Moment of Inertia of a bolt group 8

[Amar-Dj]

Hello everyone,

As a mechanical engineer working as a structural engineering, I still need to gather and understand the structural engineering theory.

I have a hard time to understand the moment of inertia of a bolt group (I_x and I_y).

Let's say that you have two columns of 3 bolts each. The zero is centered on this pattern. ''Y'' being the vertical axis and ''d'' is both the vertical and horizontal distance between bolts.

I have read that I_y= d²+(-d²)+d²+(-d²) = 4d²

Can anyone give me a reference explaining the math behind this and where does it come from?
Also, how come the units are mm² instead of mm⁴?

Thank you all for your answers.

 

[DaveAtkins]

When calculating the moment of inertia for a bolt group, it is customary to ignore the area and moment of inertia of each individual bolt. That is why the result is mm² instead of mm⁴.

Since the moment of inertia of each individual bolt is ignored, the moment of inertia is equal to one times the distance from the centroid squared. For your example, I_y would be (6)(1.5)² = 13.5 mm².

DaveAtkins

 

[TheDW]

What Dave said...you can also think of it as giving each bolt a unit area of 1mm^2, which would make the units work. The I of each individual in that case would be 0.08mm^4, which is negligible enough to ignore.

 

[BridgeSmith]

The reason the units are mm² is that the "I" of the bolt group only considers the "Ad²" portion of the moment of Inertia calculation (Io + Ad²), where "A" is set equal to 1 for convenience of the calculations (so you don't have to multiply by the area of the bolt to get stress and divide it back out to get force). The moment of inertia of the bolts themselves about their individual centroids is ignored as being inconsequential.

For the polar moment of inertia, which is what you would use to calculate the force for a bolt group where a moment is about the centroid of the bolt group, is Ix + Iy. The formula and derivation can be found in this thread.

The formula I use for the total polar I of a bolt group (that I either derived or found a long time ago) is:

(nc*nr/12)*[sr²*(nr²-1)+sc²*(nc²-1)]

where:

nc = number of columns of bolts
nr = number of rows of bolts
sc = spacing of the columns
sr = spacing of the rows

I don't have time right now to recreate the derivation, but it should just be a matter of rearranging and combining the equations for Ix and Iy given in the thread I linked to above.

Edit: In it's most basic sense, the polar moment of inertia of a bolt group is the summation of d², where d is the distance from the centroid of the group to the center of each bolt. You can add them up individually, or combine and reduce the terms to simplify it for larger groups.

Rod Smith, P.E., The artist formerly known as HotRod10

 

[rb1957]

d is the distance of the bolt area A … so your I = A*(4d^2) … in^4

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

 

[BAretired]

rb1957,

Why A*(4d^2)? Should be A*(6d^2).

EDIT: Actually, it should be Iy = A*6(d/2)² or A*1.5d² and Ix = A*4d²
The OP was asking about Iy.

BA

 

[azcats]

@BAretired

If two bolts are on the neutral axis, they don't contribute to I in that direction.

ETA - wrong axis - doht.

 

[rb1957]

@BA,

I grabbed the OP's number.

as above makes sense … 2 at +d, 2 at zero, 2 at -d. = 4*A*d^2

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

 

[BAretired]

Or, more precisely:

Ix = A(4d²)

Iy = A*6(d/2)²) = 3Ad²/2

BA

 

[3DDave]

Something looks amiss for Ix.

Ha - fixed it as I typed.

 

[dik]

The attached has a part that shows the I for bolts... uses SMath... free for download and use...




Dik

 

[rb1957]

@BA,

you're completely correct (as usual).

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

 

[Amar-Dj]

Hello,

I am reposting my question to ensure that I understand the why the inertia is expressed in mm².

I want to ensure that the ''mechanics'' in coherent even with an inertai expressed in mm².

I understand now that the I_{(x or y)} is actually expressed in mm⁴/mm².

The areas from the moment inertia (the denominator ''mm²'' in ''mm⁴/mm²'') gets canceled out in the formula of the forces due to the moment. Here is why (If am not mistaken):

From my understanding is that the P_Mx=(M_x*r_y/I_x)*A (Note that in this case, I_x is in mm⁴).....The area A from this formula is canceled out with the A from I_x= A.Sum(r²).

That's why, according to my understanding, it's enough to express the moment of inertia of a bolt group in mm⁴/mm².

Am I on the right path?

Thank you all!

 

[BridgeSmith]

Yes Amar-Dj, you've got a handle on the units. With all of the posts, you should have a pretty good handle on how to calculate Ix and Iy, and the polar I (Ix + Iy, or the equation I posted).

I wonder if we may have skipped over a more important aspect about how and when to apply the moments of inertia once you have calculated them. In my design work, I've never had occasion to need Ix or Iy for a bolt group individually; only combined for the polar I to calculate forces due to an in-plane moment (producing shear on the bolts, for web splice plates on an I-beam, e.g.). For out-of-plane moments (producing tension or prying on the bolts), I think you would typically calculate the moment of inertia about an axis closer to the compression edge of the plate. For that application of load, you'll have bolts in tension, but I believe the compression will typically be carried through contact of the plates (unless there are standoff sleeves around the bolts carrying the compression, of course).

Rod Smith, P.E., The artist formerly known as HotRod10

 

[rb1957]

the dimension of (area 2nd) moment of inertia in L^4 (eg in^4, mm^4).
it is not, repeat not, mm^2 (as you posted originally); because your original post missed the Area of the fastener.
Yes, bending stress = My/I … ie in.lbs*in/in^4 = lbs/in^2.
and the fastener load is stress*area.
Now you can say load = My/(I/A) (work through the math, it should take you 5 minutes) and I/A is your original post.
so maybe some wires got crossed somewhere ? just don't call I/A moment of inertia.

And I would suggest being very careful with this simplification as it can easily lead you wrong.
eg, what if the fasteners have different areas ? (and please, other posters don't tell me, I know)

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

 

[Amar-Dj]

Hello BridgeSmith,

I totally understand how inertia is calculated by unit of area (mm²). That's easy : I(x or y) = Sum (r_i)² * A_i [in mm⁴] or I(x or y) = Sum (r_i)² [in mm⁴ / mm²]

My question was just regarding the use of such a result in (mm⁴/mm²). My guess was that the typical calculation of a group of bolt, normally P_Mx=(M_x*r_y/I_x)*A should be expressed this way (Note that in this case, I_x is in mm⁴).....The area A from this formula is canceled out with the A from I_x= A.Sum(r²)

All this is simplified then to an inertia expressed in mm² and the P_Mx is expressed as follow: PMx=(M_x*ry/I_x) (without multiplying it with the area of the bolts). If that's reason, then everything is ''mechanically'' coherent.

Thank you all for your answers

 

[rb1957]

I(x or y) = Sum (ri)^2 * Ai [in mm4] … correct

I(x or y) = Sum (ri)^2 [in mm4 / mm2] … incorrect, this is not I

I(x or y)/A = Sum (ri)^2 [in mm2] … correct, for the special case when all fasteners are the same A (and the same material)

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

 

[BAretired]

That's why, according to my understanding, it's enough to express the moment of inertia of a bolt group in mm4/mm2.

The property called Moment of Inertia has units of mm^4. In the case of bolt groups, the property can be calculated as Σ(Ab*r²) where Ab is the area of bolt and r is the distance normal to the axis under consideration, to the centre of gravity of the bolt group.

If bolts within a group were variable in diameter, Ab would need to be included for each term in the summation. Normally, all bolts in a group are the same size, so the term you are addressing in this thread is really I/Ab = Σ(r²) with units of mm^2; it may be referred to in some texts as Moment of Inertia but that is not strictly correct.

BA

 

[DaveAtkins]

The area of the bolts is ignored when calculating the moment of inertia of a bolt group, because we are not interested in the STRESS in a bolt, we are interested in the FORCE in a bolt. Therefore, it is cleaner to assign an area of 1 (without units) to each bolt.

DaveAtkins

 

[BridgeSmith]

If anyone is interested, I found the formula for the polar moment of inertia (I_p) I posted earlier. It's in the commentary of the older versions (4th thru 7th Editions) of the AASHTO LRFD bridge design spec. (Equation C6.13.6.1.4b-3).

Rod Smith, P.E., The artist formerly known as HotRod10