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Derivation of bolt group moment of inertia

Derivation of bolt group moment of inertia

Derivation of bolt group moment of inertia

(OP)
Hello everyone,

I have a formula that I can not seem to derive. The moment of inertia for a bolt group is given about the x axis (at C.G. of bolt group) as:

Ix = 1/12*[nb^2*(n^2-1)] where n = rows of bolts, b = spacing between bolts

I tried assuming the bolt group had a unit width with a height = (n-1)*b, but I still do not come up with this formula. Does anyone know where this comes from or how to derive it?

Thanks

RE: Derivation of bolt group moment of inertia

So if n is the number of rows of bolts what represents the number of columns of bolts?

Typically the moment of inertia of a bolt group is the sum of the D^2 terms where D is the distance from the centroid to each bolt.


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RE: Derivation of bolt group moment of inertia

(OP)
JAE,

For multiple columns you multiply by the number of columns (provided they have the same number of bolt rows). There is a similar expression for Iy as follows:

Iy = 1/12*mD^2*(m^2-1) where m = number of bolt columns, and D is the spacing between the bolt columns.

I have used the formulas before, I just don't know how to derive them and was curious.

RE: Derivation of bolt group moment of inertia

If you are talking about the moment of inertia, and not the polar moment of inertia, then I agree with JAE--it is the summation of the number of bolts in each row times the square of the distance of that row from the centroid.

DaveAtkins

RE: Derivation of bolt group moment of inertia

It's a rough approximation that gets tougher as the number of bolts decrease. And I disagree slightly with one of your terms.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Derivation of bolt group moment of inertia

In your method III, I believe that the four ought to be a two. Method III should always provide a lower bound estimate of the true result predicted by method one in my opinion.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Derivation of bolt group moment of inertia

The equation is derived in Chapter 8 of "Design of Steel Structures" (third edition) by Gaylord, Gaylord, and Stallmeyer.

RE: Derivation of bolt group moment of inertia

just consider the fasteners as point area's (ie no self moment of inertia) and work though the calc ...
you can work from each fastener or you can (if you have a simple pattern, say 4 rows with n fasteners per row) generalise ... locate the centroid (in the example it's at the middle), calc the offset for each row (the inner rows are p/2 from the centroid, the outer rows are 3p/2) and calc ... I = 2*(9p^2/4*nA+p^2/4*nA) ... how does that fit with your equation ?

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

RE: Derivation of bolt group moment of inertia

Another procedure is to find I about the lowest bolt, then correct for the eccentricity to the c.g. of bolts.

Ibot = Ab.b2[1 + 22 + 32....+ (n-1)2]

Icg = Ibot - n.Ab.[(n-1)b/2]2

If n = 5, Ibot = Ab.b2[1 + 4 + 9 + 16] = 30Ab.b2

Icg = Ibot - 5Ab.4b2 = Ab.b2(30 - 20) = 10Ab.b2

BA

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