Calculation and formula of horse power of shredder
I would ask about the calculation and formula of horse power of shredder machine, does anyone know the formula of it? Please help me, thank you.
I’m curious how you came to this conclusion. I’ve seen some of the other lines and they say you can use as low as 1/2 hp. The one Dave uses is a 2 hp motor.
I think the hardest problem is finding a motor/reducer combination.
Is a 1hp usefull?
Second, It’s not only the HP or kW that matters. The interesting unit for shredding is torque which is given in Nm (Newtonmeters). It is a combination of power and speed.
If you use a 3 HP motor (2.2kW) with a reducer 1:25, you have the same torque as with a 1.5 HP Motor (1.1kW) with a reducer 1:50. You are turning only half as fast, however you have the same torque, the same “strength”. So theoretically, a 100W motor would be sufficient with the suiting reducer… So yes, as @pantheonengineeringdesigns said, the hard part is finding the right combination, giving you enough torque, because a 3HP motor with a 1:5 reducer won’t shred good.
I hope that helps 🙂
So, the questions should be, how much torque do you need?
I acctually got that question from the guys in R&D on Optinova when I asked for a motor…
Yes @ashrak ! Mine has got ~280 Nm and that’s enough for most things. Just some really thick plates (4mm+) or bricks out of solid PP were able to stop my shredder But that happens faster than you would imagine! 3 or 4 smaller pieces of a flowerpot flat ontop of eachother and it gets stuck 😀
@flo-2, @ashrak, mine has 254 Nm and the weakest point is the coupling with the motor axle. This linkage consists of two M8 bolts that go through the shredder axle.
These bolts got shear cut twice while shredding a neck of a 5L water canister made of PET😞
BUT it is good enough for shredding all the plastic we have.
is it alawys with its motor or i can buy it only ?
@adellokman in most cases its 2 individual parts, however they have to fit together, so not every reducer fits on every motor. If you take a look at my shredder and scroll down a bit, you can clearly see the two parts (post from the 20th April) where the left cubic part is the reducer and the right cylinder part is the motor. In my case they are made to fit together, which is what you want.
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 of (2.5″ x .4)/2″ (the worse case from design of my shredder) so .5″^2. Radial distance to tip is 3.5″ Diameter/2 (my design of shredder) D = 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. my area is .5″^2. I solve for Force so Force = 3500 * .5″^2. That equals 1,750 Lbs of force. Now solve for how much torque i need from my motor. Torque = Force * Distance my radial distance is 1.75″ so Torque = 1750 * 1.75 = 3063 inch-pounds of torque. lets say i have a motor that rotates at 5 rpm. For horsepower ours solves for 3063 is torque so (3063 * 5) /5252 = 2.916 hp.
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 2.916 * 2 = 5.8 HP.
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. so we need 3063 inch-lb of torque. i weigh 160 lbs. so take 3063 and divide by 160 lbs to get a 19.14 inch long bar with 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 large 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. Next post is about gearboxes and calculations for that and sizing motor with gearboxes. anyways back to class talk to you guys later. Enjoy.
what 3063 foot-lbs in grown ups (ie metric) measurements?
4152.87 Newton Meters
4152 Nm is a HUGE force!! (actually a moment, not a force)
I think you meant to convert from inch-lbs not foot-lbs. That’s still 345Nm which sounds like an awful lot. In Dave’s design with a 2kW motor and 70 rpm the theoretical max. torque is 272Nm, probably considerably less in real life allowing for losses. I think ~350Nm would be enough to shear the 20mm shaft of the shredder if the blades were to lock.
you are right i looked back through and it is in inch pounds. its just that the last guy that replied asked for 3063 FOOT-LB to N M so i didnt even look back at my solution so yes it is in inch pounds if you go back to solution.
this calculation is from another post on another thread that i did for size of cross section of shaft so it does not fail in torsion.
anyways equation for the size of the cross section is this.
Also what i am talking about is the thinner cylindrical cross section that connects to the motor not the square section. The square section is strong enough.
tau = (T*(radius))/J
J for circle cross section = ((pi)*(D^4))/32
So solving for D you get:
((T*(16))/(tau*pie))^(1/3) = D (diameter)
T = 535 N.m (3HP at 40RPM)
tau = 215 MPa * .80 = 172MPA = 172,000,000 PA
D = .025 meters
= .0828 feet
= 1 inch
so currently you are at .945 inches
so you need to increase by .055 inches
otherwise known as 55 thou
Hi Steven @thegreenengineers,
You’re obviously much more familiar with the math behind this than probably most of us. Whilst I always start with calculations when designing something, I don’t like to get too bogged down in them as I find the real world situation can differ considerably due to factors you can’t always account for. Also, if you make a small mistake in a formula or overlook something, you can be many orders of magnitude out without even realising it. So I find it useful to keep a mental ‘second opinion’ by referencing to what I know works and what doesn’t. Of course experience can be valuable here, but if you have a rough idea what works, then calculations can refine this, and your rough idea can highlight any glaring errors in the calculations.
Not to pick holes at all in your work above, maybe I haven’t fully understood, or maybe you’re not talking about the PP shredder, or maybe it’s a problem due to thinking in different units, but I was referring to the cylindrical part of the shredder shaft and said I thought this would fail at not much more than 300Nm (I haven’t calculated this, it’s just a ‘gut feeling’).
You seem to be working backwards from a figure of 535Nm, I’m not sure where this comes from? (3HP @40 rpm is approx 535Nm, but I’m not sure where these figures are from either?)
You’re using a value of 215MPa for the yield stress, which is appropriate for mild steel, but are you using 0.8 as a multiplier to get the shear stress? Isn’t it usual to use 0.57? Maybe it’s a safety factor thing?
0.945″ is 24mm, the round part of the shaft is only 20mm.
You seem to be saying that a 1″ shaft will stand 535Nm? But what torque will a 20mm shaft fail at?
Personally I wouldn’t put more than 200Nm through a shaft like this, allowing for a small safety margin. There is also a large stress riser caused by the sharp shoulder were the hexagonal (it is hexagonal, not square) part of the shaft meets the cylindrical part, this is where I would expect it to fail. It would be better to put a radius here, (there is space in the design to incorporate a 3.5mm radius).
Interested to hear your thoughts on the above,
Someone was asking me that they had a new design and they wanted to know what people thought about it and i asked if they had run any FEA on it. i then proceeded to run FEA on it and under the motor he was going to run with it the shaft was going to fail in torsion so i preceded to give him the equation to find out what size shaft to use. so the design from that calculation is not the precious plastic design.
Motor .5HP with gearbox 100 RPM useful ?
highlighting :- Motor 3 Phase and will modify it to 1 Phase
Depends on the cross section of cut when you find that out use the equations i posted.
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