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How much cross section area can cut this shredder?

This topic contains 10 replies, has 4 voices, and was last updated by  Stan 7 months ago.

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Frank Agraz thefranky

How much cross section area can cut this shredder?

24/05/2019 at 20:43

In case I want to have PET shredder with a motor of 1500 W with 3490 RPM and a pulley system of 49 : 1, that allows having 71 RPM in the shaft, how much is the bigger cross-section area that this shredder can really cut? I make some research about the topic and seems that to know that, first I have to know the torque of the motor:

Torque (N.m) = 9,5488 x Power (kW) / Speed (RPM)= 9,5488 x 1500 W / 71= 201,74 N.m

After knowing this, I look here and I find the PET’s yield Strength (since I want to make a PET shredder), which happens to be 90 Mpa. Knowing that I can make this:

T = F. b  Where:

T = the motor’s torque

F = Force needed to cut the material

b = the radio of the blades which is 59,785×10^-3 m

And knowing this:

Stress = F / A. Where;

A = cross-section area of the material to cut

Stress =  PET’s yield Strength

F = Stress . A

 

T = F. b = Stress. A. b

A = T /(Stress . b)

Which gives me:

A = 201,74 N.m /( 90Mpa x 59,785×10^-3 m)

A = 3,75×10^-5 m^2 = 0,375 cm^2

Isn’t that too little? Looks like a really small area… Maybe I’m doing something wrong? I don’t think that I’m using an inappropriate engine because I use the excel spreadsheet that has this post and looks that I have good parameters for the design. And on the other hand, If maybe someone doesn’t understand the context of my calculations you can see this post, I used it as a reference to make these calculations.

 

Please, Can someone help me?, maybe if I’m not really clear, you can let me know that :). I’m really sorry for my English.

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helper
25/05/2019 at 10:09
1

Well. You are using the wrong stress. That stress is fail in compression or tension. You want to cut the material this is called shear. so you want to use shear stress which from the first site i found is this Shear Strength, Average value: 15.3 MPA. so this is about 6 times lower than the value you are using.

helper
25/05/2019 at 10:17
1

well another place says 55Mpa usually rule of thumb is 60% of the ultimate stress. Ultimate stress is used as yield stress is not the stress required to actually break the molecules inside the bond to break them. Yield stress is just what is required to reform the structure/shape of the material. Ultimate stress is the stress required to break the bonds inside the material.

I thought that 16Mpa is pretty low. i would do some research to find a good average of the shear stress and use like the high side.

The easiest thing to do to increase the area is decrease blade radius and decrease rpm.

helper
25/05/2019 at 16:54
2

Hello @thegreenengineers!, thanks for answer my question. I had to say that I studied your post of “Calculations Required for Shredder” to reach these conclusions. I appreciate your contribution to the community.

Well, I look for the PET’s shear stress here and seems to have a value of  58,6 MPa, using that I have a resulting area of:

A = 201,74 N.m /( 58,6Mpa x 59,785×10^-3 m)
A = 5,75×10^-3 = 0,575 cm^2

And the other hand, if as you said I used the 60% of the ultimate stress (which is 90Mpa) 90 Mpa * 0,60 = 54Mpa, that gives me an area of:

A = 201,74 N.m /( 54 Mpa x 59,785×10^-3 m)
A = 6,25×10^-3 = 0,625 cm^2

 

Both cases seem to be small still, I’m a little confused about this because I follow the indication of the forums and they said that the machine will work fine with a 2 hp (1500 W)  motor and with 70 RPM in the shaft. These calculations show that shredder that I take as a reference is not capable of cutting plastic with a significant cross-sectional area, even when I’m using almost the same dimensions of the Dave hakken’s shredder since he used a radius of 60×10^-3 m for the blades.

helper
26/05/2019 at 06:38
2

thanks. Yea i have been away for a while because of universtiy. But i actually just graduated yesterday for mechanical engineering so i do plan on being on here more often and spedingmore time on stuff such as this and The Green Engineers.

so that goes to about .25″ x .25″. thats not that bad. most of everything that is shredded is waterbottles and such which have like a .040″ or up to 1/16″ (.0625″) thickness but are really tall which does not really cause any problem. But yea i would reccomend dropping the rpm if you want thicker stuff. more than 1 revolution per second is a little high. My own shredder product that i am working on mass manufacturing now will be operated by hand with a breaker bar(3 foot) first. with with me 170lbs will put down 170 * 3 = 510 ft lbs. ~710 Nm 3.5 times the torque of the electric motor with 2 hp. Then the next expansion would be a 15 rpm motor. with chained gear drive.

Another thing that i might do that i just thought of is just increase the teeth and drop if you double the teeth and drop the rpm in half then you will still have the same through put. my current verionhas 2 teeth i have been thinking about adding more.

helper
26/05/2019 at 13:50
1

Congratulations my friend!! I’m glad you’ve come to the end of a long road. 🙂

Thanks for your advice, I will try to reduce the RPM so that the machine isn’t so slow, what value you recommend me? maybe 40 RPM? , Is there a theoric way of know how much plastic can shred the machine for an hour? I don’t want the machine to be slow either, the idea of double the teeth sounds great, but are not feasible in my case :s

Although there’s a little problem, my units are in cm^2 not in inches^2. So if for example, I used the first area, in inches the value will be: 0.089 in^2  but for what I can see, it’s still big enough compared to the measurements you mentioned about the bottles. So it’s still ok, right?

Another thing, using as reference the examples above, which area I should use? the first or the second?

 

helper
26/05/2019 at 14:40
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One last thing @thegreenengineers I’d like you to know, I have been consulting this topic on reddit  too and one person told me this about my area’s calculations :

You seem to be calculating the max area that you could cut simultaneously, like every tooth bit into the side of the .375 square cm solid bar of plastic at once.

but if you’re making teeth like the YouTube video one, you’re using a very, very small fraction of that cutting area at one time

That is to say, If your total shredding area is say 4 inches across and you have 16 teeth rotated, they are interspersed with the spacers, giving cutting edges that’s 1/8th an inch of thick biting into the plastic at one time, not the entire area at one time.

Each tooth will see a combination stress depending on its direction (check out moores circle for more info) that’s a combination of its tensile stress and shear stress

But since the bottles are hollow, each tooth will be able to easily pierce the bottles and continue slicing.

Where I would think you need to worry about stalling is where the plastic flakes are completely full in the shredder, so the entire cutting surface works at once.

Her comment made me think, he tries to say that the blades never will really have contact with the total cross-area of the material to cut?, so the truth is that at the end of the day, the blades only will bite a very small fraction of the cross-area or something like that?

I don’t think I understand the idea that this person wants to transmit to me, what do you think about he told?

helper
05/06/2019 at 09:34
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So basicly these cross section are regardless of blade thickness. The cross section is with respect to the parrallel contact of the blade the cross section goes top to bottom not along material. Where blade thickness comes into play is the resistance to torsion at the hex/square shaft input.
My perticualr and precious plastics design has only 1 blade contacting at a time so you only have to take that 1 force into account. Again the question was not can it cut a waterbottle it was maximum cross section it can cut. Now about is theoretical as in your blade may not even have the spwept area to even cut that much area of a given plastic. These are just assumming that whatever you throw in may be cut. So basicly lets say you can cut .375 cc of PET but in actuallity your blades can only sweep .25cc then you are fine until you go with a harder material. If you still have a question on it i will make a video for you explaining it. Sorry about the long time between responses.

warrior
05/06/2019 at 21:01
0

Well,  I think the thickness of the blade is part of the area calculation. If you consider the ASTM D732 shear strength test configuration as an analog (an example here https://www.universalgripco.com/astm-d732 ) , The area over which the stress is applied is the perimeter of the punch tool and the material thickness. So for each shredder tooth, assuming for the moment that it contacts normal to the material, it would be the exposed perimeter of the single tooth times the plastic thickness. Now in real life, you will not get the same stress distribution as in the ASTM D372 material test. But I think, as long as you are looking at a single tooth cutting a single flat layer of plastic, using the perimeter times the thickness of the plastic should give you a conservative torque requirement.

warrior
06/06/2019 at 12:26
1

I think what thegreenengineers means is that once the point of the tooth has penetrated the plastic (assuming it has enough positive rake) then the cutting action only takes pace along the two edges of the blade, so its width is irrelevant.

 

I’m not convinced this is a good way to calculate the power requirements though, it’s not like a punch that would be perpendicular to the material, the cross section varies throughout the cut and also according to the shape of the tooth and thickness of the plastic.

warrior
06/06/2019 at 19:49
1

Yes, though depending on the geometry, the peak torque may occur when that chisel tip penetrates the plastic, especially as it gets blunt. Some of the commercial granulators have fairly wide blades and some of the insert based shredders (rubber) have blunt geometries that start approaching the punch. I think the V4 effort was looking at these at one point. As people look at variants away from the V3 design both for cost and performance, it will be interesting how the design evolves.

I agree that the punch analog is not accurate, but I think it provides an upper bound value. I would like to see someone take on the FEM analysis of a real design. It would make a great engineering project. Probably have to develop a materials property test method consistent with the details and also address rate dependency.

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