Bolt tension due to moment about a ring of bolts 2

[Polecat]

The TIA-222.G.3 standard says that the max tension on a single bolt (in a


baseplate attached to a pole) is calculated from the following formula:
1.02*pi*Mu/(n*Dbc) (assuming that there are at least 12 bolts and a minimal axial load is present)
Where:
Mu is the applied moment in inch-kips,
n is the number of bolts, and
Dbc is the bolt circle in inches

A dimensional analysis shows that the resulting answer is in kips of tension (or compression), but I cannot figure out how they derived the formula. TIA say that the bolt circle is treated like a ring of steel with a diameter equal to Dbc.

You can convert the total area of all the bolts into an equivalent area in a ring, and then calculate the section modulus of that hollow circle. Then you can do a M/S to get the max stress, which could then be applied to a single bolt to get its tension. However, I can't see where a formula that simply contains a pi/D ratio is anywere related to a section modulus.

Can anyone shed some light on this for me??

http://www.spiraleng.com

 

[JLNJ]

Maybe it has to do with the distribution of the compression force under the ring. This would lower the bolt force.

 

[KootK]

We're talking about a base plate that is nut supported rather than concrete supported, correct?

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.

 

[desertfox]

Hi

Try this site:-

http://machinedesign.com/fasteners/how-bolt-patterns-react-external-loads

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[transmissiontowers]

ASCE 48 and ASCE 113 has an equation for bolt force that is probably the same as the EIA when a monopole is used with a baseplate on leveling nuts. The general method is to find the MOI of the bolt group and find the bolt force as Teguci has done.

_____________________________________
I have been called "A storehouse of worthless information" many times.

 

[Teguci]

@desertfox - good site that actually gets into the numbers - thanks.

From the site -
"It can be shown that [2θ j, j = 1,..N] are equally spaced on the unit circle (R = 1). Therefore, their centroid is at the origin and ∑ cos(2θ j)/N is its X-axis coordinate. Consequently,∑ cos(2θ j) = 0."

http://machinedesign.com/fasteners/how-bolt-patterns-react-external-loads

Using Lagrange's cosine summation, you don't need this leap of logic. You calculate directly that the sum of the cosines = N/2.

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities

Using the cosine addition formula and Lagrange's trig identities, you can also show that ∑(j=1 to N); cos^2(2 pi/N x j + a) = N/2 where a is any initial starting angle. This tells us that the moment of inertia for a bolt circle, where you have at least 3 bolts, is constant regardless of the angle being looked at.

And I used to hate trig class.