Flattening a flight of a screw conveyor 1

[PatCouture]

I don't know if this subject had already been discussed here but I would like to be able to calculate exactly what it takes in a sheet of stainless steel to make the flights of a screw conveyor. because every time I try I can't find the proper dimensions the first time. I always have to do 3 or 4 tests before I have the right flat design. I have a screw conveyor of 12 inches outside diameter, pitch 12 inches, inside diameter 2"7/8, (2 1/2 sch 40). Thanks in advance for any help.

Patrick

 

[smckennz]

Normal practice is to develop by treating the screw as a helix. There are countless books that show the development.
I am not aware of any "official"allowances made to allow for stretching the metal while forming I used to allow 2 - 3mm diametral clearance on the inside diameter otherwise the thickness effect and any pressing errors/misalignment resulted in having to stretch the flight over the shaft which would then mess up the butt joint alignment between flights.
Try giving yourself clearance on the ID.

Cheers

Steve

 

[Grue]

There's a discussion on this topic on

http://www.thepiers.net/pn/index.php?name=PNphpBB2&file=viewtopic&t=51

I posted a sketch on

http://www.thepiers.net/pdf_manual/Conveyorflight.pdf

There's a cad version of the drawing on

http://www.thepiers.net/pdf_manual/screwexplanation.pdf

If you make a template from thin gauge material the process is relatively easy. Come back to me if you have trouble.

Glenn

http://www.metalbashatorium.com

 

[sacem1]

Gentlemen:

In screw construction the problem is not drawing or laying out the flights of the screw but actually find the correct inside and outside diameters for the disks that you are going to cut, then twist a little so that you can weld them one to the next to achive a continuos strip and then streching them.

You will have to get a tight fit to your center tube and the right outside diameter, this while having the usual "step" equal to one diameter for every 360º turn, you will find that one disk once streched accounts for less than 360º of a flight.

The flights once streched will have a thinner section on the edge than the original plate thickess due to heavy metal flow on the outer edge and that varies with the relationship between inner initial hole and outside diameter, I have found that the smaller the center hole (smaller cenetr tube) for the same outside final diameter the higher flow of material and consecuently bigger thinning of the steel plate.

The big manufacturers even offer you two different ways of manufacturing the screws one as described and the other keeping a thicker section on the outside by using laminating techniques only possible in mass production plants.

Cheers

SACEM1

 

[sacem1]

Sorry I cut my answer short in the above answer.

I have found that only trial cuts will solve your problem and keeping a good reliable record of past experiences will let you have a future help for other works or starting points for future experiments.

If any one out there has solved this and has a formula for this calcs I wii also appreciate the help.

By the way the disk thickness also has an influence in the size of the starting disks.

Cheers

SACEM1

 

[Grue]

Sorry to disagree, the layout of the flight IS a drawing problem requiring triangulation. By doing this you find the true size of the inside and outside "diameters", and as you say a little twist and a weld.

Working with grain elevators and harvester equipment (imported) one collects a number of standard patterns to suit the usual equipment one is required to repair. They are all made the same way. Layout the flight using triangulation on a pattern, and start marking out the required number for the particular screw.

Glenn

http://www.metalbashatorium.com

 

[PatCouture]

I'm sorry Grue, but by experience I have to agree with sacem1 because I try the formula by triangulation, and it was close but always wrong, so I tried doing a ratio to find the correct one and It was wrong again, so my conclusion is that there is no proportion between the dimensions of the flattened disk and the pitch that it gives. That's why I have to do 3 to 4 test before finding the right dimensions.

thanks for the help.

Patrick