Modelling a torsion spring in NX5 3

[RobLN]

Hello

I'm trying to model a torsion spring in NX5. Trouble is I can't get the spring to update correctly if I change the number of turns on the helix.

Model History Overview:

1. Datum CS
2. Helix
3. Tube
4. Sketch
5. Datum CS (for sweep)
6. Sketch
7. Sweep

Imagine the torsion spring as having two protrusions from each end of the helix. One is a simple tube. The other is a sweep to model a hook feature on one leg of the torsion spring. The latter is the sweep feature that fails when I attempt to modify the value for the number of turns on the helix.

I can't seem to get the Datum CS to remain associative to the face on the end of the helix to which it is attached...

Any advice greatly appreciated!!

Regards

Rob.

 

[JohnRBaker]

Can you provide the part itself?

John R. Baker, P.E.
Product 'Evangelist'
Product Design Solutions
Siemens PLM Software Inc.
Cypress, CA

http://www.siemens.com/plm

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To an Engineer, the glass is twice as big as it needs to be.

 

[RobLN]

Hi John

Here is the part I made. Try changing the number of turns within the helix node. (currently 2.4) the result is interesting but not what I require!

Regards,

Rob.

 

[JohnRBaker]

Give this a try.

John R. Baker, P.E.
Product 'Evangelist'
Product Design Solutions
Siemens PLM Software Inc.
Cypress, CA

http://www.siemens.com/plm

http://www.plmworld.org/museum/

To an Engineer, the glass is twice as big as it needs to be.

 

[doc1955]

Here ia a sping I use it is a deformable sping you can control mean dia, wire dia, number of coils and the pre load angle.

Doc

http://www.goodrich.com

 

[JohnRBaker]

If you wish to have a spring which will behave closer to what you would see in the real world, as you torsionally flex the spring you are in reality altering the number of Loops in the helix while the Length of the helix remains basically constant. Therefore the Radius must and will change to allow that to happen. The attached model is an updated version of my original example where I've modified the expressions such that this compensation will now take place automatically thus producing a more real-world like behavior.

Note that this same phenomenon also occurs when compressing a coil spring, but the effect is much more pronounced when dealing with torsional springs of this type and shape, and therefore should really be accounted for in the design since it could be a factor when designing the other parts of the mechanism where this spring will be used.

John R. Baker, P.E.
Product 'Evangelist'
Product Design Solutions
Siemens PLM Software Inc.
Cypress, CA

http://www.siemens.com/plm

http://www.plmworld.org/museum/

To an Engineer, the glass is twice as big as it needs to be.

 

[RobLN]

Hello John, Doc,

A big thank you to both for the feedback on the spring design. That's exactly what I'm after.

I'll take a look at the expressions, I'm new to that side of the CAD and I've seen some other springs modeled that are extremely complex. I'm pleased to see that these expressions are far easier to understand & follow.

Just one clarification though - the Base_L expression, in your model John, I assume that is the factor which adjusts the diameter of the torsion spring. This value would be different depending upon the steel grade?

Also, on your model, Doc, there does not appear to be an adjustment for the radius (in the expressions) as the legs of the torsion spring are rotated; however there seems to be a small reduction in the diameter how is this achieved?

Regards

Rob.

 

[JohnRBaker]

I was not attempting to compensate for any strain placed on the wire which made up the spring. I was only working under the premise that as the number of loops decreased/increased due to the torsional winding/unwinding of the spring, that the overall LENGTH of the wire would be basically constant (except of course for any effect that the increase in strain would have, but that would be minimal and besides, it would require FEA techniques to solve anyway) and it was only that which I was compensating for.

The 'Base_L' expression was a measured value of the length of the wire (at least the helical portion, which is all that I needed to worry about) which I used to determine, based on the original neutral diameter and number of coils, the relationship between the number of loops and the diameter of the spring. Then all it took was a simple equation relating this to the changing number of loops which allowed me to derive how much the diameter would have to change so that you would still have the same length of wire. Perhaps a bit crude but not all that different than what I did for the 14 years that I worked as a machine designer of commercial food and chemical processing equipment. Of course in those days (1965-80) we didn't have any laptops with spreadsheet programs and just barely CAD near the end, so we had to do all this with Frieden Calculators and slide rules ;-)

John R. Baker, P.E.
Product 'Evangelist'
Product Design Solutions
Siemens PLM Software Inc.
Cypress, CA

http://www.siemens.com/plm

http://www.plmworld.org/museum/

To an Engineer, the glass is twice as big as it needs to be.

 

[doc1955]

"Also, on your model, Doc, there does not appear to be an adjustment for the radius (in the expressions) as the legs of the torsion spring are rotated; however there seems to be a small reduction in the diameter how is this achieved?"

It is in the expressions.
(p10 (Helix(3)radius)
sqrt((sqrt((pi()*Mean_Diameter_10)^2+Wire_Pitch_10^2)*Number_of_Turns_10/(Number_of_Turns_10+p7))^2-Wire_Pitch_10^2) /(2*pi())





Doc

http://www.goodrich.com

 

[RobLN]

I see wehre the vales come from now. Thanks to you both for taking the time to explain it all through.

What on earth is a Frieden Calculator. I remember that my dad had an early calculator from the 70s with a bright green display. Is it something like that?

 

[JohnRBaker]

This is a Frieden Calculator:

Where I worked we had both electric and hand-cranked models (I never did learn to use the hand cranked ones). They were so noisy that if someone was on the phone the office rules were that you had to wait until they off to use them. And if you were using one of the electric ones, woe be to you if you accidentally divided by zero since it had no way to detect this, it just cranked and cranked until all the registers read 99999999. Once you noticed what was happening, it was easiest to just unplug it and start over.

I can remember when around 1971 we got our first electronic calculator with the little green screen (they only ever bought but one as they were very expensive), and then a couple of years later the company provided each department with ONE handheld calculator which we all had to share, but it was so much faster than the old mechanical ones that the productivity improvement was significant. Of course as the prices started to drop for handhelds, some of us eventually purchased our own. I know I waited until they had dropped below $100, but finally around 1974 I got my first Texas Instrument. However, I can remember years later a couple of the old-timers insisted on still using one of the two newer Frieden's that they left on a desk in the back of the drawing office. When I left in 1980 they were still there.

Note that I've still got a couple of handhelds in my desk drawer, one of which I still use, mostly when filling out expense reports. And I still have my slide-rule which is just as accurate as it ever was ;-)

John R. Baker, P.E.
Product 'Evangelist'
Product Design Solutions
Siemens PLM Software Inc.
Cypress, CA

http://www.siemens.com/plm

http://www.plmworld.org/museum/

To an Engineer, the glass is twice as big as it needs to be.

 

[hudson888]

I think we have a comptometer in the basement somewhere do I win anything!

Best Regards

Hudson

www.jamb.com.au

Nil Desperandum illegitimi non carborundum

 

[jschetl9]

HI John,
I am trying to apply vector force(unit force) at one end of the simulated 4-barlink mechanism. I used Revolute+Revolute+Universal and Ball joints at each end.
Grubler count is shwoing 0 and the dof also zero (with driver on one revolute).
Now when I apply unit vector force, when I run the for excel map the simulation is giving a torque of 0.9 where as the FBD free body diagram gives an end torque of 8.63psi.
Pl guide.
Thanks