Bolt Group Ip

[SteelPE]

I am trying to figure out how to calculate how to calculate the moment of inertia of a bolt group. In accordance with page 7-9 or AISC 13th edition Ip=Ix+Iy in units of in4/in2

Are you suppose to include the bolt area in the calculation of Ix and Iy? The way the units are written (in4/in2) I wouldn't thinks so, but then again, if you don't, you are not truly getting Ix or Iy of the bolt group.

 

[CANPRO]

I'm not familiar with that code reference, but the units seem to indicate that you use a unit area (1 in²) for the bolts, which would give you final unit of in⁴ for Ip. The /in² is added to indicate that Ip is per unit area of the bolt.

Similar to calculating the section modulus of a line of weld. When I do this I use a unit width of 1 for the weld, resulting in a value of in³/in, then when I divide the moment by the section modulus, I get a value of kip/in.

 

[SteelPE]

What you say seems to make sense. Interestingly enough, I wrote a spreadsheet both ways (taking into account the bolt area and not using the bolt area) and I get the same answer using a 1 square inch bolt vs not including the area in calculation of Ix and Iy.

 

[CANPRO]

So your spreadsheet confirms the unit area assumption, which makes sense - the way the load is distributed to the individual bolts depends only on the geometry of the bolt pattern. The bolt size would only come into play if you mixed bolt sizes within the bolt group.

 

[DaveAtkins]

When you calculate polar moment of inertia for a bolt group or a screw group, the "area" of each bolt or screw is considered to be 1.0, with NO UNITS. That is why polar moment of inertia is in².

DaveAtkins

 

[rb1957]

lots of different methods here.

I think you need to be careful if you exclude the bolt area for the calc, as dave notes the units are "odd". With "odd" units you have to use the result in a correspondingly "odd" manner, ie you'll be adding the bolt area later in the calc or else understanding that the later calcs (using this "odd" result) know that using this inertia (I got tired on writing "odd") results in bolt force rather than stress (as in the usual bending calc ... My/I).

another day in paradise, or is paradise one day closer ?

 

[Shu Jiang PEng SE]

My understanding is:
1. Ignore the Ix and Iy of the bolts
2. Assume bolt area is 1 to simplify the calculation of “Ip”. The actual Ip should be the calculated “Ip” times bolt area.

 

[CANPRO]

rb1957, I think that is why the units are expressed as in⁴/in², so that you remember it is inertia per unit area of bolt. At least that is why I always make sure I write my units like that when working with unit values.

 

[rb1957]

so long as it's clear. Personally I think it creates a difficult situation when faced with bolts of different areas. I'd rather do the full calc, taking the short cut of simplifying for a common area (as suggested above Ip = "Ip"*bolt area); then I'd know how to account for different areas.

another day in paradise, or is paradise one day closer ?

 

[SteelPE]

rb1957, I understand what you are saying but the way the equations are written, if you use the area of the bolt you will not get the correct answer. You would need to make adjustments to the other equations used to calculate the load in the bolts.

I originally included the area of the bolt in my calculation and in the end, has different answers for the loads in my bolts vs the calculation eliminating area from the equation. The way the process is written, I believe they don’t want you to include the area in the Ix and Iy calculations .

 

[rb1957]

that's fine, the equations are written assuming one fastener size and produce a fastener force.

but with that equation you're limited to that assumption (not that it is a real problem in the real world).

another day in paradise, or is paradise one day closer ?

 

[WARose]

Go to Google books and look at 'Pressure Vessel Design Manual', 3rd edition by: Dennis R. Moss. On p.317, he gives formulas for Ip, Ix, & Iy. They are independent of bolt area.

I've always treated it like just like a pile group problem. (I.e. independent of cross sectional area.)

 

[Shu Jiang PEng SE]

I agree: the force distribution among the bolts is not related to bolt area.

 

[rb1957]

"the force distribution among the bolts is not related to bolt area" ... so long as all the bolts are the same size

another day in paradise, or is paradise one day closer ?

 

[Lomarandil]

Yes, you're technically correct rb1957 (the best kind of correct).

As has been mentioned, in the realm of traditional building/bridges structural engineering, if you're using different size bolts in one connection (really, different size bolts in any given type of connection), you're just asking for trouble of a half-dozen different flavors.

----
The name is a long story -- just call me Lo.

 

[CANPRO]

I originally included the area of the bolt in my calculation and in the end, has different answers for the loads in my bolts vs the calculation eliminating area from the equation. The way the process is written, I believe they don’t want you to include the area in the Ix and Iy calculations .

If you aren't getting the same answer then I believe there is an error somewhere in your calculation. You should get the same force distribution if your bolt area is 1 or any other value.

 

[rb1957]

"I originally included the area of the bolt in my calculation and in the end, has different answers for the loads in my bolts vs the calculation eliminating area from the equation. The way the process is written, I believe they don’t want you to include the area in the Ix and Iy calculations." ... it should be really clear what they mean, and if it isn't (if they use Ix and Iy without defining these) then I'd suggest writing to them to get them to clarify.


another day in paradise, or is paradise one day closer ?