Bolt dimensions for a bolted joint

AI Thread Summary
Calculating the size of a bolted joint involving a 316 stainless steel bolt and a 6061 aluminum thread requires careful consideration of material strengths and engagement length. The aluminum has significantly lower yield strength, making the thread the weakest link, which could lead to stripping if not properly sized. A minimum bolt diameter of 6 mm is suggested, utilizing a threaded insert to enhance strength and prevent damage to the aluminum. It is also recommended to use anti-corrosion agents and lubricants to mitigate galling issues during assembly. Proper engagement length and bolt sizing are crucial for ensuring the integrity of the joint under a 2000N load.
KavaKovala
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Hi everyone,

I have a question about sizing a bolted joint. How to calculate the size of the screw? Knowing that the bolt material is 316 stainless steel and the thread material where the bolt will be bolted is 6061 aluminum. This screw will be pulled at 2000N.

Thanks!
 
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Welcome to PF.

How thick is the threaded 6061 aluminium plate?
Or, what length of the 316 bolt is engaged with thread cut in the 6061 aluminium?Stainless steel 316
316-7 = 700 MPa.
316-8 = 800 MPa.

Aluminium 6061.
6061-Grade = Max yield strength.
6061-O = 83 MPa.
6061-T4 =110 MPa.
6061-T6 = 240 MPa.

The 6061 aluminium has a significantly lower strength than the 316 stainless. It is in shear, so the thread will need to have a greater area, or the aluminium thread will be stripped from the hole. The weakest link will therefore be the thread cut into the 6061 aluminium. Using a nut and washer rather than threading the aluminium would lower cost of manufacture.

I would place a stainless steel thread insert in the aluminium. That increases the contact diameter and so area of thread contact, which will reduce aluminium damage due to galling or repeated disassembly. For thin sheet it could be a nut insert, nutsert, rivnut, etc. For deeper holes it could be a thread insert, helicoil or recoil, etc.

There will need to be some way of locking the assembly to prevent the bolt unscrewing.
There may need to be a lubricant used during assembly.

As a quick initial estimate of the minimum bolt size ...
316-8 has tensile strength of 800 N/mm².
Area required for 2000 N is = 2000 / 800 = 2.5 mm²;
Minimum inner radius = √(2.5/π) = 0.892 mm;
Minimum inner thread diameter = 1.78 mm.
That tells us, to handle the pull out tension we need at least a 2 mm diameter 316 bolt.

But the thread in the 6061 aluminium must have a greater diameter.
For 10 times lower yield in aluminium we need √10 times the diameter = 5.6 mm.
So at first guess, I would expect to use a minimum of a 6 mm bolt in a threaded insert.

The size of stainless steel bolt and type of threaded insert used will be determined by the length of engagement with the aluminium.
 
Last edited:
Beluncor,

Thank you very much for your reply, I appreciate your help.

The bolt length could be 15mm.

I did not understand why √10 x diameter (2?)= 5.6mm. If the diameter is correct (2mm), the result should be 6,32mm or 8mm Bolt.
 
Baluncore said:
Minimum inner thread diameter = 1.78 mm.
I kept inner diameter of 1.78 mm * √10 because I was only estimating approximation.

Is there any reason why you might not use an M8x1.25 bolt in a long helicoil ?
https://en.wikipedia.org/wiki/Threaded_insert
 
Yes, I can use helicoil, I just told you the bolt length that I will intend to use.
 
KavaKovala said:
I just told you the bolt length that I will intend to use.
You need to know the length of thread in the aluminium engaged by the bolt.
 
Sorry, I expressed myself wrong, my english is not good, I would like to say that the length of thread in the aluminium engaged by the bolt is 15mm.
 
Having had many unpleasant experiences with stainless steel bolts in aluminum (in a marine environment), I would strongly recommend using thread inserts as Baluncore suggests, or at the least some kind of anti-corrosion agent such as Tef-Gel.
 
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Likes DaveE
Your English is good.

Assuming the 6061 is annealed. 83 MPa = 83 N/mm².
2000 / 83 = 24 mm².

For aluminium; Shear Yield Stress = Tensile Yield Stress * 0.55
https://en.wikipedia.org/wiki/Shear_strength

Also, the bolt thread may be only 75% of full depth, in an over-size hole = 60%.
So that gives 24 mm² / ( 0.55 * 0.6 ) = 72 mm².
With a safety margin of 2 = 145 mm².

Area of the shear cylinder is, A = π * diam * depth.
Here is the short list …
Helicoil M4x0.7 x2D; A = 100.5 mm². Insufficient section.
Helicoil M5x0.8 x2D; A = 157.0 mm². Take care not to over-torque.
Helicoil M6x1.0 x2D; A = 226.2 mm². My low-mass choice.
Helicoil M8x1.25 x1.5D; A = 301.0 mm². A more robust solution.
 
  • #10
Baluncore and Sandy, thank you for your help. It is very clear.
 
  • #11
I should also point out that unlubricated stainless steel threads under high loads are subject to "galling" - essentially, instantaneous cold welding, such that the parts cannot be separated without destroying them.
 
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