Calculations Required for Shredder
Wanted to start this thread talking about the required calculations for the shredder.
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
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.
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.
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 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.
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.
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.
ask some questions for some calculations and i will do them for you. and post them here.
Yes, maybe a video will help me and others to understand more about the area calculation because I still confused about how exactly we determine the worst case of possible cutting area (the maximum size of plastic that we could possibly put in there). I still think that this has to follow the shredder geometry of some sort, since we can’t just determine the worst case by ourselves (the scrap/cutted plastic size will follow the shredder geometry), otherwise our design will be redundant.
“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)…”
Can you explain this with a sketch please? I didn’t understand the idea of triangle sweep shape of cut. Also what do these values “(2.5″ x .4″)/2″ = .5″^2)” depict?
I will try to answer your question.
1. The whole purpose of this calculation is to define the required horsepower, so you could determine which motor you gonna use. In turn, you won’t make a machine with insufficient power (the machine is not gonna work) or overpowered machine (the machine is redundant and cost higher than It should be).
2. Why should we calculate horsepower? Refer to answer no. 1. How will you know the RPM? Well, basically you can’t just determine the RPM if you’re using a direct transmission (without gearbox). With the gearbox, on the other hand, you can determine the RPM output (the one that will move your shredder), while the RPM input is determined by calculation using formula (Power input x efficiency = power output). You will obtain the value of required Power input, which value you can use to choose your motor, and using the motor specification, determine the gearbox ratio you should use (or use your gearbox ratio to choose your motor later. The choice is yours.)
3. For the answer to this question, please refer to answer no. 2.
4. The constant comes from the conversion of the power unit. This depends on what your starting units are. For this, I suggest you should google the constant for the conversion of your unit.
5. Why 80%, well It’s the rule of thumb for the yield strength that needs to be surpassed for a shear mode of failure (which is the mode used in cutting mechanism). Although from the reference book I read, It’s supposed to be 50%. I hope someone could clarify me on this one.
6. For the answer to this question, I’m still wondering tho’, but I think It depends on the blade thickness and material thickness (see my previous reply). I guess I will try to conduct some further research on It.
7. For a material to be cut (or shred), It must be in static equilibrium, so It will deform. That is why It needs 2 forces (or resultant of forces, see my previous reply), perpendicular to the plane of cut, and opposite to each other, which are active force (the cutting force), and reaction force (the resistance force) that acts as a support that holds the material/plastic/scrap in place. That is why in my previous reply I stated that when the scrap is less than the blade thickness (plus the gap, if that exist), the scrap won’t be able to be cut. That is why the cutting area will depend on the blade thickness (plus gap) as It becomes the defining parameters for cutting capability. I still wondering if this is right tho’, hope someone will have a better explanation for this.
Hey, I wanna ask. You said that the cutting area is found by the cross-section area of the material of cut that the shredder blade contacts. Is this actually means the product of blade thickness and thickness of the material/height of material?
Or, is It means that the cross-section is the area of cut that I want to make, regardless of the knife geometry?
In my opinion, the plastic to be cut must have a dimension more than the blade thickness for the plastic to receive support from the counter-blades (if I referring to the design by Dave Hakkens).
Attached below is the schematic of how cutting actually works in my opinion. Please give me your thoughts.
Hello to everyone!!! I’m new in this world of design. I read all the post but I have several questions about it:
1._Which is the first thing that I should calculate to make the shredder?, Is to know the horsepower just like thegreenengineers did it?
2._ What is the purpose to calculate the horsepower? , I mean, It is to know which motor I should use? , if that so, I’m a little confused when comes the moment to put the value of the RPM, because, how I will know the RPM if I’m still not supposed to know what engine I’m gonna use??
3._what’s the purpose of the second formula? (where thegreenengineers talk about the engines and the gearbox). It is known which motor I should use too? , if that so, why thegreenengineers start its example knowing the motor from the beginning? “Lets say we have a 1.5KW(~2 HP) Spindle motor for engraver/router” That doesn’t have sense for me…
4._ “The equation to find the horsepower is (RPM * T) / 5252”. From where this formula comes from? Why have that constant? I’m making a project and I need to justify that formula
5._ Why I have to use the 80% of the yield stress? where this rule comes from? Does have some type of name to google it?
6._ thegreenengineers says: “The area is found by the cross-section area of the material of cut that the shredder blade contacts”
How I can calculate this area in my specific case? thegreenengineers use a hypothetic triangle, but what area I should use in my case? It is difficult to me think in the cross-section area of a bottle, glass or another non-perfect object. Maybe I should use a hypothetic cross-section area like thegreenengineers?, if that so, I don’t know what figure I should choose and what measures to put on it.
7._ thegreenengineers says: “The shredder apply a perpendicular force to the material”, two perpendicular forces that are equal and opposite in direction
Does someone have an image of this? I don’t really see where these forces are when the material is cut. I supposed I need a little more of imagination…
I’m sorry if my english is not the best…
If you have more than one blade (tooth) cutting at the same time then you will need to multiply the torque needed by the number of teeth cutting simultaneously, eg 2 teeth = 24Nm, 4 teeth =48Nm etc. Note that you can have multiple blades but the teeth don’t have to cut at the same time if you stagger the rotation of the blades on the shaft. It’s probably also good to have a safety factor in case 400N is not the maximum force needed, so multiply the total torque needed by 1.5-2x to give some margin.
Hi andyn, Good day to you,
If involving at least two blades to cut the bone, should I use more power for the motor or less power? What do you mean by multiply the figures accordingly? It is hard to use low power motor and high torque like this 12Nm. Should I use gearbox? If so, I just take the rpm of the lower power motor divide by the 40rpm to get the gearbox ratio? Thank you.
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