# Misc. Project: Single Shaft Shredder

#### Henk Trekker

Summary
Single shaft copper shredder
Summary: Single shaft copper shredder

I design a rotor of a shredder, but i need to find a motor for the rotor of the shredder.
My questions are:

1. What is the best rpm for a copper shredder ( what kind of effects wil the shredder have on high rpm 1200-3000, lower rpm 300-1200 )

2. what kind of material is usefull for a rotor ( aluminium is strong and light, but g/cm3 of aluminium is lighter than copper. will it break in rotation contact? )

3. hydraulic / electric / benzin / diesel motor, what is the best choice for this project?

4. hydraulic is strong in nm low rpm, wil it make problems if i don't use a gearbox?

The rotor:

Diameter = 242.706664725 mm
Radial = 121.3533323625 mm
Circumference = 750 mm
Length = 610 mm ( with driveaxle +- 750 mm )
it has 120 knives / cutters ( length Width= 17.678*17.678mm )

Why i think 300nm, F= m * v^2 * r,

m * v^2 / r, = F
steel ( 7.8g/cm3 ) = ( 300 * 7.6026542216873^2 / 121.3533323625 = 143 * 2 ( for safety ) = around 300 nm

aluminium ( 2.755g/cm3 ) = 110 * 7.6026542216873^2 / 121.3533323625 = 52.4 nm * 2 = around 110 nm ( very big difference with steel )

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#### trurle

It is difficult to see what are you trying to calculate. Please specify.

Your equations seems to calculate torque from the moment of inertia of (accelerating) rotor which is not significant for typical shredder.
You should put in calculation shearing strength of copper (scrap) multiplied by axis to tooth distance to get actual torque.

Regarding shredder speed, the slower speeds are generally better. You need to break material, not to make it flying - at 3000 RPM some loose scrap may end up flying up to distances 75m.

#### some bloke

I've never seen shredders operating fast. You want it geared down for maximum torque (too much torque isn't a problem, provided you have a torque-clutch which prevents it destroying the shredder if a block of steel was dropped in). 20-50RPM would be my estimate - you're aiming to shred it, not to grind it down, right?

#### Henk Trekker

@ Trurle, Thank you for the support!
@ Some Bloke, Thank you for the support!

25000 psi ( shear stress of copper )
0.025 m ( axis to tooth distance )

25000 * 0.025 = 625 nm

I want to granulat the scrap copper in pieces of 3 - 6 mm ( filter controls it ), the scrap copper has an little vinyl coating. the coating need to be shred off from the copper.
60 rpm is more than enough? ( every tooth set has design at 36 degrees of the rotor, it wil make at 60 rpm, 600 times contact at 1 minute :) )

The setup:

hydraulic motor ( reason: low rpm, very high torque )
Gearbox ( low rpm high torque )
torque-clutch

rotor:
60+- rpm
750-1000+ nm ( i will choice for the biggest torque in my pricerange )

If I made a mistake I would love to hear it.

#### trurle

25000 psi ( shear stress of copper )
0.025 m ( axis to tooth distance )
25000 * 0.025 = 625 nm
You messed up units.
N/(m*m) * m = N/m, while torque is N*m

#### Henk Trekker

@trurle, thanks for improving!

Yes, we all learned from mistakes.

#### jrmichler

The torque is the force to shear the material times the radius of the cutter from the axis of rotation. The shearing force is, very approximately, the area sheared times the shear strength of the material. The diagram below shows a sheet metal punch process, which is similar to your shredder:

The diagram shows a circular punch with diameter D shearing sheet metal with thickness t. The area sheared is PI X D X t, and the punch force is that area times 0.7 times the material tensile strength. A conservative estimate of the total work per tooth is the punch force times the material thickness. Multiply total work per tooth times the number of teeth actually cutting per revolution times revolutions per minute to get power.

The force per tooth is also used to design the tooth attachment. The force per tooth times the peak number of teeth cutting at one time is used to design and select shafts, bearings, and gear reducers. The peak power is used to size the motor. The teeth on the shredder rotor cut against the teeth of the stationary anvil. The forces from the above calculations are used to design the anvil bar and attachment of the cutting teeth to the anvil bar.

After making sure that everything is strong enough, the next step is to check stiffness. The peak deflection under peak cutting forces should be several times smaller than the clearance between the rotor teeth and and the anvil teeth. That clearance is normally 2% to 10% of the material thickness when shearing sheet metal.

That is the method for estimating forces, work, and power for shredding sheet metal. If you are chewing up solid blocks, the force calculation is different.

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