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Simple moment applied to shaft, all rotational calcs = 0.00000

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Simple moment applied to shaft, all rotational calcs = 0.00000

I'm out of options here. I'm running a simple static analysis on a crankshaft (just imagine a solid round bar). I've created a cylindrical coordinate system. I constrain one end in R, T, and Z as fixed. The other end I apply a 1Nm moment. I've created two measures at points on the end where the moment is applied. One is at 12 o'clock and the other is at 6. The measures are set as rotatonal (radians).

The results are fine and reasonable for everything except the measures and the rotational calculations. See below:

Measures:

max_beam_bending: 0.000000e+00

max_beam_tensile: 0.000000e+00

max_beam_torsion: 0.000000e+00

max_beam_total: 0.000000e+00

max_disp_mag: 5.520417e-04

max_disp_x: -1.152828e-05

max_disp_y: -5.520416e-04

max_disp_z: -3.781698e-04

max_prin_mag: 1.840552e+00

max_rot_mag: 0.000000e+00

max_rot_x: 0.000000e+00

max_rot_y: 0.000000e+00

max_rot_z: 0.000000e+00

max_stress_prin: 1.840552e+00

max_stress_vm: 2.288784e+00

max_stress_xx: 4.405939e-01

max_stress_xy: -3.826951e-01

max_stress_xz: 4.070771e-01

max_stress_yy: -7.137716e-01

max_stress_yz: -1.207598e+00

max_stress_zz: 1.763673e+00

min_stress_prin: -1.327123e+00

strain_energy: 6.989292e-03

Measure2: 5.031145e-04

Measure3: 2.162827e-04

Measure4: 5.520417e-04

Rotational1: 0.000000e+00

Rotational2: 0.000000e+00

Ultimately I'm trying to compare the rotational stifness between two different shaft designs. Thanks

1 ACCEPTED SOLUTION

Accepted Solutions
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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Shouldn't that be in radians?

No, because we've defined a displacement measure; for this, the displacement is the magnitude of a vector, and the direction is a unit vector that is in the theta direction with respect to the initial configuration of the system.

EDIT: I thought about this a little more and there is a way you could directly measure the rotation (provided you know the axis about which the rotation is occuring).

  1. Create a datum point along the axis of rotation and offset it such that the point does not fall within the geometry.
  2. Create a weighted-link between the cross-section of your shaft (i.e. a surface) and the datum point you just created.
  3. Define a measure for rotation on the datum point.

This should give to the rotation value you're looking for.

View solution in original post

25 REPLIES 25
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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I assume you're model uses solid elements; if so, then the reason you're getting 0s for your measures are because solid elements don't have rotational DOF. If you want to measure the rotation of this shaft you'll need to do the following:

  1. Create a cylindrical coordinate system oriented such that the twist of your shaft is in the theta direction.
  2. Create a measure to calculate the displacement a point on your geometry with respect to the theta direction of your cylindrical coordinate system.
  3. Create a custom measure that calculates the inverse tangent of the value of your previously defined displacement measure divided by the distance from the point of rotation to the point that the distant measure is define at.

You could also link a dummy beam element to the geometry and measure the rotation of the beam element itself, but there are a bunch of caveats to this method.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Thanks Shaun. I'll look into this. What you're saying makes sense

But if I create a measure relative to theta it still is in a length unit (mm, in my case). Shouldn't that be in radians?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Shouldn't that be in radians?

No, because we've defined a displacement measure; for this, the displacement is the magnitude of a vector, and the direction is a unit vector that is in the theta direction with respect to the initial configuration of the system.

EDIT: I thought about this a little more and there is a way you could directly measure the rotation (provided you know the axis about which the rotation is occuring).

  1. Create a datum point along the axis of rotation and offset it such that the point does not fall within the geometry.
  2. Create a weighted-link between the cross-section of your shaft (i.e. a surface) and the datum point you just created.
  3. Define a measure for rotation on the datum point.

This should give to the rotation value you're looking for.

View solution in original post

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Ok. I did this also. The result is .00136 rad, which is the same as what I got with your "custom measure" method from your original response. I guess this has to be right. I just have to figure out why the it's not matching the hand calculations.

Just to be clear: I created a point on the axis of rotation, 0.01mm offset from the surface that I applied the moment to. Then I created a weighted link between that same surface and the new point. It's referenced back to WCS and free in all degrees, And my new measure is rotation (rad) at the new point.

See anything wrong there?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Based on how you're applying the moment, you should be able to place a rotation measure on the 0.01 mm offset datum point.

What is the value of your Shear Modulus?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

That's what I did.

I'm using 79.3 MPa for my shear modulus since all my dimensions are in mm. The part is forged steel.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I'm using 79.3 MPa for my shear modulus since all my dimensions are in mm. The part is forged steel.

I'm guessing you meant to say 79.3 GPa (not MPa). With this assumption, and the following properties, I get an angular twist of ~1.284e-3 rads:

theta = (32*L*T)/(G*pi*D^4)

where:

L = 100 mm

D = 10 mm

G = 79.3 GPa

T = 100 N*mm

I created a corresponding Simulate model and got ~1.278e-3 rads. Now, I had to make an assumption on your Possion's Ratio (0.3) to calculate the Young's Modulus from your Shear Modulus. I suspect there is a decimal error in your hand calculation and potentially a minor setup error in your FEM.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Your question got me thinking and the Shear modulus is 79.3GPa actually. I was using 79.3 MPa because of my mm units. But I believe now that I should have been using 79.3 kPa (N/mm^2).

When I change this I get 0.00128 radians with my hand calculation (excel sheet), while Creo is giving me 0.00136. I think we got it.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

OK, although I'm a little surprised that the difference between your hand calc and Simulate is that much (~6%) given how simple the problem is.

BTW, N/mm^2 is MPa not kPa (N/mm^2 = N/(1e-3m)^2 = N/1e-6m^2 = 1e6N/m^2)

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I had 79.3e9 N/m^2. Converting it to N/mm^2 should be 79.3e3. That's what I was calling kPa. Sorry for the confusion.

I thought it was close enough. You have any ideas why we're getting different values?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Can you post your model?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Sure thing. Thanks again.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

OK, I found your mistake. The material properties in your FEM are E = 200 GPa and nu = 0.33, which results in G = 75.19 GPa (not 79.3 GPa). If I use 75.19 GPa instead of 79.3 GPa in my previous calculation, I get ~1.354e-3 rads (which matches nicely with the ~1.36e-3 from your FEA).

If your shear modulus needs to be 79.3 GPa and you Poisson's Ratio is 0.33, then you FEM needs a Young's modulus of 206.18 GPa. However, if your Young's modulus is 200 GPa and your Poisson's Ratio is 0.33, then you need a shear modulus of 75.19 GPa.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Great. Thanks...again!

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Hello Shaun. Sorry to bug you again, but do you have a link to some documentation on how a weighted link works and what it's generally used for. Thanks. I'd like to understand it better and the help file isn't cutting it.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I believe there is a thread in this forums on weighted-links that should provide you with a little more information. As far as how weighted-links work, they're based off of an energy minimization method such that dependent side of the link goes through the same amount of work as the independent side. Weighted-links are typically used to "connect" mass points and loads to other geometric references on elements (points, edges, and surfaces), but they can also be used to connect a point on an element (like a beam or spring) to other elements (a classic example is using a WL to connect one side of a beam element to a radial surface that matches the diameter of a bolt head to mimic a fastening joint).

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

First off. Thanks a ton, Shaun. I haven't worked with torsion in ages.

I'm working with a dummy model for now. It's a solid rod 10mm diameter by 100mm long. I'm putting 1000 Nmm on one side (end surface) and fullly constraining the opposite face.

The cylindrical coordinate system and the WCS are both concentric to the constrained face. The moment is applied in the positive Z.

I created the custom measure ( function looks like this: (atan(PT0_T))/5 ) and I get 0.00136 rad. My hand calculations (using THIS FORMULA) gives me 1.285x10-6.

So I'm still struggling.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I created the custom measure ( function looks like this: (atan(PT0_T))/5 ) and I get 0.00136 rad.

Your equation should be: atan(PT0_T/5), not atan(PT0_T)/5.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I changed it and I still get 0.00136. So your 2 methods are still agreeing.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Just occured to me that, while your previous equation is wrong, it won't show a difference due to small angle approximation (I think).

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

10-4

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Yes. Angular deflection.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

My hand calculations for moment of inertia do not match what Creo is giving me either. Any thoughts on why? I was trying to correlate hand calcs back to the machine.

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

Are you sure both calculations (hand and Creo) are with respect to the same origin?

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Re: Simple moment applied to shaft, all rotational calcs = 0.00000

I got them to work out when measuring the cross section. So we may have had the origins off. I don't that is holding me up though.

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