Hey Lochnagar

Ok so I was away last week with work and didnt get a chance to look at this until. I have also received some more information and the plot thickens.

So this is the bolt that was used:

Steel (STKM11A)
Diameter 11mm
Length 20mm
No. of threads 16
Material 10B21

So this gives me a bit more info.


Firstly I dont have the torque value yet (which I should have soon).

So at the moment when we are looking at T=kFD

k = 0.2
D = 0.011
F =

This is where I am stuck. Where can I get the proof load of the bolt?

Cheers

 

Hey Lochnagar

Ok so I have some more information.

Contratry to yesterdays post it is an M8 Bolt. Length 23mm, No. of threads 16.
I have been told the torque applied was 185KG.

So looking at what you have talked about above.

T=cDF

F = T/cD

F = 18.14/0.2x0.008
F= 11337.5N

Looking at the clamping bracket:

11337.5 x 51.7 - P x 28.67 = 0

P= 11337.5 x 51.7/28.67

P = 20444.67N

Friction force on the handle bar:

20444.67 x 0.2 = 4088.9N

So torsional resistance is:

4088.9 x 2 x 0.011 = 89.96Nm

So we have a weight limit on the scooter of 100KG. Lets take a guess that up to 50KG could be applied on the handle bars. The grips are 88mm above the pivot point.

How does that look?

Thanks

 

Hi FLM,

I am just looking at your last posting on 3rd February - and you indicate the "torque applied was 185Kg". There is a mistake here - because torque has the definition of (force times distance) and units of Nm - not the units of mass Kg - as you have indicated.

So maybe you can look into this - and let everyone know what the torque that was applied?

Additionally, are you sure that the tubing (the handlebars) are not made from steel STKM11A - as opposed to the bolt as you suggest?

Just out of curiosity - what is the "big picture" to this posting - has there been some sort of accident with this scooter?

Hope this helps.

 

Hi Lochnagar

Yes sorry you are right that was a typo on my behalf - the handlebars are STKM11A and the bolt is 10B21 (heat treated).

I was supplied with the torque as 185kg so I converted it to Nm which is how I got 18.14 in the above equation. I have gone back and asked the factory to confirm the torque value in Nm.

So the big picture here is that the factory claim a stem with 1 bolt wont work as it can not generate enough clamping force on the bars i.e. the handle bars will slip. So they have moved to a 4 bolt stem. Before any physical testing I want to show to them mathematically that it is possible which I believe it is - there has been no accidents!!

 

Hi FLM,

There is something missing in your explanation of the 3rd February regarding this mass of 185Kg because you can't convert a mass to a torque.

For example you could have a mass of 30Kg acting on a spanner of length 0.25m - and if the mass was acting perpendicular to the length of the spanner - then you would compute the torque as follows:-
N m
(30 x 9.81) x 0.25 = 76.6Nm

Hope this helps.

 

Hi Lochnagar

Yep I realise this and am waiting for the answer. Its Chinese New Year so all the factories are on holidays so I may have to wait a bit. I will post here when I get the answer as may need your input to ensure I am calculating it the right way - if thats ok?

Cheers

 

Hi Lochnagar

Ok I am back in action in this and nearly have all the info. Can I go back to one of your earlier values - where did you get the 6730N in the following sentence:

So at 8.5Nm tightening torque - this equates to a clamping force of 6730N.

Thanks

 

Hi FLM,

Just going back through this thread (no puns intended) - we had assumed that the bolt was an M6 X 1P and we had also assumed it was a stainless steel bolt grade A4-70 - since at the beginning of this thread - you were not sure of either the bolt size - or the grade of material - so to put some numbers to this problem - I had to assume something.

So what we are trying to do is to torque the bolt until the stress in the bolt is 75% of the yield strength. So for a grade A470 bolt - the yield strength is 450N/mm^2 - so 75% of 450N/mm^2 = 337.5N/mm^2 - which I rounded up to 335N/mm^2 - in the earlier postings - which I have copied below, for reference.

1) So you know the XSA for the M6 x 1P (defined on page 412) = 20.1mm^2.

2) So you need to compute the pre-load force - so that we end up with an axial stress of 337.5N/mm^2 or 335N/mm^2 (if you use my rounded up number). Stress = F/XSA, or F = Stress x XSA, F = 335 x 20.1 = 6730N.

3) So you now need to compute the torque to achieve this pre-load force of 6730N. I use the more accurate equation on page 437 - but you can use T = 0.2 X F x D (which gives a minor error). So T = 0.2 X 6730 x (6/1000) = 8Nm. (Remember we assumed the bolt was an M6 - hence the 6/1000 above).

4) See the book reference I have referred to above - with the link below.

https://books2all.files.wordpress.com/2013/12/shigley-machine-design-9th-edition.pdf
Hope this helps.

P.S. It must have been a long Chinese New Year

 

Hey - yes it was for the factory....typically after Chinese New Year a factory can lose up to 50% of its workers so they have been very slow this year.

So I will just do a recap of all the info I have:

M8 Bolt (XSA from pg 412 = 36.6)
Length 20mm
No. of threads 16
Material 10B21 (heat treated)

Torque Applied: 185KGf = 1814N
Yield Stress of bolt: 400KGf = 3923N (3923 x 75% = 2942)

Now I think they have made a mistake here on the yield stress of the bolt - I have asked them to have a look again and make sure it is minimum yield stress.

But anyway using those figures:

- Stress = F/XSA, or F = Stress x XSA, F = 2942 x 36.6 = 107677N.

- T = 0.2 X F x D So T = 0.2 X 107677 x (8/1000) = 172Nm


That doesnt seem right to me? Given that they are saying they apply 1814Nm

 

Hi FLM,

The bolt grade that you have detailed above is AISI 10B21 - is an American grade - which according to my book - is an SAE grade 8.2 bolt - which has a proof strength of 120Ksi = 825N/mm^2, and an ultimate strength of 150ksi = 1035N/mm^2.

With 16 threads over a length of 20mm - this equals a thread pitch of 1.25mm. So you appear to have an M8 x 1.25P bolt.

So the XSA = 36.6mm^2.

So 75% of yield strength (or proof strength) = 0.75 x 825 = 615N/mm^2.

So the pre-load force F = Stress x XSA = 615 x 36.6 = 22500N.

So the torque required to generate this force T = 0.2 x F x D, = 0.2 x 22500 x 8/1000 = 36Nm.

Hope this helps.

(Unfortunately you are mixing up forces with torques in your numbers in your posting above - which you can't do - and I think you have also maybe got some wrong data from the Chinese.
The only thing I am a little unsure of is why the Chinese have made a metric bolt with an American grade of steel - but since I am not an American - I can't answer that question).