10-15-2013 03:40 PM

I'm trying to figure out a calculation to get the thickness decrease of the outside edge of a flat bar that is rolled into a flange. Imagine a 50*5 bar that is rolled per the below picture. In theory it's obvious that the thickness (5) stays the same at the inside diameter of the ring and that there is a reduction in the thickness of the material on the outside edge. Currently I'm basing my reduction on the ratio of the "inside edge circumference" divided by the "outside edge circumference". So the new thickness is something like 5-(5*ratio). In the model I have the "inside edge circ" is 313.287 and the "outside edge circ" is 626.573. So the ratio is 0.5 and the end result is 5 - (5 * 0.5) = 2.5.

And from the mass end of things Inventor calculates in just fine as volume * material density.

My problems arise when I apply this calculation to the same ring rolled as a helix, like in a screw conveyor. I can calculate the lead angle of the flite to base the initial sketch on and can model the flite (using coil) just fine (there is a quirk in the lead angle plane but I am ignoring that for now). And applying what I think is the correct "thickness reduction" calculation (above) works. The problem is that the mass is way off. As an example using OD of 228, ID of 48.3, PITCH of 228 gives me a length around the inside helical edge of 273.877 for 1 pitch of flite. Converting this to a volume I get length=273.877, width = 5 and height is 89.95 ((OD-ID)/2) which equates to 0.000123176 m3. Multiplying this by density of steel (7850 kg/m3) and this should equal 0.9674 kg. Inventor reports this 1 pitch of flite at 1.335 kg. A very big difference and I can't find the culprit apart from thinking the thickness of the outer edge is wrong.

I have attached the part file (2013) if anybody can help me with this problem.

Brendan Henderson

Web www.blh.com.au

Twitter @BLHDrafting

Windows 7 x64 -64 GB Ram, Intel Xeon E5-1620 @ 3.6 GHz

ATI FirePro V7800 2 GB, 180 GB SSD & 1 TB HDD, Inv R2013 PDS Premium SP2 Update 3 (Build 200), Vault 2013 Workgroup Update 2 (Build 17.2.9.0)

Web www.blh.com.au

Twitter @BLHDrafting

Windows 7 x64 -64 GB Ram, Intel Xeon E5-1620 @ 3.6 GHz

ATI FirePro V7800 2 GB, 180 GB SSD & 1 TB HDD, Inv R2013 PDS Premium SP2 Update 3 (Build 200), Vault 2013 Workgroup Update 2 (Build 17.2.9.0)

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10-16-2013 11:35 PM in reply to:
BLHDrafting

Brendan,

The "length" of the helix is not the length of the inside edge. I used the length at the centroid of the swept area but it resulted in a calculated volume that was off by about 2% (volume too large compared to reported volume in IV). Not sure where the error is but maybe it might point you in some direction.

Neil

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10-17-2013 05:25 AM in reply to:
nmunro

Ignore the previous advice, a lack of sleep <g>. The swept volume is the sum of the area of thin vertical slices of your cross section multiplied by the length of the helix at the center of the slice. Taken to the limit the volume could be determined by an integral involving the formula for the sloped line that defines the boundary of the cross section, and the equation governing the helix. A numerical method with reasonably thin sections might be easier to implement.

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10-17-2013 01:51 PM in reply to:
nmunro

Thanks for the input Neil. More grist for the mill.

The machine that makes this helicoid (flite) can be seen here :- http://www.lenham-uk.com/production.htm

For anyone having any input to this, note that there is no material removal or addition in this process. The coiled material is fed into 2 offset cones that are also slighlty angled to each other. An assumption is that the cones don't squash any material on the inside edge of the helicoid (the spiral that is welded to the pipe/tube) but from the inside to the outside there is a continual decrease in material. I have measured the material thickness at the helicoid inside edge and there is no reduction. This decrease (squashing) produces the radial curve in the material, and the offset in the cones produces the axial curve (spiral).

With the above in mind (no material is removed or added) the volume of 1 pitch of the helicoid hasn't changed. I can't help but feel that the gradual thickness reduction from the inside ot the outside is a function of the length difference of the inner and outer helix. maybe this is where I'm wrong and I need to move on from that as all the tinkering I have done based on that thought still produce drastically wrong results (should be 0.676 kg/pitch

Scott Wertel thinks there may be a function of Poisons ratio at play which could be the case. While there is obvious stretching of the material it is practically impossible to tell if any combination of compression/stretching occurs at the helicoid inside edge.

This helicoid (flite) is cut off the machine in a 3 metre length. When weighed it's mass varies about 2-5% of the theoritical calculated mass. Usually the finished product is heavier which may indicate that there is some material compression.

Brendan Henderson

Web www.blh.com.au

Twitter @BLHDrafting

Windows 7 x64 -64 GB Ram, Intel Xeon E5-1620 @ 3.6 GHz

ATI FirePro V7800 2 GB, 180 GB SSD & 1 TB HDD, Inv R2013 PDS Premium SP2 Update 3 (Build 200), Vault 2013 Workgroup Update 2 (Build 17.2.9.0)

Web www.blh.com.au

Twitter @BLHDrafting

Windows 7 x64 -64 GB Ram, Intel Xeon E5-1620 @ 3.6 GHz

ATI FirePro V7800 2 GB, 180 GB SSD & 1 TB HDD, Inv R2013 PDS Premium SP2 Update 3 (Build 200), Vault 2013 Workgroup Update 2 (Build 17.2.9.0)

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10-17-2013 03:16 PM in reply to:
BLHDrafting

to me that sounds like you are on the right track, if the material is being stretched, there's a good chance that some of it is also being compressed.

I guess in my mind you might want to start thinking about it more along the lines of a sheet metal bend, there is no real 'k' factor going on in your equations. How I'm not exactly sure...

Maybe rework your numbers to assume that the material undergoes even compression and stretching, and then slowly skew that towards a higher ratio of stretching.

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