Bolt loads for beam rigidly connected to overhead support

[bsmet95]

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.

 

[KootK]

You'd need to post a sketch of your connection for us to say for sure. That said, a simply supported beam is generally assumed to deliver only vertical shear, equally distributed to all bolts, at each end.

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.

 

[Teguci]

Read through this thread -

http://www.eng-tips.com/viewthread.cfm?qid=402110

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.

 

[KootK]

I had not realized that we had two columns of bolts at each end here. I agree with Teguci's response. Once you know the forces coming through the connection, you ought to be able to use the instantaneous centre of rotation method to distribute those forces to the bolts,

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.

 

[SlideRuleEra]

What are the properties of the beam? I see that it is 59.25 (inches?) long with the 15,000 lb. load at the (center?). How the beam tends to deflects under load will either complicate (flexible beam) or simplify (rigid beam) the estimate.

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[bsmet95]

W6x25#. Ix = 53.4 in^4, Sx = 16.7 in^3

 

[rb1957]

i'd start with 15000/n, i think n = 8.

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).

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

 

[SlideRuleEra]

Here is what I would do (and have done in similar situations). For analysis, assume a simple support at the center of each bolt group. I'll continue to assume the dimensions are in inches and load is at the center of the beam. That makes the W 6 X 25 simply supported with a span of 54.25" (say 54"). Bending stress in the beam is ok, so is shear. I calculate maximum deflection from the 15 kip load at 0.032" (1/32"). Deflection = L/1700. And that is at mid span - deflection at the bolt groups would be much less.

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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[bsmet95]

I appreciate the help, but am not quite clear.

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?

 

[Teguci]

SRE's method assumes that the bolts will yield and the reaction at each end can safely be distributed equally to all 4 connections. He then checks that assumption by doing a deflection sniff test and notes that the anchors will need to yield much less than 1/32" to allow for the pinned connection assumption.

 

[jayrod12]

I'm with SRE on this one. if you had a longer beam with a more significant deflection then it may warrant further analysis, but it's likely that local yielding of the plates would allow enough rotation that the bolts equally shared the load even in those cases.

 

[bsmet95]

OK, thanks.

 

[rb1957]

i'd suggest reducing your span (L) to be to the CL of the end bolt pairs (like SRE says, about 54").

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 ?

 

[bsmet95]

Thanks, everyone.

 

[SlideRuleEra]

bsmet95 - I reviewed your calcs, and one thing "jumped" out that will make a difference. This is, your assumption that M = PL/8 (Beam Fixed at Both Ends). It takes a lot to truly "fix" a beam. Note that the spacing of the bolts in a group very small (3.5") compared to 54" span.

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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[rb1957]

umm, err, the moment of a simple support is mid-span, so it'd design the beam, not the bolts.

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 ?

 

[SlideRuleEra]

rb1957 - I believe we are looking at the problem in different ways. I agree with you that selection of the Moment equation denominator (4 to 8) concerns the beam, not the bolts.

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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[rb1957]

i agree it's a simple approach (as SS beam).

your "Fixed Supports: M = PL/8" confused me ...

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

 

[NS4U]

To back up a little, it was mentioned that these bolts should be designed considering prying. I disagree. If the beam is designed to be simply supported and does not require some level of fixity at the ends, then the prying action, although present is self limiting. Or in other words as the bolts are 'pryed' the bolts will stretch or the flange will bend which in turn reduces the effect of prying.

Now, if you're relying on end fixity, then you need consider prying.

Any thoughts?

 

[rb1957]

prying will limit itself with plastic behaviour of the end ... if you can live with that ...

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