When does elastic failure occur

[blubee]

Here's a ccx file applying axial rotation forces on this shaft.

I would like to understand at what point does the material elastically fail. Here's a link to the file with results: https://filetea.me/n3wScs9a4cGScGwRdN8SOlrHA

When I check on the results, then show the displacement, on my screen a factor of 30, the mesh is still aligned with the model, at 31 or above the mesh starts to deform.

I am using a standard material for testing but would like to be able to understand at what force does this material fail? What if I wanted to add a motor to this shaft in the real world, what would be the maximum torque that i could apply since the diameter of the shaft is 10mm and the total length is 250mm.

[thschrader]

Hello blubee,
according to your thread
https://forum.freecadweb.org/viewtopic.php?f=3&t=23992

In the thread above we did a static calculation of your model.
But I think its a dynamic problem. If the rotation starts and goes up to high rpm,
the shaft/disk begins to vibrate (resonance). Second problem is, that you must
do a superposition of the stresses calculated by static analysis and from the dynamic
effects. In freecad you can do a modal-analysis (=frequency-analysis) of the shaft.
Sorry for my english (native german speaker) and for the technical slang...
regards thomas.
further info see here:
https://en.wikipedia.org/wiki/Critical_speed
http://www.ewp.rpi.edu/hartford/~ernest ... Shafts.pdf
http://www.cncroutersource.com/critical ... lator.html
http://sandv.com/downloads/0505swan.pdf

EDIT 22.08.2017:
I calculated the first two critical eigenfrequencies at approx 1100 Hz and 3400 Hz.
That means:
The first vibration will occur when the shaft/disk rotates (theoretically) with 1100Hz x 60 sec = 66000 rpm.
This value is so high, no engine would reach it.

1100_Hz.JPG (57.49 KiB) Viewed 2142 times

3400_Hz.JPG (73.27 KiB) Viewed 2142 times

[blubee]

Hello blubee,
according to your thread
https://forum.freecadweb.org/viewtopic.php?f=3&t=23992

In the thread above we did a static calculation of your model.
But I think its a dynamic problem. If the rotation starts and goes up to high rpm,
the shaft/disk begins to vibrate (resonance). Second problem is, that you must
do a superposition of the stresses calculated by static analysis and from the dynamic
effects. In freecad you can do a modal-analysis (=frequency-analysis) of the shaft.
Sorry for my english (native german speaker) and for the technical slang...
regards thomas.
further info see here:
https://en.wikipedia.org/wiki/Critical_speed
http://www.ewp.rpi.edu/hartford/~ernest ... Shafts.pdf
http://www.cncroutersource.com/critical ... lator.html
http://sandv.com/downloads/0505swan.pdf

EDIT 22.08.2017:
I calculated the first two critical eigenfrequencies at approx 1100 Hz and 3400 Hz.
That means:
The first vibration will occur when the shaft/disk rotates (theoretically) with 1100Hz x 60 sec = 66000 rpm.
This value is so high, no engine would reach it.
1100_Hz.JPG
3400_Hz.JPG

@thschrader
Thank you for the great explanation as well as all the links and technical speak. It helps me when searching know what terms to look up.

This problem isn't about high RPM but very low RPM and would the shaft or the disk fail and how to calculate those stresses.

With a 15hp motor and 10RPM that would give about 94537ft-lb of torque or 10680 Nm.

What would be the process to test how much force would it take to cause any part of the system to fail?

For example the force starts at 1 P and continue until either the shaft fails or some part of the disk fail?

[EDIT]
static test file: https://ufile.io/fjthd
frequency test file: https://ufile.io/dmufy

I did another test of a simple cylinder with similar constraints. The static results look just like before but the frequency results are a bit confusing to me.
There are 10 results returned from the frequency simulation.

Under the results of any of the displacement, when you change the factor the mesh deforms.

You said:
I calculated the first two critical eigenfrequencies at approx 1100 Hz and 3400 Hz.
That means:
The first vibration will occur when the shaft/disk rotates (theoretically) with 1100Hz x 60 sec = 66000 rpm.
This value is so high, no engine would reach it.

so you calculated 2 instead of 10 and in the calculix settings you set the Eigenmodes count, low and high limit? Could you walk me through that process?

[thschrader]

Hello blubee,
I had a quick look at your files.
Some hints:
I was interested in the first 2 eigenfrequencies as shown above, so I have set
number of modes in ccx to 6. In modal-analysis you must double the number of
eigenfrequencies you are interested in. For example, if you want to get the first
6 eigenfrequencies, you must set number of modes to 12 (minimum).

Your given torque of 10860 Nm cant be true for a 15 hp engine. Even a diesel-engine
produce much lower (Audi A4 TDI 2.0 with 150 hp: 350 Nm torque)

This load-case can not produce torque in the shaft. Please the
the post from UR_:
https://forum.freecadweb.org/viewtopic.php?f=18&t=23778

no_torque1.PNG (35.89 KiB) Viewed 2109 times

There is a much easier way to get the first eigenfrequency:

first_eigefrequency.PNG (5.47 KiB) Viewed 2109 times

Can you give some more info about the project?

Please upload your files without results and a cleared mesh. Reduces file-size.
regards thomas

[blubee]

Hello blubee,
I had a quick look at your files.
Some hints:
I was interested in the first 2 eigenfrequencies as shown above, so I have set
number of modes in ccx to 6. In modal-analysis you must double the number of
eigenfrequencies you are interested in. For example, if you want to get the first
6 eigenfrequencies, you must set number of modes to 12 (minimum).

Your given torque of 10860 Nm cant be true for a 15 hp engine. Even a diesel-engine
produce much lower (Audi A4 TDI 2.0 with 150 hp: 350 Nm torque)

This load-case can not produce torque in the shaft. Please the
the post from UR_:
https://forum.freecadweb.org/viewtopic.php?f=18&t=23778
first_eigefrequency.PNG


There is a much easier way to get the first eigenfrequency:
first_eigefrequency.PNG


Can you give some more info about the project?

Please upload your files without results and a cleared mesh. Reduces file-size.
regards thomas

Hi I did see those threads about the torque limitation in FreeCAD. I think caculix does support this feature though but ccx is a massive program and writing those files by hand is a bit over my head at the moment.

Here are my files;
disk: https://ufile.io/gyf87
shaft: https://ufile.io/yry6g

I have a real world problem that I am trying to solve for.
I have some medium carbon steal that's has a yield strength of 490 MPa : http://matweb.com/search/DataSheet.aspx ... f6c&ckck=1

I have a 15HP 1200RPM motor with a reducer that brings it to 85RPM so it's about 926ft-lbs or 1250Nm of torque. I am using 1ft-lb == 1.35Nm for the conversion.


Okay so here is my problem, with that material linked above, would this motor and reducer be too strong and cause failure if the maximum force is applied to the shaft and the disk stays stationary.

I'd like to see the maximum force that the shaft and disk can bear in case the shaft is spinning the disk, the disk gets stuck and the motor running fully at 85RPM would the disk or shaft fail?

If the force at 85RPM is too high for this material, then what's an RPM that would work?

[thschrader]

I have a real world problem that I am trying to solve for.
I have some medium carbon steal that's has a yield strength of 490 MPa
I have a 15HP 1200RPM motor with a reducer that brings it to 85RPM so it's about 926ft-lbs or 1250Nm of torque.

I'd like to see the maximum force that the shaft and disk can bear in case the shaft is spinning the disk, the disk gets stuck and the motor running fully at 85RPM would the disk or shaft fail?

Hi blubee,
I am not an engine designer and I must confess that I am not really understanding the whole problem.

But here is my approach:
As we discussed in
https://forum.freecadweb.org/viewtopic.php?f=3&t=23992
the 20 mm transmission (the weakest part of the model) can be loaded with a maximum torque of 231 Nm,
valid for steel S235 with yield stress 235 MPa (235/sqrt3 gives max shear stress = 136 MPa)

When you use steel with 490 MPa, the maximum allowable torque rises from 231 Nm
to 231*490/235=482 Nm. By a loading of 1250 Nm as you said above I think the transmission will break
This is only from a static point of view (I am a structural engineer, I live in a static world... )
Please keep in mind the centrifugally forces, and maybe impuls-loading when the system suddenly stopps.
It is very difficult to simulate all these problems in one FEM-model.
Can you do a live test with the model? If so, you can determine the max loading until break.
With this value you can calibrate your FEM-model for further calculations.

It is not a good idea to rely on FEM-analysis ALONE when you design critical parts...
ups, I forgot the video...
https://www.youtube.com/watch?v=pDwqPaQjjLY
regards thomas

[blubee]

I have a real world problem that I am trying to solve for.
I have some medium carbon steal that's has a yield strength of 490 MPa
I have a 15HP 1200RPM motor with a reducer that brings it to 85RPM so it's about 926ft-lbs or 1250Nm of torque.

I'd like to see the maximum force that the shaft and disk can bear in case the shaft is spinning the disk, the disk gets stuck and the motor running fully at 85RPM would the disk or shaft fail?

Hi blubee,
I am not an engine designer and I must confess that I am not really understanding the whole problem.

But here is my approach:
As we discussed in
https://forum.freecadweb.org/viewtopic.php?f=3&t=23992
the 20 mm transmission (the weakest part of the model) can be loaded with a maximum torque of 231 Nm,
valid for steel S235 with yield stress 235 MPa (235/sqrt3 gives max shear stress = 136 MPa)

When you use steel with 490 MPa, the maximum allowable torque rises from 231 Nm
to 231*490/235=482 Nm. By a loading of 1250 Nm as you said above I think the transmission will break
This is only from a static point of view (I am a structural engineer, I live in a static world... )
Please keep in mind the centrifugally forces, and maybe impuls-loading when the system suddenly stopps.
It is very difficult to simulate all these problems in one FEM-model.
Can you do a live test with the model? If so, you can determine the max loading until break.
With this value you can calibrate your FEM-model for further calculations.

It is not a good idea to rely on FEM-analysis ALONE when you design critical parts...
ups, I forgot the video...
https://www.youtube.com/watch?v=pDwqPaQjjLY
regards thomas

Thanks for all the help, but I am not designing an engine. The fact that that stress will break the steel is good enough to know for me.

Can you explain what formula are you using to calculate the max shear stress for the steel S235?
when you did 235/sqrt(3) to get 136MPa is that because you are familiar with S235 steal?

Couldn't I have started with the 490MPa and worked from that?

[thschrader]

Can you explain what formula are you using to calculate the max shear stress for the steel S235?
when you did 235/sqrt(3) to get 136MPa is that because you are familiar with S235 steal?
Couldn't I have started with the 490MPa and worked from that?

Hi blubee,

green colour: yes
blue colour: divide your yield stress (in your case 490 MPa) through sqrt(3) to get max allowable shear stress.
picture 1 source:
https://en.wikipedia.org/wiki/Von_Mises_yield_criterion
picture 2 source: my own static calculation . German text means "max allowable shear stress in welding",
beta/gamma are security-factor/correction-factor.
blubee: this is a CAD-forum. We are "visual"-people. Please let us know what you are designing...

shear_stress1.PNG (11.18 KiB) Viewed 2035 times

[blubee]

Can you explain what formula are you using to calculate the max shear stress for the steel S235?
when you did 235/sqrt(3) to get 136MPa is that because you are familiar with S235 steal?
Couldn't I have started with the 490MPa and worked from that?

Hi blubee,

green colour: yes
blue colour: divide your yield stress (in your case 490 MPa) through sqrt(3) to get max allowable shear stress.
picture 1 source:
https://en.wikipedia.org/wiki/Von_Mises_yield_criterion
picture 2 source: my own static calculation . German text means "max allowable shear stress in welding",
beta/gamma are security-factor/correction-factor.
blubee: this is a CAD-forum. We are "visual"-people. Please let us know what you are designing...
shear_stress1.PNG

Thanks for the formulas, I was also doing some research and found some formulas as well but I am not exactly sure how to fully apply them.
Here are some FreeCAD project files: https://ufile.io/x4jdq
I used the assembly2 workbench, if you don't have it, it's located here: https://github.com/hamish2014/FreeCAD_assembly2

I will use a medium carbon steal that has a yield strength of 490MPa. I want to do static analysis to get within the ballpark of acceptable forces that I could apply to this material.

Some formulas that I came across while researching.
max shear strength for a material = torque * radius / J
J = pi/2 * radius^4

There are 2 main pieces that I wanted to test the max force that they can withstand.
I have a 15HP 1200RPM motor that I will reducer to 85RPM this gives 926ft-lb or 1255/m2 of force.
The first 1 is the blade; it's Diameter is 300mm with that in mind a radius in meters is 0.15 using that I get:
1255*0.15/((pi/2)*0.15^4) = 236728.24128
236,728 Pascals or 0.24 MPa

The shaft has three different parts, the cylinder that will connect to coupling, the smaller hexagon and the larger hexagon.
cylinder diameter = 40mm
small hexagon diameter = 60mm
large hexagon diameter = 70mm

I can calculate the shear stress on the cylinder with the above formula
1255*0.02/((pi/2)*0.02^4) = 99869726.7902
99.86972679 Pascals or 100MPa

Doesn't this mean that the shaft could hold almost 5 times more force?

I don't quite understand using 1255MPa in the formula I though that the shaft would be pushed to the limit but it seems not.

All the files are in the link.


OS: Ubuntu 17.04
Word size of OS: 64-bit
Word size of FreeCAD: 64-bit
Version: 0.17.11904 (Git)
Build type: None
Branch: master
Hash: 88e2b6de9a256896de2069fe1b55c66296b75ee1
Python version: 2.7.13
Qt version: 4.8.7
Coin version: 4.0.0a
OCC version: 7.1.0
Locale: English/UnitedStates (en_US)

[thschrader]

Here are some FreeCAD project files: https://ufile.io/x4jdq

Doesn't this mean that the shaft could hold almost 5 times more force?

Hi blubee,
thanks for the linux-files, but I am working with Windows 10.
Please have a look at the calculation, done with smath-studio, see
https://en.smath.info/view/SMathStudio/summary
That means: under 1250 Nm torsion-loading the 20 mm transmission and the small hexagon will fail
solution:
1. change geometry if possible
2. or reduce loading
3. or use high tensile steel (check your bugdet...)
http://www.interlloy.com.au/our-product ... ile-steel/

calculation.PNG (51.97 KiB) Viewed 1969 times