Sheet metal flat pattern showing incorrect flat pattern length

Sheet metal flat pattern showing incorrect flat pattern length

lancerichmond
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Message 1 of 38

Sheet metal flat pattern showing incorrect flat pattern length

lancerichmond
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Autodesk Inventor Professional 2022.jpg

 

I know it's an old problem but, did anyone actually find a solution that doesn't involve playing around with code, re-writing bend tolerances or reinventing the K-Factor? I have a simple steel pipe in sheet metal that I have "ripped" to obtain the flat pattern. The correct length should be 314.16 mm when made flat, however, Inventor says it should be nearly 4 mm shorter! This is definitely incorrect......

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Message 2 of 38

SharkDesign
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How thick is your rip, because you'll lose some material there?

 

Are you simply doing a flat pattern or you're using code to measure this?

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Message 3 of 38

lancerichmond
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Hi,

Pipe:

1. Outside diameter: Ø100 mm
2. Wall Thickness: 1 mm
3. Rip Gap: 0.0001 mm
4. Length/Circumference = Diameter x Pi = 100 x 3.142 = 314.16 mm
5. Inventor's Calculated Length = 310.641 mm
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Message 4 of 38

lancerichmond
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Simple flat pattern Vs Circumference=Pi x Ø

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Message 5 of 38

SharkDesign
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It'll take material stretch into consideration as well so it won't be identical. 

 

 

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Message 6 of 38

SharkDesign
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Also you are calculating the OD at 314.1593, but remember the ID is 307.8761.

The centre line of this is 311.0177.

 

4mm difference is probably right. 

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Message 7 of 38

lancerichmond
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I understand the material fundamentals especially when it comes to theory Vs practical however on such a small piece of material, 4 mm is a lot! I only did this this example because I was using a pipe that was roughly Ø800 and a wall thickness of about 40 mm. The true length vs Inventor's flat pattern was a very large difference......

 

T.I.A.

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Message 8 of 38

lancerichmond
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I tried working out the mean diameter in hand calcs and still, the answer is incorrect......

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Message 9 of 38

JDMather
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@lancerichmond wrote:

 I was using a pipe that was roughly Ø800 and a wall thickness of about 40 mm.


That is a very thick wall thickness and doesn't match your Post #3?

Can you Attach your *.ipt file here?


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Message 10 of 38

lancerichmond
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Hello,

 

I said "I only did this this example because I was using a pipe that was roughly Ø800" because when I flat patterned the Ø800 mm pipe (hollow bar), I noticed that the length was greatly incorrect. That's when I modeled a smaller pipe to see if I would also get the incorrect sheet metal length when it was "ripped" and flat patterned. Which is indeed what happened.

 

Excuse my ignorance (I don't post a lot) but, how do I upload my file?

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Message 11 of 38

JDMather
Consultant
Consultant

@lancerichmond wrote:

Excuse my ignorance (I don't post a lot) but, how do I upload my file?


The process to Attach files is currently broken.

Attaching files to posts - Autodesk Community - Inventor

 

The work-around is to Edit Reply.

Now you should see the Attach.


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Message 12 of 38

lancerichmond
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Lol, ok I thought I was a bit slow......

 

I've attached my file in the edited reply....

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Message 13 of 38

JDMather
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Consultant

You calculated the OD of finished pipe.

What do you calculate as the ID?

What is OD-ID=


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Message 14 of 38

lancerichmond
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The length based on the ID should be 307.88 mm however Inventor says it should be 310.641 mm......

 

The length based on the OD should be 314.16 mm however Inventor says it should be 310.641 mm......

 

Whether I "Define A-Side" using the OD or the ID of the pipe, Inventor still gives me the same answer........

 

My previous post did specify a wall thickness of 1 mm and an OD Ø100 mm.

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Message 15 of 38

JDMather
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Consultant

@lancerichmond wrote:

The length based on the ID should be 307.88 mm however Inventor says it should be 310.641 mm......

 

The length based on the OD should be 314.16 mm however Inventor says it should be 310.641 mm.......


You are answering questions that I did not ask.

OD=314.16

ID-307.88

 

OD-ID=6.28 mm

Where did that difference come from.

 

Now, let's imagine that you have a flat sheet 100mm x 314.16mm.

The 314.16mm is what it is - measurable and verifiable.

 

Now, let's bend that with the intention of having it a round circle tube 100mm long by 100mm diameter.

 

If the OD-ID=6.28mm, where did that extra 6.28mm go???

The 314.16mm is 6.28mm too long for the ID?

Did the material compress 6.28mm?


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Message 16 of 38

SharkDesign
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What k factor are you using when hand calculating the flat pattern size?
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Message 17 of 38

SharkDesign
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If you think about it like this:

 

If you have an L shaped piece. There is one bend and a slight amount of stretch/compression.

 

In a cylinder, every single part of that metal sheet is stretched/compressed to form a circle. This is going to = a much bigger difference.

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Message 18 of 38

JDMather
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Consultant

@SharkDesign wrote:
What k factor are you using when hand calculating the flat pattern size?

I am not using any k-factor.

I am not attempting to calculate a flat pattern size yet.


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Message 19 of 38

JDMather
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Consultant

JDMather_0-1628595949083.png

Now how can that be?

Examine the Attached file.

If we start with a 100mm*PI flat sheet

how can it be three different arc lengths when bend?

OD is different than ID and both are different than mid-surface length???

Where did the material go?

(Don't look at next post until these questions are pondered.)


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Message 20 of 38

SharkDesign
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Yes, but what I'm saying is that if you use the default K factor as in the real world you get the 310mm that the OP says is wrong. If you set the K factor to 1, you get the 314.159mm that is obtained from simply measuring the circumference. 

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