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Calculating max torque for bolt

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310toumad

Mechanical
May 12, 2016
63
If I had a steel bolt being threaded into a tapped hole of a softer material, aluminum for example, how would I go about calculating the maximum allowable torque on the fastener assuming I have enough thread engagement? In this scenario, the only failure I can see is the internal threads in the aluminum hole eventually shearing. My approach was going to be to calculate the shear area of internal threads using the formula (6) on page 1537 of the machinery handbook. You can also find it at the bottom of this page:
Then, if I know the yield strength of the aluminum I can calculate the shear yield strength (Ssy = .58Sy) and use that as my threshold for finding what maximum allowable axial force in the bolt can be:

.58Sy = F/As where F is the axial force and As is the shear area derived from the formula above. Once I solve for the axial force I can use the general formula T = KFD to find the corresponding torque value. Does this strategy sound like a reasonable approach?
 
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presumably your bolt has seated somewhere, so the torque is not driving the bolt along the thread, so what'll happen first …

1) will the thread strip (where you're going), or
2) will the bolt bear into the Al plate ?

another day in paradise, or is paradise one day closer ?
 
My guess would be the head of the bolt would dig into the aluminum as the bearing stress begin to exceed the yield strength. although at some point I would think the aluminum would reach some elastic limit where it could not be pushed in any further, at which point the stress in the internal threads would rise sharply and they would break. I suppose depending on where the hole is located, for example if it was near an edge, then the bearing stress might create a failure in the form of a crack before any sharing of the threads occurs.
 
so the initial failure has "little" to do with the threads ? Though at some stage you'll strip the threads in the Al plate.

another day in paradise, or is paradise one day closer ?
 
If you have, as you describe, enough thread engagement why could you not exceed the bolt strength before stripping threads?
Did you make the comparison of thread shearing capacity to bolt tension capacity?

Ted
 
Consider posting in a more appropriate forum.... "Welding, Bonding & Fastener engineering"

MJCronin
Sr. Process Engineer
 
Hello Note that if you google Alum propeties it will list the bearing stress of aluminum at more then the Yield. Does not make intuitive sense, but that is what is listed. For Alum 6061 T6 this book lists yield at 35KSI and bearing yield at 58. Ult tension at 42 and Bearing Ult at 88.

I believe it has to do with the lower E Elasticity spreading the load out over more area beyound the contact area.

I have a real problem with designing to these higher values and most always assume yield = Bearing Yield. Just to be safe.
 
All flavours of Al I'm used to have Fbr > Fty and by a sizeable margin.

another day in paradise, or is paradise one day closer ?
 
Bolt planet has an online calculator that has many cool output results including engagement to protect the bolt and "nut" threads from stripping.

I can not vouch for it's accuracy.
Personally I question the importance of Ultimate Tensile Strength over Yield Strength in these, and many other day-to-day analyses as well.

============.

" assuming I have enough thread engagement? "

1.5 diameters of engagement is "enough" to develop the full strength of a SHCS in some pretty weak materials according to pages 64-67 here -
 
" assuming I have enough thread engagement? "

1.5 diameters of engagement is "enough" to develop the full strength of a SHCS in some pretty weak materials according to pages 64-67 here -
Just to sum up what you said well:
What umbrako is saying is that the first few threads take most of the load (I is I think commonly accepted)
And therefore taking credit for thread engagement beyond the first few threads to prevent stripping would be questionable.
(And yet seems to be a standard calculation in lauded books like Machinery's handbook... questionable).


=====================================
(2B)+(2B)' ?
 
" And therefore taking credit for thread engagement beyond the first few threads ... "

Generically aluminum is ~ 3X stretchier than steel, and E for steel is darned close to 30,000,000 psi regardless of alloy or hardness.

I have not done any FEA modeling to help understand/prove/disprove the thread stress gradient, but by inspection a steel fastener engaging a low modulus material would tend to share the load with more threads down the hole.
This would tend to explain how more thread engagement can develop full fastener strength in some lesser materials.
In the extreme case, if the female threads were rubber, I believe they would all get into the act ~ equally.
 
Yes good point. Sorry, I had misunderstood your previous post.


=====================================
(2B)+(2B)' ?
 
In a case like this there are too many unknowns to be able to use calculations to really get close to an accurate answer. I would highly recommend testing to failure and then taking a % of that as your seating torque. If we try to estimate the effects of the head embedment on the aluminum bearing surface, as well as everything else going on in this joint, there will the potential for a massive amount of error. Just test some parts to failure and then you'll have no doubt about what the reality actually is.
 
The OP's method looks correct with a few moments' glance, however without calculating contact stress through the head as hinted above they're only solving half the problem. Within the context of the joint itself there's really four failure modes - 1. male thread failure, 2. female thread failure, 3. fastener head failure, 4. material failure of the part under the fastener's head. 1-3 result in an obviously failed joint, the parts literally separate. #4 often results in a less-than-obvious failure as an undertorqued joint ready to fail.
 
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