The resisting torque should be same or slightly greater than 25Nm to maintain shaft in equilibrium of stopping condition.

A conservative approach would be to assume that all the rotational kinetic energy of the rotor is used to "wind up" the shaft, calculate the amount of torque required to deliver that degree of "wind up" and then add 25 Nm (because the motor will still be providing torque) to that.

A.

Rocks are easy. I've trashed two mower engines running over lengths of heavy gage wire.

Given the usual assumptions, you can't accurately estimate the peak torque. zeusfaber's approach is an upper bound estimate (a good thing). A wise person might put a slip joint between the rotor and the shaft such as a pair of nylon discs under compression.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Thanks all for your responses. I'm trying to get some guidance on how to approach this calculation to increase my knowledge.

I'm new here. I think I'm not violating any rule. Do you think so Greg?

I seem to recall they used to put a shear key between the blade mount and crankshaft. Before they decided crankshafts were *cheaper* than shear keys :)

The problem with sloppy work is that the supply FAR EXCEEDS the demand

I think the shear key is a bit of internet myth. For a key to operate close enough to its shear strength to be protective of the shaft one would introduce far too many unnecessary failures. Also consider that the key doesn't shear cleanly and still damages the shaft as it fails.

A shear key does not have to fail completely. A "soft" key that allows 10 degrees of rotation before a complete stop of the crank compared to 1 degree, will reduce the impact force on the crank and other parts by a factor of ten.

However, the key still needs to be replaced, and if you have to pay someone to fix the mower, the economics often leads to replacing the mower. After all, there may be other damage and wear.

Well, I dunno about internet myth, but I seem to remember my (very much pre Net) B&S mower engines having an aluminum square key.

Neither here nor there these days :)

The problem with sloppy work is that the supply FAR EXCEEDS the demand

The man question here is how would I know the max torque reached by the spin of the blade, the data I have is that the motor gives 25 N.m at stall, but I would like to know how much torque the moment of inertia adds to the shaft.
Great comments by the way.

To be clear there are keys on both the cutting blade hub and on the flywheel. The flywheel key was aluminum, and if it deformed, the spark timing went off.

Quote (jil1969)

I think I'm not violating any rule.

Greg's not having a go at you - that's just a piece of helpful advice that appears in the signature block at the end of all his posts.

A.

Initial momentum divided by 0 time for deceleration = "E" as an old school calculator would say. Or "division by zero error". Or "infinity", as you prefer.

The (more-or-less steady state) torque that the motor develops is irrelevant relative to what happens in the impact.

In reality, the time for deceleration won't be zero but will be extremely small. The blade will bend some, the shaft will twist some, the rock will move a bit, etc. But, predicting what this is going to be, is not possible with the information supplied.

One would hope that the means by which the blade is attached, has intentional design features to accommodate this. The blade certainly has some flexibility to it. Maybe there is a shear-pin that's designed to snap. Maybe the blade is secured to a drive flange with a couple of bolts that are intentional weak spots, designed to break off in this situation. Maybe the drive flange isn't one-piece with the motor shaft and there's something that amounts to a friction-clutch or overload-clutch in there.

Impact situations are not easy to analyse, and the outcome may involve something changing shape. Your (presumably modern) car is designed to protect its occupants if it is driven at 50 km/h straight into a solid immovable concrete wall. But it's designed to change shape in doing so, thus absorbing the energy of the impact.

I was just making sure, Zeus. thanks for clarifying.

The basic equation is torque = inertia * angular acceleration and angular acceleration is change in velocity / time. So hypothetically, if the blade stops instantaneously, then the torque is essentially infinite. So, beyond that, you have to make some assumption about the stopping time to get anything useful, but in any case, your motor's torque is essentially irrelevant, since you've stipulated that the blade stops, which means that any practical torque from the motor is overcome by the fact that the blade stops.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
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Quote (zeusfarber)

A conservative approach would be to assume that all the rotational kinetic energy of the rotor is used to "wind up" the shaft,

My emphasis.

"The rotor" is part of the motor. Not the entire rotating mass.

A 17 mm/.670" Ø shaft? with the blade hung out in space about 1.7 inches in space?
A 40 inch blade spinning 5000 rpm? 300 feet per minute tip speed?
How is the external rotor retained on the .67" shaft?
How is the blade affixed to the shaft?
What is the shaft made of?
A gasoline 17 HP engine ( much lower rpm and proportionally higher torque ) would have a 1.0 inch or larger output shaft.

https://www.youtube.com/watch?v=B7EhCsUE3-s

I see sheared keys, bent shafts, pierced, torn mower decks, rotors set adrift, maybe even broken shafts in your future.

I'd try to pick my battle.

Tmoose,

It's actually an stepped shaft
17mm-22mm-17mm
The rotor is press-fitted to the shaft.
The blade is yoked to the shaft and a nut is installed.
Carbon Steel shaft.

"yoked to the shaft"??

What does that mean?

Anyway that looks like your weak point and if this blade does stop dead then it will almost certainly snap there. The consequences for a 1m diam blade spinning at 5000 rpm could be very bad indeed.

This is not a calculatable thing as there are too many unknowns and possible options for what breaks first. IMHO.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

One thing that will happen is that the rotational inertia of the blade will have to go somewhere if one end of the blade is suddenly stopped, likely bending that thin motor shaft.

Reality is that you have a highly dynamic system and static stress analysis will only allow you relative estimates of durability. The cheapest and best solution to get a rock-solid design (oh my) is to build a prototype or two or base yours off of a known design.

If you need to convert this to a static torque limit, perhaps an experiment will give good data. One idea is to torque the bolt that holds the blade to varying degree and measure the torque required to rotate the blade relative to the mower shaft. Start with a static torque-slip measurement, match mark the blade to the shaft, whack a rock, repeat static torque-slip measurement. Increase/decrease torque until you are in the vicinity of where the blade does/does not slip. Might be wise to add a clutch material under the washers to improve the consistency of the static friction.

In any case I have struck many rocks with a 60" zero turn mower and it's safe to say you can't stop a mower blade by running over a rock - you can only scalp it. The blade will not stop rotating. It will repeatedly glance off of it because the mower isn't moving fast enough to get the rock in between rotations of the blade.

3DDave's point renders any discussion of torsion moot in my opinion, if the blade is locked to the shaft.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Even if it isn't locked - the angular momentum will change its rotation center from the shaft axis to the rock location - this shift will be resisted by a direct side load on the shaft even if there was a ball bearing mount for the blade on the end of the shaft.

Yup. My ZT mower runs 500mm blades and the shafts are 17mm . No sign of them bending despite being used to scalp rocks and trees for 20 years (bearings have exploded). So if we double the blade diameter what size shaft is needed?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

"Can somebody guide me on how to calculate the max torque the shaft of the shaft in the image below?"

if the blade is driven with a torque of 25 Nm, and it stops (when it hits the immovable rock) then the torque is 25 Nm, no ?

Does the rock add torque to the blade (and spin it backwards ?

Does the blade hit the rock and continue ? no, as per problem statement.

If the blade stops the the torque applied is 25 Nm. Even if the mower was running at less than 25 Nm, 'cause the rock will demand the maximum output from the motor (as it tries to rotate through the rock).

another day in paradise, or is paradise one day closer ?

One word. Inertia.

If an aeroplane isn't moving what is the maximum force on the wheels if they are locked in position [Engine thrust]
If the aeroplane has just landed and the wheels hit a solid low wall, what is the force on the wheels? [A lot more than engine thrust]

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

https://www.mfgsupply.com/32-1709.html

https://www.stens.com/751-032-crankshaft-straighte...

https://www.youtube.com/watch?v=w2Ro9GmEyOE

yes, but the problem is bounded by the motor driving the blade. There's no need to talk about airplanes and such.

As I write this I realise my error. The question was "oddly" worded. The torque the rock applies to the blade can be anything this side of infinite. Applied for an infinitesimal time it would do a finite amount of work (stopping the blade). The force the rock applies to the blade (and so it's torque) can be very high as in any impact problem ... impact force is determined by the time interval that the impact lasts.

another day in paradise, or is paradise one day closer ?

Long long ago, I was mowing with our Craftsman gas lawnmower with aluminum deck.
I hit a stump about 6" diameter that was cut ALMOST low enough to mow over.
The impact broke the bolt mounts that attached the motor to the deck, broke the motor side in a couple of places, broke the deck side in a couple of other places. The result was that the motor did a very rapid 180 degree spin on the deck and kept running. I quickly killed it.
The upside of this was my mom and dad concluded there was no salvaging that mower, so they let me disassemble the engine just to see how it worked inside. And that was fairly educational for a kid.

On another prior occasion with that same mower, I ran over a bucket handle and shot it back towards my ankle, left a little bloodied spot on my ankle.
My mom took me down to the doctor's office, they X-rayed it, and found there was about a 2" long piece of that bucket handle completely in my ankle. So the family pediatrician got to fish the shrapnel out of my body, which was a little bit out of his normal line of work. That also chipped a bone, but didn't break it all the way through, but it got me out of PE for 6 weeks. I had that little piece of wire for many years, but haven't seen in several years now, so it may be gone. And at some point after that, they invented those little flexible flaps at the back of a mower to avoid the ol' shrapnel-in-the-ankle trick.

I see people mowing in sandals and flip-flops. Ignorance, bliss..

The problem with sloppy work is that the supply FAR EXCEEDS the demand

nice though the side bar has been, returning to the OP's question ...
torque in the shaft is due to the force impacting the rock,
the force impacting the rock depends on how long the blade takes to come to rest. If it comes instantly to rest, then the force is infinite.

now what is the maximum loading the shaft can carry ? well, that is quite a complicated question.
Do we take the impact loads and treat them as static ?
Are we trying to ask what is the maximum pure torque that can be reacted by the shaft ? that is much easier to answer.
But in the case at hand, the rock imposes a shear force on the shaft (reacting the load applied by the rock) and this complicates the loading in the shaft.
How much shear force can the blade carry to the shaft ??

another day in paradise, or is paradise one day closer ?

jil1969,

Is this your design, or someone is directing you to verify this design¿
I guess you need to see the grass cutter (lawn mover) design for the same purpose. You need to reduce the inertia first ( this requires the the rotating circular plate diameter as low as possible, you call this plate), and attach pinned blades to this circular plate. This way, if one of the blade hits a rock the shaft will not be effected, perhaps that blade only will be damaged. I trust this will simplify you design.

I realised that geesaman.d mentioned the same.

It's all about the attachment details of the blade to the shaft. A mysterious "Yoke" doesn't tell us much.

If you're concerned about breaking the shaft or other unintended consequences then I would make this some sort of shear pin arrangement. As from all the responses above sudden and complete stop can create huge forces / torsion magnitudes above the torque from the motor.

I can recall an old cylinder mower we had having one of these for the same reason if you accidentally tried to mow a stone or a large branch you didn't destroy the whole machine. Had to replace it once which is when I discovered it in two pieces.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

What is the purpose of this theoretical torque? I would think if you're trying to justify motor design or ensure the motor doesn't fly off it's housing and turn in to a bladed top by a strike then turning down a small section of the shaft between the blade and motor would achieve that. (or better; thickening the shaft in the motor and using the turned down portion where the blade attaches as a bearing set point) Then calculate the necessary torque to break the turned down section. That is an arguable 'this side of infinity' torque value.

My mower blade is held on by a bolt. If I were developing a test to determine torque from the blade then I would take a spare motor shaft, put it in a vice, attach a mock blade and add weight till the bolt gave. In an unmovable object meets unstoppable force scenario, the material values can help determine limits.

I think mower blades have anti-rotation features in addition to axial bolting.

Blades on mowing decks with individual belt driven spindles often have curious round toothed internal splines to match the spindle shaft.
http://mobileimages.lowes.com/productimages/e43c6a...

Blades that attach to the engine shaft often attach to a keyed hub on the shaft. the hub often has either two drive pins at a larger radius, or has the edges folded up to cradle the sides of the blade.

If you stop the blade, Newton says the rotor will want to continue rotating unless acted upon by some force, torque. The torque necessary to stop the rotor depends on the rotating inertia of the rotor, not the blade. You could have the same effect without the blade by suddenly stopping the shaft rotation with some external brake suddenly applied at the blade end of the shaft. The force on the rock and blade will be that of stopping both blade and rotor inertias.

Ted

From a motor manufacturer:

What is inertia and how is it calculated for a rotary motor?
Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force. Celera Motion standard rotors look much like a cylinder. The inertia formula for a cylinder = density*1.57*Length*(Ro^4-Ri^4), where density is the material density, length is the length of the rotor cylinder, Ro is the outer radius of the rotor and Ri is the inner radius. Since our rotors are combined of magnet material and steel, the density of each could be averaged, or the two inertias can be calculated separately, and added.

Ted

sad, isn't it ?

another day in paradise, or is paradise one day closer ?

There is the torque and also the bending moment to consider. The force imparted on the blade by hitting the rock will cause an equal and opposite force at the shaft. This will be acting some distance below the bearing of the motor causing a bending moment in the shaft. Having bent many lawn mower engine shafts, it is a thing.

rebuilding many small engines , hundreds of lawn mowers, typically one does not want to hit a big rock, bad things happens. but most of the time, on gas motors the fly wheel key will shear and thats the just of it. it needs to be pulled and repaired. these old lawn mowers were built very tough and could take years of abuse. on and electric motor the blade hitting a rock would just stall the motor.
so calculate the torque produce by the motor at the shaft, then the inertia of the blade. and the torque it will produce. but to me is all mute due to the motor stalling.

I think the possible solution for this case is a hub that slides over the shaft od the motor with an aluminum key. thus calculate the force to shear the key on impact.
much easier to remove and replace. held in pace with a single fine pitch high strength bolt. shaft gets bent it's scrap, or if the mounting is damaged. blades can easily be replaced.

It's the rotor rotating inertia that drives the torque through the shaft to the stopped blade. (WR^2 * n)/(308*t) W rotor weight in pounds R rotor radius in feet n change in rpm t change in seconds

Ted