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Impact load on rotating shaft
- Thread starter ban7rjg
- Start date
Hi
Firstly you can calculate the stress in the normal way but take each step independently.
That said the only torsion stress that will occur is between the windings and the point at which you apply this force, how many step diameters do you have between motor windings and where your applying the halting force?
Firstly you can calculate the stress in the normal way but take each step independently.
That said the only torsion stress that will occur is between the windings and the point at which you apply this force, how many step diameters do you have between motor windings and where your applying the halting force?
Ban7rjg:
The motor will apply some max. torque when it is locked up (stopped). This torque is applied to the shaft btwn. the motor and the reaction point, the point which applies the locking force. Of course, the smallest diameter will see the highest torsional stress and the details of the various steps in diameter need some special attention as to their stress concentration affects. Take a look at some of your Strength of Materials textbooks and some Machine Design textbooks for the details. I am continuously distressed by the fact that no one seems to have a textbook to look at any longer. They used to have some really well vetted and valuable info. for helping us understand engineering concepts and design problems. And, the reading and studying process seemed to make the concepts stick with the reader much better than the spoon feeding of the internet.
The motor will apply some max. torque when it is locked up (stopped). This torque is applied to the shaft btwn. the motor and the reaction point, the point which applies the locking force. Of course, the smallest diameter will see the highest torsional stress and the details of the various steps in diameter need some special attention as to their stress concentration affects. Take a look at some of your Strength of Materials textbooks and some Machine Design textbooks for the details. I am continuously distressed by the fact that no one seems to have a textbook to look at any longer. They used to have some really well vetted and valuable info. for helping us understand engineering concepts and design problems. And, the reading and studying process seemed to make the concepts stick with the reader much better than the spoon feeding of the internet.
With a sudden stop at the pinion the inertia of the rotating motor parts will be of most concern. If the braking force is transferred thru the gear mesh, it is quite possible the pinion gear teeth will shear before the shaft fails. In this case the torsional stress in the shaft will be limited by the shear capability of the pinion teeth.
tbuelna,
The system will fail at the weakest point, which could be weird. I worked on a system that had a couple of big output gears all driven by a 500:1 micro-gearmotor. It was impossible to rotate the gearmotor by rotating the shaft, but it was fairly easy to do so by grabbing the assembled system. My calculations showed that the weak link of the system was the very tiny armature of the motor. Gear ratio squared and all that.
--
JHG
The system will fail at the weakest point, which could be weird. I worked on a system that had a couple of big output gears all driven by a 500:1 micro-gearmotor. It was impossible to rotate the gearmotor by rotating the shaft, but it was fairly easy to do so by grabbing the assembled system. My calculations showed that the weak link of the system was the very tiny armature of the motor. Gear ratio squared and all that.
--
JHG
drawoh-
I would agree, and just add that the system will fail at the weakest point between rotating motor masses and the point in the drive that is seized up. Which based on the description provided appears to be the pinion gear at the end of the motor shaft. The inertia of the rotating parts coming to an abrupt stop will create torsional and bending strains in the shaft and gear structures. The shaft connecting the motor to the pinion gear likely will not be as highly stressed as the 1.5 meshed pinion teeth reacting the same load.
I would agree, and just add that the system will fail at the weakest point between rotating motor masses and the point in the drive that is seized up. Which based on the description provided appears to be the pinion gear at the end of the motor shaft. The inertia of the rotating parts coming to an abrupt stop will create torsional and bending strains in the shaft and gear structures. The shaft connecting the motor to the pinion gear likely will not be as highly stressed as the 1.5 meshed pinion teeth reacting the same load.
- Thread starter
- #8
Thanks everyone for your feedback.
The reality (as observed practically) is that the shaft shears at the smallest diameter.
In all cases of failure, the pinion has been in tact. only the shaft shears off near a neck region.
I tried to use the formula : STRAIN ENERGY = KINETIC ENERGY ; to find the shear stress.
Is this approach right?
If yes, then how do i incorporate all sections of the shaft (stepped) into the problem?
The reality (as observed practically) is that the shaft shears at the smallest diameter.
In all cases of failure, the pinion has been in tact. only the shaft shears off near a neck region.
I tried to use the formula : STRAIN ENERGY = KINETIC ENERGY ; to find the shear stress.
Is this approach right?
If yes, then how do i incorporate all sections of the shaft (stepped) into the problem?
Hi
You calculate the stress in each diameter based on the formula below for each section
Shear stress= 16*T/pi * d^3.
You just take each step and apply the formula as though it were a single shaft as stated in my first post.
If you want the deflection of the shaft that's a slightly different story.
You calculate the stress in each diameter based on the formula below for each section
Shear stress= 16*T/pi * d^3.
You just take each step and apply the formula as though it were a single shaft as stated in my first post.
If you want the deflection of the shaft that's a slightly different story.
- Thread starter
- #11
Hi
I am not sure you will get the impact time from that formula, because when the motor suddenly jams the motor initially will try to raise the torque to some value in the order of 2 to 3 times the running torque of the motor, it is this time interval where the torque increases to the above 2 to 3 times the running torque before stopping you require for the impact calculation.
The torque your proposing in the formula is the running torque I assume and not the overload torque and I*w is the is the inertia of the components and w is the angular velocity at normal running.
I am not sure you will get the impact time from that formula, because when the motor suddenly jams the motor initially will try to raise the torque to some value in the order of 2 to 3 times the running torque of the motor, it is this time interval where the torque increases to the above 2 to 3 times the running torque before stopping you require for the impact calculation.
The torque your proposing in the formula is the running torque I assume and not the overload torque and I*w is the is the inertia of the components and w is the angular velocity at normal running.
The best way to protect all the rotating components is to provide a weak point. A friction clutch if you want the system to endure multiple seizures without maintenance. A shear-pin or similar if maintenance is inevitable after each seizure anyway.
je suis charlie
je suis charlie
- Thread starter
- #14
- Thread starter
- #15
When I did the calculation as above, the result is almost matching practical data.....
Desertfox -
Δω = ω2 - ω1; here ω2 = 0, since motor comes to halt upon sudden brake
ω1 = angular velocity just before braking (there will be 2 different speeds - one of the pinion&drive train, next of the armature, since this speed is multiplied by an epicyclic gear train)
In accordance, there will be 2 moments of Inertia.
Calculation proceeds for this.
Desertfox -
Δω = ω2 - ω1; here ω2 = 0, since motor comes to halt upon sudden brake
ω1 = angular velocity just before braking (there will be 2 different speeds - one of the pinion&drive train, next of the armature, since this speed is multiplied by an epicyclic gear train)
In accordance, there will be 2 moments of Inertia.
Calculation proceeds for this.
Hi
When I read your first post I thought you were considering some kind of jam while the machinery was operating, its taken you five posts to inform us that its actually a braking problem, a fact that you probably knew all along, had you revealed this fact in your first post it would have a difference to the answers you received.
When I read your first post I thought you were considering some kind of jam while the machinery was operating, its taken you five posts to inform us that its actually a braking problem, a fact that you probably knew all along, had you revealed this fact in your first post it would have a difference to the answers you received.
- Thread starter
- #17
Hello
I mentioned 'In case the pinion is suddenly locked'. So in fact it is a braking phenomenon - not a problem.
Reading the posts by other contributors (in particular drawoh & tbuelna), I felt all understood my question.
Nonetheless, sorry if there was any confusion.
Thanks!
I mentioned 'In case the pinion is suddenly locked'. So in fact it is a braking phenomenon - not a problem.
Reading the posts by other contributors (in particular drawoh & tbuelna), I felt all understood my question.
Nonetheless, sorry if there was any confusion.
Thanks!
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