# Understanding 'simple' splines - help!

• 09-09-2020, 10:10 AM
YorkshireDave1
Understanding 'simple' splines - help!
Hello everyone.

Unfortunately, I am an 'instinctive' inventor. As such I fully realise just how much I do not know. That said, invariably 'things' work and are reliable.

[ATTACH]2489[/ATTACH]
This image is of an 'adapter'. Through its centre runs a stainless shaft of that profile. Another metal object slides over (around) the adapter so the 'adapter' becomes the linking mechanism between the two metals.

For reasons of cheapness and adaptability, the 'adapter' has to be plastic. Unfortunately though I'm finding with my fine detail 3D prints that the adapters blow out far too early i.e. they do not allow enough torque to be applied before failing.

After some uneducated analysis, I've concluded that the design of the inner shaft with its 6 (much tougher) splines is probably causing too high a point load (apologies for my terminology) on the plastic adapter so causing it to shatter. It further occurs to me that as the 'shear'? strength of the internal stainless is soo much higher that of the plastic the area of plastic it pushes against should be correspondingly higher!

I'd like to understand what's happening so I can redesign and get more torque transferred between the two metals.

Sorry for being thick! I can confirm that advancing age doesn't always guarantee increasing wisdom
• 09-09-2020, 10:37 AM
Kelly Bramble
To start... and the most important question for the moment, how much torque is the coupling seeing?
• 09-09-2020, 12:16 PM
YorkshireDave1
Hi Kelly.
Under 10 Newton's. A screw thread on the stainless centre fails at 7.1N
Hope that helps
• 09-09-2020, 01:22 PM
Kelly Bramble
Unknown size and I'm guessing 10N (2.25 lbf) is applied at the tooth? Without knowing the size of the coupler and where that force is being applied I'm guessing a bit.

Torque is given in lb-in, or maybe N-mm BTW.

What I'm after is whether the design materials, geometry and maybe the 3D print parameters are adequate or not. I have plastic gears on my 3D printer that have held up with a couple of thousand hours on them.

Engineering and design most correctly is evidence based. Meaning that we design our minimums to a requirement not a gut feel.

WE need size, materials, known forces, 3D print specified (fill, wall thickness, etc..) to better understand if there is a design, manufacturing or other challenge.

[ATTACH]2490[/ATTACH]
• 09-09-2020, 03:30 PM
YorkshireDave1
Kelly

The internal stainless spindle has an OD of 5.6mm but it has an M3.5 threaded hole in it. The splines on that spindle are just 0.4mm deep in their centre and 1.33mm wide.

The OD of the 'adapter' (the triangular splines) is 10.5mm and root dia of those splines is 9.42mm. Each external spline on the adapter is 90 degrees.

The rotational force is applied to the part that slides over the outside of the adapter. The adapter then transfers that force to the stainless spindle running through its centre.

Due to the small dims and tight tolerances involved, we had no choice but to use what Shapeways (the 3d printers) call their high detail plastic. Other types we had tried were simply too poor at delivering the tolerances and were too sloppy to be even usable. They are however solid printed with the print orientation such that the layers are placed at right angles to the rotational force applied.

Currently, the plastic parts fail at approx 5N. They shatter and I suspect that happens because the plastic used is so hard and not so 'plastic'.
We have also tested a different design of adapter. That one used similar 90 degree splines on the inside & outside and failed at 7N. We know it was made of acetal. Unf we cannot use that going forward.

My objective in coming on here is to understand whether it is the design in the centre of the adapter which is reducing the ability to transfer as much torque or the material we are using for our adapters or, indeed, both!

I do hope that makes thing clearer ;-)
• 09-09-2020, 03:37 PM
YorkshireDave1
OOps
This image shows how the adapter fits over the spindle
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This image is a modification and has my question.
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Sorry posted pre-edited image :bash:
• 09-09-2020, 05:26 PM
Cragyon
There are many - many plastics that can be 3D printed. Like ABS, PLA Nylon - what the materials? Kelly seems to be using a crowbar to get useful information out you and still not getting what (he/she) needs...

The spines look low area to me as well as geometrically hard to machine and/or print.

What tolerances do you really need and why?
• 09-10-2020, 06:38 AM
jboggs
The key to failure analysis is to identify the very FIRST point of failure, the initial crack, and why it failed AT THAT POINT. Many times we find a "stress concentrator" at that point, usually in the form of a sharp corner. Many times simply eliminating that sharp corner increases the load bearing capacity of the part significantly. This approach is more microscopic than macroscopic.

So, where is the very first point of failure? Can you set up your trials so that the load application ceases when that very first crack appears? Or set up a series of gradually increasing trials, in which you apply a small torque and examine the piece, then a slightly larger torque and examine it again.

By the way, torque (what you call a rotating force) is always expressed in terms of force and distance, ft-lb, oz-in, N-m. To try to express it in terms of force only ("the plastic parts fail at approx 5N") is to leave out a critical piece of information. How far from the center of rotation is that 5N force applied? 3mm? 5mm? 50mm? It makes a difference. For example 5N force applied at 20mm distance from the center is a torque of 100N-mm.
• 09-10-2020, 02:39 PM
YorkshireDave1
JBoggs many many thanks. That makes so much sense and is very enlightening.

We are using a meter placed inline with the centre spindle, so would I be right in presuming the measured force is actual? Now I've written that I'm not sure how that force changes as it moves away from the centre i.e. at the edge of the splines. Does it reduce or increase?

What I'm going to do is redo the tests in stages and examine at each stage. We'll report back.

In the meantime, can you help me pls? I'm thinking that a force of 7N applied in the centre to 6 splines means that each spline has a load of 7N divided by 6. If I increase the spline count to 10, does that mean the force on each reduces as it becomes 7N divided by 10? If so, then assuming that 'point load' is insufficient to break the plastic at that point, can the plastic take more overall? I do hope you can understand what I'm saying!?
• 09-11-2020, 08:24 AM
jboggs
A force of 7N applied at the center is simply a force of 7N. Force has a linear direction and a single location. A rotary force is a torque. And torque is always expressed in terms of force @ distance. If the "force" you are applying at the center of the splines has a rotary nature to it rather than linear, if you are twisting the drive shaft, it is a torque. (Find a text book on the relationship between force and torque for a better understanding.)

So, you are applying a torque to an arrangement of 6 splines. Is that torque equally divided between all six? In theory, yes. In reality, never. Because of small differences in geometry and fit one or two of those splines will actually see much more of the torque load than the others. Which ones? Whichever one fails first.

Can the performance be improved with more splines? Maybe. Maybe not. Lots of factors involved. I would look closely at the actual detailed geometry of the area where the torque transfers from the shaft to the spline.
• 09-11-2020, 09:00 AM
YorkshireDave1
[QUOTE=jboggs;16829] Can the performance be improved with more splines? Maybe. Maybe not. Lots of factors involved. I would look closely at the actual detailed geometry of the area where the torque transfers from the shaft to the spline.[/QUOTE]

Thank you very much JB. Very enlightening.

The measure of turque we are seeing is from an instrument running through the whole thing's centre line and is displayed in Nm. Rather than 'pester you', do you know of a ref (kind of 'Idiots Guide') to gaining simply a better understanding of the impacts of geometry on torque transfer? Having Googled I'm very aware it's a hellishly complex subject.

Would it be a 'rule of thumb' that having more 'mating surfaces' equates to better transfer of torque overall?
• 09-11-2020, 12:52 PM
jboggs
In a way you just answered some of your own questions. You said the instrument displays "Nm". That is a different animal than "N". One is a unit of force. The other a unit of torque. And there is a big difference between "Nm" and "Nmm", in fact a thousand times difference.

Don't let my direct approach intimidate you. Once you understand it, it really isn't "hellishly complex". In fact, its very simple. Its just a matter of wrapping your brain around it. I would bet Kelly has some simple explanations somewhere on this site.

Think of a seesaw. You have two forces acting about (or around) a fulcrum point, two torques. A 200 lb man 10 ft from the fulcrum creates a torque of 200 * 10 = 2000 ft-lb. A 100 lb child on the other side 10 ft from the fulcrum creates an opposite torque of 100 * 10 = 1000 ft-lb. Obviously the imbalance goes to the side with the greater torque. Here's how torque works: if the 200 lb man moves to a position 5 ft from the fulcrum he now creates a torque of 200 * 5 = 1000 ft-lb. The seesaw is balanced now because the two opposing torques are equal. Instead, what if the 100 child moved to a position 20 ft from the center. Then his torque would have been 100 * 20 = 1000 ft-lb. That would have balanced the seesaw too.

Drop a long straight wooden stick that weighs 1 pound on the ground. Pick it up in the middle. Your hand is zero inches from the center of gravity so the torque (also called moment) you feel is ZERO. The force is 1 lb but the torque is zero. 1 lb * 0 in = 0 in-lb. Put it down. Place your hand about 20 in. from the center and pick it up again. What you feel this time is torque because your hand is some distance from the center of gravity of the stick. Its the same weight as before but the torque created by that weight very different. 1 lb * 20 in = 20 in-lb.
• 09-11-2020, 01:08 PM
Cragyon
Use industry standard splines, like from the following. Predictable and understood.

[url]https://www.engineersedge.com/gears/involute_spline_and_serration__13650.htm[/url]
• 09-11-2020, 03:07 PM
jboggs
What he said!
• 09-13-2020, 05:06 PM
YorkshireDave1
[QUOTE=jboggs;16833]What he said![/QUOTE]

As I hinted at in my original post, but no one seemed to notice, a standard would be fine if I were not looking for an element of patentable novelty...

JB your direct approach is far from offensive. You have always tried to help rather than belittle and I really appreciate that. I shall remove myself from Engineers Edge permanently so you are all no longer disturbed.

Thank you all again.
• 09-14-2020, 06:27 AM
jboggs
No need to remove yourself. Hang around. There is always more to learn.
• 09-14-2020, 07:18 AM
Kelly Bramble
[QUOTE=YorkshireDave1;16836] would be fine if I were not looking for an element of patentable novelty...

[/QUOTE]

Not sure how to take your statement - but inventing a spline geometry is a waste of effort as well as unpatent-able..

Also, while we're on the subject, don't waste time inventing the following:

Bearings
Hinges
etc.
• 09-14-2020, 09:58 AM
YorkshireDave1
[QUOTE=Kelly Bramble;16838]Not sure how to take your statement - but inventing a spline geometry is a waste of effort as well as unpatent-able.. [/QUOTE]

I didn't say I wanted to invent a new spline Kelly. What I said was that I'm 'looking for an element of patentable novelty'. In this instance, using what is already out there in spades reduces my opportunity for novelty.

• 09-14-2020, 10:28 AM
Cragyon
[QUOTE=YorkshireDave1;16840]I didn't say I wanted to invent a new spline Kelly. What I said was that I'm 'looking for an element of patentable novelty'. In this instance, using what is already out there in spades reduces my opportunity for novelty.