Bolt loads for beam rigidly connected to overhead support
Bolt loads for beam rigidly connected to overhead support
(OP)
I must support a load with a beam as per the attachment. There are two pairs of bolts at each end. This appears to be a statically indeterminate situation for bolt reaction calc's. is there a way to determine or estimate the tensile load in each pair of bolts?
Thanks.
Thanks.






RE: Bolt loads for beam rigidly connected to overhead support
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: Bolt loads for beam rigidly connected to overhead support
Your connections will exhibit considerable prying action and you may want to reconfigure your connections a little to reduce the impact of the prying action.
RE: Bolt loads for beam rigidly connected to overhead support
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: Bolt loads for beam rigidly connected to overhead support
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RE: Bolt loads for beam rigidly connected to overhead support
RE: Bolt loads for beam rigidly connected to overhead support
if you want more detail, this is a simple doubly cantilevered beam ... a standard textbook problem. this will give you the fixed end moment that you can apply the the bolts as a couple.
if you want to minimise the prying mentioned (it looks like you're clamping the beam upper face to the support), add some washers between the beam and the support.
if you really want to simplify the analytical problem, put more washers under one set of bolts so that only one pair is working at a time (then bolt load = 15000/4).
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RE: Bolt loads for beam rigidly connected to overhead support
Based on the above, I would estimate the force in each bolt to be: (15,000 lb. + 5 ft. x 25 lb/ft) / 8 = 1890 lb. Say, 1900 lb. For bolts supporting an overhead load, a generous safety factor is called for. IMHO, differences in loading among bolts cannot be calculated accurately... for a real project and in this particular case.
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RE: Bolt loads for beam rigidly connected to overhead support
Page 1 of my attachment is how I worked it according to rb1957, as I understand it. Page 2 is a different method, which I'm not sure about. Both methods yield results quite different from SlideRuleEra's method.
Am I in the ballpark?
RE: Bolt loads for beam rigidly connected to overhead support
RE: Bolt loads for beam rigidly connected to overhead support
RE: Bolt loads for beam rigidly connected to overhead support
RE: Bolt loads for beam rigidly connected to overhead support
representing this end moment as a bolt tension load and a triangluar compression is fine, though i'm not sure this is what you did. for me, i'd say the couple, the tension bolt load would be M/(2/3*5) on a pair of bolts; in addition to the 15000/8 direct tension load. My moment load is about 50% bigger than yours ... your comparable bolt load (20384) is about 111000/5.
SRE's approach, which i think is close to 15000/8*SF, is fine.
An interesting part of this problem is the preload you're going to apply to prevent the joint from gapping under load.
another day in paradise, or is paradise one day closer ?
RE: Bolt loads for beam rigidly connected to overhead support
RE: Bolt loads for beam rigidly connected to overhead support
Fortunately, there is a way to account for "partially" fixed supports, but it is often overlooked, I'll explain:
Fixity of beam supports is a continuum, it is not a case of either:
Simple Supports: M = PL/4
or
Fixed Supports: M = PL/8
The denominator can be any number that is greater than or equal to 4, or less than or equal to 8. To accurately pick a reasonable number is an experience based judgment call. Look at all available info. On this problem there is quite a bit to suggest a low number is appropriate:
1. The beam is very rigid, for the applied load.
2. Spacing of the bolts in a group is quite small.
3. The beam overhang beyond the last bolts is small (1.5").
Based on my opinions on these issues, I judged that a denominator of 4 was justified. Using that number, the problem is much simplified, but still a reasonable approximation of reality. In the "pre-software" days, solving problems this way was common.
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RE: Bolt loads for beam rigidly connected to overhead support
but i do agree with you, that this is probably closer to a SS beam.
one thing to notice (I think this is your point SRE) is that the for the SS beam the moment curve is 0 >> PL/4 >> 0 along the span and for the double cantilever it is -PL/8 >> PL/8 >> -PL/8 so a partially fixed beam is somewhere between these two limits, say -PL/16 >> 3PL/16 >> -PL/16
another day in paradise, or is paradise one day closer ?
RE: Bolt loads for beam rigidly connected to overhead support
The assumption that I forgot to explain is that if the beam can reasonably be assumed to be acting as simply supported, then concluding that the beam has simple supports is reasonable, too.
From that point, the only remaining question is where should these (assumed) simple supports be located.
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RE: Bolt loads for beam rigidly connected to overhead support
your "Fixed Supports: M = PL/8" confused me ...
another day in paradise, or is paradise one day closer ?
RE: Bolt loads for beam rigidly connected to overhead support
Now, if you're relying on end fixity, then you need consider prying.
Any thoughts?
RE: Bolt loads for beam rigidly connected to overhead support
another day in paradise, or is paradise one day closer ?
RE: Bolt loads for beam rigidly connected to overhead support
I assumed the beam to be ridged and took moments from one end of the beam. In addition I assumed a unknown bolt load per unit length and called it µ, so each bolt pair was given a length from one end of the beam i.e. la, lb, lc ld.
la=1.5
lb=5
lc=54.25
ld=57.75
My equation ended up looking like this:-
15000*59.25"/2 = µ*2* (la^2 + lb^2 + lc^2 + ld^2)
solve for µ
µ= 35.25
so the maximum bolt load occurs at 57.75 therefore 57.75*35.25 = 2035.6875
now for a pair of bolts at 57.75 then the force is doubled to 4071.375 and the next nearest pair at 54.25 would be 3824.625.
So because the beams symmetrical I ignored the bolt forces close the the pivot where I took moments and considered the pair of forces I calculated above to check for equilibrium
ie forces at each end of beam = 3824.625 + 4071.375 = 7896
so if I double this I get 15792
upward forces = downward forces 15792=15000 well not quite but a 5% error is acceptable to me and it also agrees pretty closely with just dividing the total downward force by the number of bolts.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein