Preload and Losses

[Daparojo252]

Dear All,

I am after guidance, and would like someone to check to see of I am on the right lines.

I have need a fastener that needs a preload value 70% of its yield.
Material has a yield of 600 MPa, therefore 70% is 420 MPa.
The Cross-Sectional Area of the fastener is 1000mm2 (smallest undercut).

Therefore the Pre Load required is 420 x 1000 = 420 kN

Pre Load = Clamp Load

The Clamping length is 90mm, the nominal thread of 40mm diameter.

If I was to use tensioning rather that torque tightening for a more accurate way of stretching and minimising losses, then the load losses approx. :-

40/90 + 1.01 = 1.45 x Pre Load = 609 kN Of Applied Load to the Bolt to retain 420 kN Clamping (Preload) over a short length of 90mm.

Now, that would mean that the stress in the undercut 609 kN / 1000 mm2 = 609 N/mm2

Stress in Bolt is greater than the yield of the material, therefore it would fail.

I would have to specify a greater material yield than originally specified.

Am I on the correct lines, or am I missing something.

 

[MintJulep]

Length is not area.

 

[desertfox]

Hi Daparojo252

I believe you are correct with the calculation, the formula you are using can be found on the tentec site, basically you have to stretch the bolt further to retain the preload you require after losses are taken into consideration

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[RolMec]

I think you need to consider the losses from torqueing (torqueing tool impact) before applying the stress calculation, i.e. to go down on the 420 MPa. However, the influence from torqueing method / tools can go either way, so you'd need to check on the minimally achievable pre-tension.
Overcritical (above YS) tensioning exists, but is a very special way to be done and perhaps not applicable with your job?!
Shigley chapter 22.3 ff or else VDI 2230 (if you're in metric area) might help.
Regards


Roland Heilmann
Lpz FRG

 

[daparojo]

Cheers guys for the help.

What I have trouble getting my head round is that the joint/bolt must have a preload of 70% of it's yield. However, to create that clamp force (taking losses into consideration), the applied load takes it higher than the materials yield - therefore failure.



If I had 2 identical bolts, one to be Torque tightened and one to be tensioned hydraulically - both must have the same clamping load - would I be correct to say that the Torque tightened Bolt would be subject to a higher initial load to generate the same Pre Load as the hydraulically stretched bolt?

 

[BrianE22]

For most joints the applied load is shared by the parts and the bolt. So not all of the applied load is added to the preload for the screw. For a typical joint about 70% of the applied load is taken up by the parts leaving only 30% going through the bolt.

 

[MintJulep]

40/90 + 1.01 = 1.45 x Pre Load

Seems completely arbitrary to me. Can't make any sense of it.

See here for how to calculation elongation vs. Load.

http://www.engineeringtoolbox.com/bolt-stretching-d_1164.html

You need to calculate the unthreaded and threaded lengths separately, then add them together to determine the total elongation.

joint/bolt must have a preload of 70% of it's yield

It's a rule of thumb. If you are to the point of actually calculating elongation then this should not be a design input.

1. Determine the clamping load needed to do whatever you are trying to do.
2. Divide that by the number of fasteners to get the load per fastener.
3. Take a first guess at fastener diameter.
4. Calculate stress and confirm no yield. If yield, bigger fastener.
5. Calculate stretch.
6. Confirm that the stretch is more than the possible embeddment.

 

[daparojo]

Again, thanks for your help.

The problem that I am looking at states that the Bolt must be 40 diameter and have a preload of 70% of the material yield, hence that is where I got the 420 kN pre load from. To achieve this, then I must apply a larger load/stretch to achieve this retained loading in the bolt.

There are 4 other bolts that would carry the same loading.

The 40/90 + 1,01 equation was something I was looking at to determine the applied load - like desertfox stated, it was aquired from Tentec's website as a quide.
The 40mm diameter divided the clamped length of the application plus a factor of 1.01

 

[desertfox]

Hi

Here is the tentec ref

http://www.tentec.net/pdfs/tips on tensioning 2006.pdf

I ran into a similar problem as you are facing but we could live with less than 70% of yield

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[MintJulep]

Can you explain the physical meaning of (D/L) + 1.01?

Hint: You can't because there is none.

It's an empirical relationship best fit curve from a bunch of test data.

What your calculations are telling you is that the either the diameter is too big, or the length is too short.

 

[Ingenuity]

Interestingly, from the same tentec.net website, this dated 2004: Link

 

[AlexSK]

My 2 cents. There is difference between tensile yield and permissible bearing stress, or contact pressure, which is your case. Permissible bearing strength is typically 1.5 of the material tensile strength.
You can find bearing stress for different materials in the engineering handbooks

 

[drawoh]

Daparojo252,

The standard cross sectional area for bolts is known as the tensile stress area. For standard bolts, this is shown in handbooks. The Handbook of Bolts and Bolted Joints by Bickford and Nasser, shows you how to calculate this for both English and metric bolts. Does your bolt have a separate undercut feature?

By "tensioning", do you mean turn of the nut? You can work out the bolt's maximum stress. From that, you can work out the resulting strain. Knowing the length, you can work out [Δ]L. Knowing the thread pitch, you can work out the turn of the nut. Now all you have to do is get everyone to agree what finger tight means!

--
JHG