Equivalent Rectangular Load

[jreit]

I have a triangular load perpendicular to the supports on a precast plank that I'm not sure how to analyze. I'm treating the plank as a beam and converting the triangular load to a rectangular load but I'm concerned about being non-conservative. The plank is sitting on concrete beams right now. The load will be applied before concrete is poured to make it composite. Sketch attached for reference. Thank you in advance.

http://files.engineering.com/getfil...-ded6-484f-b728-534f6e5c61dd&file=Loading.pdf

 

[KootK]

No problem

1) Calculate the peak moment due to the triangular load.
2) Calculate the rectangular load that would produce the same peak moment.
3) Do the same for shear.
4) Do the same for deflection.


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.

 

[jreit]

Thanks KootK, but how exactly do I account for the fact that the load decreases along the length of the beam in the case of the triangular load?
It's simple enough if the load was spanning in the same direction as the plank/beam but that's why I wanted to use a rectangular load to simplify the analysis.

 

[SlideRuleEra]

jreit - Exactly what do you need to determine?

1. Will the plank support the load? (Yes - No)?
2. Deflection of the plank? (How much) or (Acceptable - Not Acceptable)?
3. Load distribution on the supports?
4. Something else?

Since this is a three dimensional problem, either the structural properties of the plank need to be accurately defined or various simplifying assumptions will have to be made to get an approximate answer.

Other questions:
Will the weight of the (fresh) concrete being placed be in addition to the 13 KSF and 6.50 KSF loads?
More accuracy on the supports is needed. Based on my reading of the sketch, the span could be anywhere from 8 feet to 12 feet.

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[jreit]

It's a short term load, so I'm only concerned with strength. I'm trying to determine the maximum shear/bending the plank will experience so I can analyze it against its rated capacity which I've already calculated.
The plank spans 12' between the supports.
The load is a triangular load from 0-13 ksf over a length of 12' and width of 4' perpendicular to the supports placed in the center of the plank to check the worst case loading scenario.
I attached another sketch, bit rougher than the first one.

 

[dcarr82775]

Your sketch has the plank being 18" thick. I might take a 36" width of plank with the average value of load over that same width and look at it is a stand alone beam. I think your 18" plank will spread things out more than that.

 

[dhengr]

Dcarr82775:
Exactly, although he might also take a look at 13ksf on a 1' wide by 18" deep beam spanning something less than 12', depending upon the support beam width. I would look at this because the 13ksf load happens right at the free edge of the slab, and then realize that there would be some help and redistribution from the next foot of slab length, as you suggest.

Jreit:
Your idea of a 6.5ksf uniform load to replace the triangular load would be way over doing it at the zero load slab edge, and would be seriously unconservative at the max. loaded edge of the slab. On a problem like this you might also consider the stiffness of the material (pile?) causing the load. Obviously, a pile of sand weighing 13ksf would load my 1' wide slab/beam element differently than 2'x2'x 4' long blocks of concrete (making up the 4' dim. in your sketch), stacked to 13ksf, which would tend to point load (line load) my 1' wide slab/beam element at your 4' span points.

 

[SlideRuleEra]

jreit - I endorse dhengr's revision of dcarr82775's approach. The heavily loaded near-free-edge portion of the panel is where a problem would occur. This is one of the rare times when establishing a realistic upper bound of loading conditions may be preferable to an exact solution for the entire panel.

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[dcarr82775]

dhengr,

I agree, any number of widths could be picked. I picked 2* thickness. That feels conservative for an 18" thick plank spanning only 12-ft with a rather "small for its size" load of only 13-ksf. Ultimately nothing more than 'engineering judgement'of course

 

[jreit]

Thank you for your replies.
The load has a width of 4', so shouldn't the minimum width of plank that I analyze be at least 4'?
As per my understanding, I should analyze an 18" thick, x' wide plank with an "average load".
@dcarr/dhengr when you say average load, are there any rules of thumb for how I should treat a triangular load? I figured the 6.5ksf rectangular was imperfect and rather crude but wasn't quite sure of a more refined approach.
The load is that of a crawler crane track. There will be timber matting for distributing the load over the entire slab, but these tracks tend to be tapered so I have some doubt over just how much redistribution will occur.

 

[dhengr]

Jreit:
With that kind of load and that size and shape, “crawler crane track” did cross my mind while I was typing, and I almost asked if that’s what it was. That load shape is due to the fact that you are lifting off the front or the back of the undercarriage, and it might be even worse if you are lifting off of one corner of the undercarriage. Then finally, you could have a high percentage of the total load (2)[(13ksf )(12' x 4')/(2)] = 624k on only a few rollers and a few grousers on one end of one track, or one end of two tracks, particularly on a fairly hard surface like timbers on concrete. Your 0-13ksf load representation would be more true of the track on soil. That’s why I brought up the point that with these kinds of loads, you should also make some judgement about the stiffness of your actual load, or its true load application pattern. I had no quarrel with Dcarr’s judgement that 2 times the slab thickness was a reasonable slab width for an imaginary beam to analyze. But, now that we know what the load is, I think I’d want 20-25ksf over a 3' beam width, that also compares with 2 or 3 grousers in track length. Some of your lift planning should involve this kind of thinking. If you rotated the undercarriage 90̊, would you treat the slab structure and the crane better, by having the ends of the tracks over your blue support beams?

 

[SlideRuleEra]

jreit - I believe we need to revist the loading assumptions. What you are describing now sounds different than the load distribution I see in the sketch. For starters, what brand and model crawler crane is this? For a single track to have a 4 feet wide footprint, it must be sizeable.

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[dhengr]

SRE:
It ain’t like we aren’t thinking alike, and it appears to be in unison too.

 

[SlideRuleEra]

dhengr - Agreed!

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[jreit]

I used the Manitowoc crane proprietary software to come up with the loads on a hard operating surface, which to be fair, is a bit of a black box, but it is the most accurate approach I could think of. Only a certain proportion of the tracks see the load as per the software which is where the 12' and 4' dimensions come from. I am fairly confident of the loading being representative of the actual loading.
@dhengr the original plan was the 90 degree rotated which I analyzed to work, but as you know, things change.
If I understand the approach dhengr/dcarr are advocating, the beam width and loading are assumed to best represent how the load distributes itself out over the plank?

 

[KootK]

I'd just throw the loads into a quick FEM run. It'll take you 30 min, you'll be able to be less conservative, and you'll understand the load distribution, or at least a version of it. There's really no great, time saving, hand calc answer to this. We've all been there. PCI manual does have some tips for load distributed that are used for generic plank systems

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]

jreit - Ok, I found the software: Manitowoc Ground Pressure Estimator
Like all software, it needs input information, but we don't know what values you entered.

If you really want to do this by yourself, I will offer a suggestion:
Span = 12', Panel Depth = 1.5', therefore Span / Depth Ratio = 8
The panel will almost certainly be overloaded and possibly fail in shear long before bending becomes a problem. If this is true, it does not matter how the load is distributed on the panel - except that a realistic load distribution will make overloading happen quicker. Also, using typical crane mats will not help.

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[dhengr]

Jreit:
Again, SRE and I are on the same page, but I’ll try saying it in a slightly different way. Some fancy distribution into the whole slab is not important in this particular problem, even though it will exist in some form. Certainly, a uniform load of (13ksf/2 = 6.5ksf) would be unconservative where things would start happening. I wanted a fairly small width beam, with max. load in k/sq.ft., near that free edge, because I don’t want a small area of slab (particularly a free edge which can distribute in only one direction, the direction of a progressive failure) to be a failure starter, which would most certainly lead to a progressive failure mechanism. Of course, check shear and bending in the 18" slab. Do you have any form of shear reinf’g. in this slab? I also agree that the timber mat won’t buy you much, an abrasion cushion btwn. the slab and track, and it will gain you a few inches in dimensions to critical shear surfaces/planes/cones.

Without a doubt, the ‘Manitowoc crane proprietary software’ stays within their safe working loads/radiuses charts, etc. They probably can’t go beyond that without opening a whole new liability can of worms. How wide is a track at the ground, how long, and what’s the c/c width btwn. the two tracks? My analysis would start slightly beyond their max. load, just in case. Over the years, I’ve collected a couple videos of cranes tipping over. We weren’t anticipated the tipping, but after the fact, and during legal discovery, we found that someone was videoing the lift when things went south.