How to calculate the Max torque on a Mower shaft when the blade hits a rock. 2

[jil1969]

Can somebody guide me on how to calculate the max torque the shaft of the shaft in the image below?
It's an brushless outside rotor motor, the max torque the motor gives is 25 N.m
I would like to calculate the max torque in the shaft if the blade a 5,000 RPM hits a rock and stops.

Thanks in advance.

 

[3DDave]

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.

 

[geesaman.d]

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.

 

[GregLocock]

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

http://eng-tips.com/market.cfm?

 

[3DDave]

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.

 

[GregLocock]

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

http://eng-tips.com/market.cfm?

 

[rb1957]

"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 ?

 

[BrianPetersen]

One word. Inertia.

 

[LittleInch]

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.

 

[Tmoose]

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

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

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

 

[rb1957]

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 ?

 

[JStephen]

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.

 

[SnTMan]

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


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

 

[rb1957]

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 ?

 

[saplanti]

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.

 

[LittleInch]

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.

 

[Rogue909]

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.

 

[Tmoose]

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/e43c6a54-0642-4d25-8c67-3d608b4da280/09983152.jpg

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.

 

[hydtools]

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

 

[hydtools]

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

 

[rb1957]

sad, isn't it ?

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