Triangular Frame Base Reactions 1

[RFreund]

See attached calculation. This is a triangular frame (plan view). Like a small lattice tower.
Can someone explain the statics behind the equation: 4M/nD
The equation is used to find the vertical force from the horizontal loads however I cannot figure out the statics behind this.

Xbar is the height to the horizontal force.
V is the horizontal force.

M = moment (11200 lb_ft)
n = number of vertical posts (3)
D = distance between vertical posts (17.5 ft)

The '4*M/d' reminds me of plastic section modulus but I can't get this to work out.

Thanks in advance!

EIT

http://www.HowToEngineer.com

 

[dylansdad]

I seem to recall that this is an "approximate" equation to determine forces in anchor bolts arranged in a circular pattern where D is the diameter of the bolt circle and n is the number of bolts in the pattern.

 

[RFreund]

Ah, yes that was the hint I needed.

basically F=Mc/I
c=D/2
I=sum((D/2)^2) = n*(D/2)^2

F=M*(D/2)/[n(D/2)^2] = 4*M/(n*D)

Thanks!


EIT

http://www.HowToEngineer.com

 

[msquared48]

This loading does not seem very realistic for a carrier constructing a new three-leg cell Tower.

Usually there is a provision for up to three carriers at three different levels on the tower with a maximum of four antennas in each of the three sectors. Additionally, there are ueually a couple of microwaves thrown in foor good measure too, plus the cable strings and climbing ladder. All this will add to the vertidal, lateraql and icing loads seen by the tower, affecting the foundation.

With only three antennas in a sector it seems like you are going to limit any future expansion on this tower. Just a thought.



Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[msquared48]

Then again. at only 2100#, this is a small tower.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[TLHS]

Huh. I was going to say that I thought that approximating the base as a circle like this when it was only three legs seemed overly optimistic. Then I did the math my way and came up with exactly the same number. Live and learn.

Note, however, that D shouldn't be the distance between posts. I should be the diameter of the bolt 'circle'

 

[IDS]

Note, however, that D shouldn't be the distance between posts. I should be the diameter of the bolt 'circle'

OK, the centroid of the triange (which is also the centre of the circle) is 2D/3 from the apex, so the diameter of the circle is 4D/3

If the diameter of the circle is Dc and the height of the triange is D then:
F = 4M/(nDc) = 4M/(3Dx4/3) = M/D

Which we could have got by taking moments about a line through the two supports in tension.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[RFreund]

Mike - this is actually a triangular tower on top of a water tower, thus the light loading.

IDS - yes I should have noted that D is the diameter that is encompassing the triangle. Its funny because the calculation that I posted is actually part of a different (newer) calculation package which references this 'original' calculation. They must have followed the original calculation blindly because they call out 'D' as the distance between vertical posts which is incorrect.
However I find the diameter to be:
D= side*side*side/(2*area triangle) or Dia = Lside^3/(2*0.5*Lside*Lx)
Because this is an equilateral Dia = 2 x Side/sqrt(3)
Which yields a different result?

Another thing I've noticed in reviewing past calculations which reference ASCE 7-05 is that it appears many of the design wind loads along the east coast have actually been reduced. It seems that they went from using 100mph to 120mph with 0.6 ASD factor. (10,000 vs 8,640) that's almost 14% reduction in wind pressure.

EIT

http://www.HowToEngineer.com

 

[IDS]

RFreund - sorry, in my calculation I assumed that D was the height of the triangle, rather than the side length.

If we have triangle height = H, side length = L, diameter of circumcircle = D then:

D = 2L/sqrt(3)
L = 2H/sqrt(3)
D = 4H/3

so F = 4M/nD is eequivalent to 4M/3 x 3/4H = M/H, which is the same as we would get by taking moments about a line through the two legs in tension (or any line perpendicular to the line of action of the applied load).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/