[Amps] Another metalwork question

[Will Matney]

Martin, I dont think much really changes as those factors are derived from ratios. You should be able to add a multiplier of 6.45 to the formula to get metric sums. Try that to see what you get and substitute MM in place of the inch measurements then multiple the sum by 6.45. I think you'll get the same just different measurment systems that way.

BA = (0.0078 * T + 0.0174 * R) * No. of deg. in bend * 6.45

Best,

Will


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On 8/2/05 at 5:57 PM Martin Sole wrote:

>Will,
>
>Thanks for the info, very useful. Do you have the formula in a suitable
>form
>for metric material? Haven't worked in imperial measurements for over 30
>years and only see it now and again on odd bits of US made kit. Actually I
>am going to make a new plenum for my second Alpha and the original is most
>certainly made to imperial measurements but nobody here would understand if
>I tried to replicate it that precisely so the new one will be made to
>metric
>dimensions. Would definitely appreciate the drawing and picture, mail away.
>
>Thanks
>
>Martin
>
>
>-----Original Message-----
>From: amps-bounces at contesting.com [mailto:amps-bounces at contesting.com] On
>Behalf Of Will Matney
>Sent: 02 August 2005 17:29
>To: amps at contesting.com
>Subject: Re: [Amps] Another metalwork question
>
>Martin,
>
>On bending steel, you use 1/2 the material thickness to figure how much to
>add in length. In other words you divide the material thickness in half
>where there would be an imaginary center line (its neutral axis) going
>trough it. Then when the steel is bent, a radius is formed on this
>imaginary
>center line even though the inside bend is a sharp 90 deg bend. So whatever
>the distance is around that small radius in the middle of the steel is the
>material to be added. Now aluminum is a different story and there is a
>formula for it too. What happens in aluminum, there is a shrinkage on the
>inside radius and a stretching on the outside different than steel. For
>aluminum see the formula and example below;
>
>BA = (0.0078 * T + 0.0174 * R) * No. of deg. in bend
>
>14 Ga = 0.0641"
>
>For 0" radius, 90 Deg bend in aluminum;
>
>(0.0078 * 0.0641 + 0.0174 * 0) * 90
>
>(0.00049998 + 0) * 90 =
>
>0.00049998 * 90 = 0.045" or about 3/64" or for ending pieces of flanges
>make
>it 1/16" from bend line to end.
>
>BA = Bend Allowance
>R = Radius of bend on the inside, not the neutral axis.
>T = Material thickness
>
>I have a pic with this and a drawing if needed I can e-mail it to you. Hope
>this helps.
>
>Best,
>
>Will
>
>
>*********** REPLY SEPARATOR  ***********
>
>On 8/2/05 at 2:46 PM Martin Sole wrote:
>
>>Well it is actually amp related, or will be at some point I hope but 
>>seeing how there is a great wealth of resource here it seems a good 
>>place to start.
>>
>>Some time back I recall seeing an article, might have been in Radcom, 
>>might have been in QST. Think it had to be either one of those two 
>>though. It addressed the process of marking out metalwork for making 
>>enclosures and explained how to allow the correct amount of material 
>>for bends etc. Was within the last year or two I think.
>>
>>Just hoping that somebody might recall where this was or maybe point me 
>>to another resource with similar information.
>>
>>Tks
>>Martin HS0ZED
>>
>>
>>
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>
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