Calculations Required for Shredder
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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.
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.
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.
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…
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.
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.
“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?
Hey guys I have been in and out of the forum. I am going to try to be around more often as I am trying to move into this space and away from my day job hopefully soon. So in shear mode of failure the thickness of blades makes no difference in the determination of the cut area. The thickness of the blades determines if the blades will survive the stress they are putting on the plastic without exploding. If the blades are to thin and you try to put to much force on them by trying to cut to big of a piece of plastic the blades will rip off. That is a separate issue when it comes to these calculations this is assuming that you are making the blades adequately thick and out of steel thus that regardless of the force applied they will not fail. That is the assumption you are making you will have to check this later on.
The area you are checking and thus the cutting area that you are looking for is the cross section of the plastic that you are cutting. not the cross section of the blade but the cross section of the plastic. Why I say that it is a triangular cross section as this is the worst case scenario for the shredder. This is the maximum size of plastic that you could possibly put in there. so lets say you produce a design of the shredder and you do not know what size material will be thrown in there you can design for the biggest possible piece that can fit into the shredder. A triangle is not necessarily the biggest geometry that will fit inside the shredder the shape will be way more complex but the area of a triangle is easy to calculate by hand.
To figure out what motor to get I would go from the end to then size the motor. what is the biggest cross section of the plastic that can possibly be cut inside the shredder. What is the worst case stiffest plastic that that could be. Then solve forward to figure out the motor and the gear box.
However be aware that sometimes the specifications of motors are of 2 different clasifications either max power they are capable of which is 60% efficiency and 50% maximum rpm or they are classified by maximum efficiency or rated power which is 80% rpm and 80-90% effieciency so from there you will have 2 motors to look for one that could have ____ rating for max power or a ____ rating if it uses rated power.
I will probably make a video about this later.
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.
Ok so i had one question asked. How do you calculate the area for a bottle. i have a image that i made for that there is a average amount for calculation. basically you want to take this length that is in black which is where the blades pinch and the furthest down a water bottle can go (i suggest crushing bottles before putting them into shredder). This is the same for single axel shredder. just measure from there the blade starts cutting to the furthest down the material can go. Then multiply this length by 2 thicknesses of the water bottle skin. As you have 2 sides because it is basically a flatten rectangle. then procceed with the calculations at the beginning of the post.
please am trying to build this hand made shredder, but am finding it difficult with the calculations. please can you help me with a sample.
am using the cutter below in the image
can you please write posts in a way that everybody understands? Just as you would do writeup-s or a summary (see also how to ‘merge images’)
thank you, nice work; hand shredders are very welcome
ok i will perform calculations tomorrow morning. I go through forms once a week. so sorry about the wait.
Well for the hell of it lets say you have a 28mm cross section circular bar of HDPE that fits exactly in that scoop of the blade. That fits in that hook of that blade. First determine that area .014m * .014m * 3.14 = .00061544 m^3. ok and again lets say we are shredding HDPE. Lets say the 3500 PSI for HDPE Shear Stress just like i said in first post. Metric equivalent of pressure is pascal 24,132,000 Pa or N/m^3. lets use worst case scenario which is the line between blade tip and center or rotation which is the shaft in this case 60mm or .060m. So with a Cross section of .00061544 m^3 and 24,132,000 Newtons of Force Per M^3 of area we need to cut (shear) HDPE. so we must multiply the 2 together to get Newtons force so .0061544 * 24,132,000 = around 15,000 Newtons ok now for torque the units are Newton-meters so now we need to multiply our newtons by distance to blade tip so 15,000*.06m = 900 N-m. Ok now we need to pick rpm for our motor this is where you can come up with your own option here. My personal choice and the speed i like is 30 RPM reason being is 2 sided blade means at 30 RPM you have 1 cut per second that is plenty fast. Equation for power is <b>Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488. the 9.5488 is some conversion factor</b>. It probably has to do to do with conversion of unit and turning minutes into seconds and such. so (900* 30) / 9.5488 = 2.83kW or 3.79 HP. that is massive.
I think that is because you have a cutter tip that is to long and a blade that is to big in diameter. thus increasing the torque required. If you gear down to 15 rpm your HP requirement goes down to 1.9HP or 1.41kW. Again alot of the calculations here are worst case. But HDPE is not the hardest plastic in the world.
naw thats fine ppboys i look at cad drawings all day at work. It is good enough to make a rough estimate but usually you need both “blades” in the case of this design it uses stationary blades. It would be nice to see where those blades are relative to the rotating blades in the assembly.
I have a question regarding the initial calculations. I think we need to divide with 33000 instead of 550 because hp is 33000 lbsft/min and 550 lbsft/sec? we multiplied with rpm which is in minutes.
I’m not quite sure about this, can you clear this confusion. Cheers!
hello, I am designing a plastic shredder for my project.
my question is
is the size of the shaft has no relation in solving the torque or the rpm?
how will you design the shaft?
is the weight of the blade affects the design?
how to determine the dimensions of the blade? or is it a designer’s choice?
Hi, I’m from Ecuador and i’m working in a design Shredder, i used this blade but i want to know your opinions or suggest or if i’m wrong, please give me your opinios.
josette: the design does come into account. Once you find out what force it requires you need to design the shaft such that it does not fail in torsion. Generally in a shredder if not designed properly this is the first place it will fail if you have a motor strong enough.
sail: this: Solve for Torque = (HP * 5252) / RPM = 10.5 Ft-lbs (pounds) is correct equation.
Romero sorry it took so long.
With cross section of 619.28 mm^2 of cut which is circle of blade there. as max cutting of material
Using HDPE as material with 24MPA Ultimate Yeild Stress
Force to cut required is 3,340.5 Pounds of Force
thus torque with your blade radius length of around 1.1 you are left with 500 ft lbs of force. so in this case if you are 160 lbs you would need to put a handle on it that is 3 feet long. But the most important thing about this is this is per contact tooth to material. If you have 6 teeth engaging the material at once then that goes up by x amount of time per extra tooth engagement so if 6 teeth are engaging you have 3000 ft lbs required etc. from my studying of my own shredder cutting by hand just cant really be done unless you have massive amount of mechanical advantage. Pulley system. giant breaker bar (10 ft long). Or hydraulic or something.
However, with these same calculations at about 30 rpm of the motor you need approximately .45 hp or 340Watts of engine power per tooth engagement. so if you had a 3 hp motor geared down to 30 rpm then you can have almost 7 teeth engaged at once no problem. Now will you mechanical components survive 3hp that is probably not going to happen.
This calculation is again worst case scenario if you tried to cut a 1″ diameter piece of HDPE in your machine. So these calculations would be fore the worst possible scenario you machine will encounter.
i think that the number of blades is to high. Reason being if you have 12 blades on a shaft lets say the chances that almost all the blades are trying to cut the material is very high. Thus the force required is multiplied by the number of blades that are engaging. I see that this might be preliminary design as you have yet to design some way to hold the blades in the center and have some way to forcibly turn it. But other than that your blade shape should work. Usually they are pitched a little forward to get more of a shearing action. As that means the entire area of the material is not trying to be cut at once. This drops down the force a little bit. Other than that it looks pretty good.
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