# Calculations Required for Shredder

This topic contains 26 replies, has 12 voices, and was last updated by Hariramkumar 1 week ago.

Wanted to start this thread talking about the required calculations for the shredder.

Horsepower:

The calculation for torque is from the shear stress required to cut a specific material. The shredder uses a mode of shear to cut material which is a perpendicular force applied to a material. In this case 2 perpendicular forces that are equal and opposite in direction. the 2 perpendicular forces are parallel to each other but opposite in direction.

Shear is calculated from a rule of thumb percentage of the yield stress(its something like 80% of yield stress) of the material you are trying to cut. (google yield stress of your material). stress is the Force applied divided by the Area that the force is applied over =F/A. The Area is found by the cross section area of the material of cut that the shredder blade contacts. Then to determine the power of the motor required is from the torque the motor puts out at a given rpm. Torque is Force times Distance F*D. So to solve for force at the cutting area you will divide the torque by the RADIAL(Radius of cutter) distance to the cutting TIP (worst case) . This solves to (F*D)/D = F. Now Equation to find horsepower is RPM (rotation per minute) * Torque all divided by 5252

Example

So lets say the contact cross section area(A) is calculated from a Triangle sweep of shape of cut

(2.5″ x .4″)/2″ = .5″^2 (the worse case from design of my shredder)

Radial distance to tip is 3.5″ Diameter/2 (my design of shredder) R = 1.75″

I googled Yield Strength of HDPE and came up with 4350 PSI (F/A aka pressure).

So i am going to take 80% of that for rule of thumb i get 3500 PSI.

So 3500 PSI = F/A.

I solve for Force so Force = 3500 * .5 = 1750 That equals 1750 Lbs of force.

Now solve for how much torque i need from my motor.

Torque = Force * Distance so Torque = 1750 * 1.75″ = 3063 inch-pounds of torque.

lets say i have a motor that rotates at 5 rpm.

For horsepower (3063 * 5) = 15315 in-lbs/sec / 12 = 1276.25/ 550 =2.32HP

Then we have something called S.F. (safety factor) which is basically the needed power multiplied by the Safety Factor which is basically a allowance for unaccounted variables or harder materials that are trying to be shredded etc. in this case lets say S.F. of 2 so 2.32 * 2 = 4.6HP.

The reason why this is so high is because we are cutting the max cross section of HDPE that my shredder can handle.

HumanPowered:

So what if it is human powered. how long of a bar “moment arm” would you need to shred that max cross section area. so we need 3,063 inch-lb of torque. i weigh 160 lbs. so take 3063 and divide by 160 lbs to get a 19.14 inch long bar witch is almost 2 feet which is very manageable. Meaning that the machine will be way cheaper and easier to power if it was just a long bar and you hang on it.

Conclusion:

So when you do your calculations plug in the design of your shredder and use the yield stress of the HARDEST material you are trying to cut. Then find your HP Minimum then multiply it by at least 2 to allow for some other variables to get the motor you need.

Motor Selection:

Motor selection is very complex topic because you have to take account motor efficiency at different rpm and torque ranges. Brushless motors are most efficient at 80-90% their max RPM a efficiency of 80-90%. However at that RPM range torque is at 1/8 its maximum output torque. Luckily when a Motor is rated a specific HP rating that rating is at its maximum efficiency. Maximum efficiency of the motor is found at approximately half its MAXIMUM output HP. So that means when you load it even more the RPMs will drop to compensate for the required torque needed.

Gearbox:

That is where gearboxes come in. The purpose of gearboxes is to try to have your cake and eat it too. In other words to run your motor at its max efficiency RPM and horsepower rating but “gear down” the rpm so that you also get the desired torque you want. So you go to your gearbox to trade RPM for torque. To do this you have a ratio such as 10:1 which means that your input gear is 10 times smaller than the output gear. Meaning the input gear has to turn 10 times to turn the output gear once. This cuts down your rpm by 10 times but increases your torque 10 times (not exactly). The higher the ratio the lower the output RPM but the higher the torque. And the reason why i said not exactly above is that gearboxes are not 100% efficient and you will lose power as heat from friction from the gears meshing (most gearboxes are around 90% efficient). So for a shredder the main question you have to ask yourself is mainly how low of RPMs can i live with that will determine your motor and gearbox selection.

Example:

Lets say we have a 1.5KW(~2 HP) Spindle motor for engraver/router. Reading off this spec sheet

rated Power: 1.5KW(~2HP)

Rated Speed: 1,000 RPM

Max RPM: 2400 RPM

So what is our torque. (RPM * Torque)/5252= HP

We have Rated RPM and Rated Power

Solve for Torque = (HP * 5252) / RPM = 10.5 Ft-lbs (pounds)

Alright so gearbox solve now

Lets say we are ok with 5 RPM for output

so from 1,000 RPM to 5

1,000/5= 200 so we would have a 200:1 ratio

So what is output torque

10.5 * 200 = 2100 Ft-lbs

Massive Torque. Low RPM. Easy Shredding. Or might tear machine apart.

Another thing you can do is run your motor at like 90% its maximum HP (dont want to run at 100% you will burn motor up) rating that will give you even more torque. However you will lose a lot of energy as heat as your motor will be running at 60% energy efficiency rather than 80% meaning you are spending 20% more energy than you need to.

Enjoy.

ask some questions for some calculations and i will do them for you. and post them here.

this is awesome… you r a genius bro. my question is : i am struggling to find low rpm motors here in australia. most of the motors here are quicker than 1800 rpm. hmm.. which is too fast i believe. i found some rpm reducers…..but not sure if this is the only solution. could u help me plz.

Good question. I will add this to the end of first post as motor and reducer selection. but here is what goes there.

Motor Selection:

Motor selection is very complex topic because you have to take account motor efficiency at different rpm and torque ranges. Brushless motors are most efficient at 80-90% their max RPM a efficiency of 80-90%. However at that RPM range torque is at 1/8 its maximum output torque. Luckily when a Motor is rated a specific HP rating that rating is at its maximum efficiency. Maximum efficiency of the motor is found at approximately half its MAXIMUM output HP. So that means when you load it even more the RPMs will drop to compensate for the required torque needed.

Gearbox:

That is where gearboxes come in. The purpose of gearboxes is to try to have your cake and eat it too. In other words to run your motor at its max efficiency RPM and horsepower rating but “gear down” the rpm so that you also get the desired torque you want. So you go to your gearbox to trade RPM for torque. To do this you have a ratio such as 10:1 which means that your input gear is 10 times smaller than the output gear. Meaning the input gear has to turn 10 times to turn the output gear once. This cuts down your rpm by 10 times but increases your torque 10 times (not exactly). The higher the ratio the lower the output RPM but the higher the torque. And the reason why i said not exactly above is that gearboxes are not 100% efficient and you will lose power as heat from friction from the gears meshing (most gearboxes are around 90% efficient). So for a shredder the main question you have to ask yourself is mainly how low of RPMs can i live with that will determine your motor and gearbox selection.

Example:

Lets say we have a 1.5KW(~2 HP) Spindle motor for engraver/router. Reading off this spec sheet

rated Power: 1.5KW(~2HP)

Rated Speed: 1,000 RPM

Max RPM: 2400 RPM

So what is our torque. (RPM * Torque)/5252= HP

We have Rated RPM and Rated Power

Solve for Torque = (HP * 5252) / RPM = 10.5 Ft-lbs (pounds)

Alright so gearbox solve now

Lets say we are ok with 5 RPM for output

so from 1,000 RPM to 5

1,000/5= 200 so we would have a 200:1 ratio

So what is output torque

10.5 * 200 = 2100 Ft-lbs

Massive Torque. Low RPM. Easy Shredding. Or might tear machine apart.

Another thing you can do is run your motor at like 90% its maximum HP (dont want to run at 100% you will burn motor up) rating that will give you even more torque. However you will lose alot of energy as heat as your motor will be running at 60% energy efficiency rather than 80% meaning you are spending 20% more energy than you need to.

Be creative to source something. I will give you some ideas.

go to car scrap yard. Find cheapest flywheel and starter motor set. That gives you super high torque and low rpm gearing cheaply.

Use gas lawnmower engine with pulleys and belts. pulley out of engine is smaller than pulley at shredder.

use pulleys instead of gears.

etc.

In case anyone is interested on reading more about other member experiences with motors, there’s an sticky thread here: https://davehakkens.nl/community/forums/topic/almost-definitive-guide-on-motors-wip/

Bro how did you get that triagular cross section area? And do you have a design for the shaft?

Hello,

**The Green Engineers. **

I want to calculate the torque and power required to drive a shredder.

Here I am mentioning the details.

Cutter Dia. =650 mm

No. of teeth on cutter = 2

Cutter Thickness= 75 mm

Cutter Teeth Depth= 80 mm

Spacer Dia. (It is placed between two cutters)= 400 Dia.

Spacer Thickness= 76 mm

Cutter mounted on a shaft having hex = 225 mm (Across Flat)

Total no. of Blade or Cutter on shaft = 13 Blades & 13 Spacer on a Shaft.

The shaft is supported from both shaft double spherical bearing.

RPM of shaft = 15 Revolution per minute

RPM will be managed by Gear Reducer.

Input RPM to Gear Reducer =1440

**Material to cut or shred is metal (mild steel scrap) **

I want to know about the how much power I required to drive and shred the Mild Steel material.

**Please tell me the method to calculate the **

**Power required to drive the shredder machine.**

**Total Power required to shred the material.**

Thanks

That would be a BIG shredder!

I don’t think you can just scale up the Precious Plastic shredder and expect to be able to shred cars with it. The hex shaft in particular would be a poor choice for a machine of this size, it’s used on the PP shredder because it’s convenient, cheap and readily available.

alright done with school for now and on summer break i will try to run the numbers for you tomorrow.

Hey mate, first of all, great post(? (Don’t really know how to call it)

Would you happen to have some sort of drawings? Because I get lost when you mention that much. Thanks in advance.

Yea i will start doing the calculations on my tablet and just uploading the hand calculation image. its also way faster and easier for me as usually i do the calculations on my tablet then have to come back here and type it out.

it is actually simple, if you are looking for long and efficient run, go with a 4kw, 5Hp motor (1440), a good reducer (SITI, 20:1 for not under 500 Euro, service class 1) and a 4Kw inverter, you can run 2 shredder units and 2 extruders on one drive shaft without any real problem, as long you feed it at the right rate of plastic which requires quite some more work on the hopper. I’m sure i can do more with that setup if i ditch the v3 shredder box for a double shaft shredder design (which can double the torque as well).

I am building such machine right now, using a lego like approach so the tables can be extended/connected since i use easy to dispose elements and layouts.

it shreds wood and PVC too :-). this setup currently draws 10 Amps with one shredder and extrusion connected, crushing average plastic (hard). it’s important of course to run a 3 phase motor, way more efficient.

Enjoy ;p

any questions or other calcs you need let me know.

Spoiler alert:

it takes about 100 hp to cut a 3mm thick / 80mm piece of steel. Pretty long piece of steel.

few things to drop the hp required:

bigger gear ratio

smaller diameter blade/teeth

smaller cross section of material needed to cut

and cutting softer material.

Hello All,

I am designing a shredder machine for soft rubber. Knives shaft look like the attached images. Please don’t care about “weird” things in these images, the design is under construction.

My question is: how to determine cross sectional cutting area ?

Thanks,

Bo Do

@bodo2018, could you please share a solid-works or fusion-360 compatible file ? i can have a look at it, we’re also interested in a pneumatic/rubber crusher.

The cross section is calculated using the Area of the material you are going to cut. lets say it is a square cross section or square piece of plastic the area is the width the the plastic times the height.

So a way of thinking about it is say you cut the piece of plastic with a band saw what is the area of the cross section of the cut. I will do a drawing on my tablet and post it.

Hi @anne-barbier, I am designing the machine so once it is completed I will share with you all here! Thanks for your interesting!

Thank you for your answer!

I understand what you explained, I also thought about the same thing before.

What happens if the waste is a bulk of thin rubber like gloves?

I have read a paper and it showed me some equations and results which made me confused as shown in the attached image.

For more details you can refer to the attached pdf file, the image is from page 5 of pdf file.

Thanks,

Bo Do

Hello all,

I am interested in designing a shredder machine for organic waste like leaves,wood.

I don’t know how to start my design calculation, and how to assume the required parameters.please help me friends…

How to select the cutter thickness, cutter diameter, cutter Material, cutter teeth depth?

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