Load Distribution Using Spreaders [Archive]

john_tullamarine

7th Mar 2014, 02:22

First point - not enough information -

(a) what are the spreader physical characteristics ?
(b) what realistic effective area of the spreader can be used to distribute the load ?
(c) what are the aircraft floor characteristics and limitations ?

The problem is that we are trying to determine what the floor structure is seeing in terms of

(a) intensity of loading - kg/sqm
(b) running load - kg/m

I think we might want to see some more information ... ?

underfire

7th Mar 2014, 04:49

JT is correct, more data is needed...

The spreaders must be of suff. strength to spread the load. The spreader material and thickness has its own load distribution characteristics. If the spreader were to flex under the load, the load is not uniformly distributed, and one cannot use the full area of the spreader. (being factious, cardboard spreaders) If the spreaders are marginal, then you can only distribute the load using the area of the spreader that is the same as the load, not extended. So now you may have another component which is the actual dimension of the load.

In looking at the running load, I look at this as dynamic load. Dynamic load has a moment arm component due to the height of the load. This is unknown.

Flooring of the ac is designed to distribute the load over that surface area. Flooring is typically shown with a bearing capacity per sq m of dead load. This is important as dead load only assumes dead weight as a vertical component. Dynamic loading takes into account the height of the objects force at vectors, such as the vertical component of takeoff, as well as any sloshing component of liquid loads.
Much like the spreaders, if the load bearing kg/m2 of the flooring is exceeded, then it becomes a point load, not a distributed load.

For calculations sake, in your example, if the flooring could handle 1000kg/m2, the flooring would span the gap, and if the spreaders were of sufficient strength to not deflect under the load, you would use the entire surface area of the spreaders, or 444kg/m for the dead load.

The running/dynamic load is a function of the height, so...

HIGHFLY33

7th Mar 2014, 06:43

Question was posed during recurrent Training, this is the information we have..

Instructor is adamant that the outside dimensions of the area defined by the spreaders disregarding the space inbetween is the area to use for the calculations thus giving a floor load of 444kg/m2

I disagree to this and say in order to calculate the area load in order to ensure floor load limitations are not exceed, you only calculate the area in contact with the floor.. thus giving 667kg/m2

Spreaders assumed to be suitable...

Does this change anything ??

john_tullamarine

7th Mar 2014, 07:05

Instructor is adamant that the outside dimensions of the area defined by the spreaders disregarding the space inbetween is the area to use for the calculations thus giving a floor load of 444kg/m2

I doubt that the aircraft OEM would have a similar view ... Effectively, we are concerned with localised penetrating or crushing stress loads so the actual contact area is that which counts. To take your trainer's view to the extreme, I presume that he would come up with a similar view if the shoring were, say, 4 sheets at each corner with negligible contact under the box but still with a few square feet extending outwards ? Basically doesn't hold water ...

Spreaders assumed to be suitable...

Therein arises my question .. there is suitable and not quite so suitable.

Reference to the OEM Weight and Balance or Loading Manuals would be the first port of call to resolve what the OEM might have desired. If these documents don't exist or don't resolve the question, then one flicks an email to the OEM Tech Support folk with the question ...

I might disagree a little with underfire's comments. Running load, I suggest, is more concerned with fuselage bending capability rather than dynamic loading. The OEM will have designed the thing to suit the design G-loads which should cover most of the likely to be encountered in-service loading.

Chu Chu

7th Mar 2014, 22:31

This is probably what John is saying, but I don't think the portions of the spreaders that project beyond the load width-wise (i.e. the parts at the top and bottom of the illustration) should count at all. Imagine you had a trench the width of the load, and placed spreaders overhanging the trench on both sides, then the load on top. The spreaders would just flip up and follow the load into the trench.

underfire

8th Mar 2014, 05:20

Structurally, not to overthink this, at all... :hmm:

Structurally, if one uses distributed load method of calculation of kg/m2, you use m2, not parts of M2 (hope that makes sense)
There is the entire surface area of the load on the deck. The area of the load is 1.5x3 or 4.5 m2. 2000/4.5= 444 kg/m2

If you used 667 kg/m2, that 4.5m2 surface area taken by the load would equate to a little over a 3000kg load (50% more)

Thats no way to load an aircraft!

That is why, if you do not have to, you make the spreaders the same size as the load, only make them larger if you are over the psm loading

HIGHFLY33

8th Mar 2014, 08:32

UNDERFIRE...

what if the floor load limitation is 500 kg/m2
wouldnt we be exxceeding the limit ??

Structurally, not to overthink this, at all... :hmm:

Structurally, if one uses distributed load method of calculation of kg/m2, you use m2, not parts of M2 (hope that makes sense)
There is the entire surface area of the load on the deck. The area of the load is 1.5x3 or 4.5 m2. 2000/4.5= 444 kg/m2

If you used 667 kg/m2, that 4.5m2 surface area taken by the load would equate to a little over a 3000kg load (50% more)

Thats no way to load an aircraft!

That is why, if you do not have to, you make the spreaders the same size as the load, only make them larger if you are over the psm loading

you are referring to the load on the deck, ie the area in contact with the floor right ???

john_tullamarine

8th Mar 2014, 09:02

The spreaders would just flip up and follow the load into the trench

Perhaps an extreme example of the point. However, the underlying consideration remains valid - shoring stiffness.

The area of the load is 1.5x3 or 4.5 m2. 2000/4.5= 444 kg/m2

'fraid not. The floor knows naught of the box as it isn't touching the floor. All the floor knows about are the two shoring planks .. which are. Providing that the shoring stiffness is adequate to spread the load the floor only is aware of the shoring planks' loading stress. ie 2000/3 or 667 kg/m2. By a process of deduction, the floor then figures that the load is 3 x 667 ie 2 tonne.

If you used 667 kg/m2 ..

'fraid not. Respectfully suggest you revise Mechanics Lesson One.

Looking again at the sketch again, running load is kg/m rather than kg/m2. Missed the error first time around ..

Chu Chu

8th Mar 2014, 10:45

I think see where I was going wrong. I was thinking of the spreaders as cantilevering out width-wise, and therefore not doing anything in that dimension unless rigidly connected to the load. But I now think the correct model is spreaders loaded in the center width-wise (by the edge of the load), and thus being effective across their full width. (Or potentially being effective if they are sufficiently stiff.)

HIGHFLY33

9th Mar 2014, 00:32

thank you John for your comments and insight,

I have corrected the erronous unit on for running load... :ugh: my bad..

Is anyone familiar with a good solid refrence for this.. ie in an IASA loading manual or in any icao or aircraft manufacture documentation in order for me to demonstrate that my calculation is correct ie 667 kg/m2 .

appreciate all contributions , keep it coming if you think you have further that can add to the discussion..

underfire

9th Mar 2014, 03:22

Sorry, but you are cherry picking the constants. One cannot use point loading numbers with a uniform distributed load equation.

The deck is rated by a Uniform Distributed Load. If the deck is rated per sq foot or per square meter, structurally, it is assumed that the load is distributed within that span.

http://i62.tinypic.com/2rrqlp4.jpg
It is the aircraft deck assembly, not just the point of contact of the skin.

It is not a concentrated load as you are describing.

A very good example would be a parking structure. Is is structure rated for the contact patch of the 4 tires? Do you have 4 small squares which you must park on? You do not, parking structures are rated at 50psf, 130kg/sqm because the structure spans the load distribution.

Using your analogy, what would be the required loading for a parking structure, 1360kg/sqm?

john_tullamarine

9th Mar 2014, 10:16

Negative ... never really was keen on cherries ...

Shall we engage in a little game ...

(a) first, for convenience, put the 2T load on a single stiff (and, presuming an idealised situation .. massless) spreader of 3m by 1.5m ... are we happy that this alters nothing so far as loads are concerned ?

(b) let's cut the two load spreader planks of the original sketch into four by the expedient of cutting then at their midpoints .... are we happy that this alters nothing so far as loads are concerned ?

(c) let's replace the original two planks (now four) by four smaller items of, say, 0.5m by 0.5m and locate them under the corners of the single stiff spreader. Has this altered the loads at all ? By your original logic, evidently not.

(d) continue this until we have four little buttons of, say 1cm by 1cm. Has this altered the loads at all ? By your original logic, evidently not.

Over to you, good sir ....

underfire

9th Mar 2014, 12:09

no matter how you slice it, I have shown how we actually do it..

HIGHFLY33

9th Mar 2014, 15:00

Underfire...

Thanks a million for your input, albeit it confused me further,

Surely we are trying to avoid exceeding the floor load limitation,

As per your assumption for this to be true UDL will be transferred to the whole span of the floor... So why is there a need for spreaders ?? Is that not to ensure that the load is distributed :confused:

For our example lets say I have an airplane with the following limitations as per the OM
RL 560 Kg /m and AL 450 kg m/2 (made up figures)

As per your method we are within limits, as the UDL 1 within limitations...
However here is whereabouts my understanding comes crumbling down..

How can we assume the load is uniformly distributed as the area of contact is limited.. a good example was as John described, when you reduce the footprint of the spreaders.

I am really struggling to come to terms with this assumption.. is there a specific value or special dimension for changing the calculations and consider what you describe as point load... :confused:

If you are telling me that the whole floor is actually acting as a spreader already, then what I'm achieving by using "additional" spreaders... Apart for floor protection and added traffic load..

I really want to get a full appreciation of this..

For curiosity when would you then consider the point load or are we disregarding it altogether ?? (surely the floor panels have a design load limit ... ??

A lot of questions arise from what appeared to be a simple question, please continue to provide your input s, as this is greatly appreciated..:ok:

underfire

9th Mar 2014, 22:32

In the example above, the load is distributed to the supports by using half of the span between the support. Each tributary area of each joist is half the span to the next one on each side. In a system like the one illustrated, you can also use a load factor reduction, assuming that if one member deflects too much, the load transfers to adjacent members.

An aircraft floor deck is a composite system, meaning everything is attached in such a way so the that everything works together, it acts like a giant plate truss, distributing the load among all of the stringers and purlins/sub-purlins through the members and the decking. That is why it is so strong.
http://www.boeing.com/assets/images/companyoffices/gallery/images/commercial/767400-k60492.jpg

Pallets are considered sacrificial. They are used to transport the load, but also to make sure there is an even contact area for the load. Deliberately rigid and flexible. Note that when loading, the pallet jack wheels concentrate the load on a very small area, and this is acceptable

You need to start looking at the load being a concentrated load when it is large load near the loading limits in a small area, usually at or near the rating.

HIGHFLY33

10th Mar 2014, 00:19

Hi Underfire,

This really puts things into a whole other perspective,
Thank you for the illustrations and picture

I guess I can sum it up and say : the floor structure itself transfer the loads and absorb the forces :D

I will Spread the word (no pun intended):E

thanks a million,

john_tullamarine

10th Mar 2014, 08:13

no matter how you slice it, I have shown how we actually do it..

If we have OEM material (Loading Manual, etc) then we follow that. Otherwise, I think we shall have to agree to disagree .. I will leave it for the specialist stressmen to enter the fray in due course ...

Your profile is rather less than informative .. might we enquire who is "we" ?

underfire

12th Mar 2014, 08:11

JT, c'mon now...'we' always refers to me, myself, and I...

Hyfly, I would suggest you download the Airbus document "Getting to grips with Weight and Balance" and passing that direct information around.

aside from that, this may help clarify...

Area Load:

The contour area is the external contour of the contact points on the floor.

http://i59.tinypic.com/66kk8y.jpg

Running Load:

The length to take into account is the length of the contact points on the floor. (specifically, length in flight direction)

http://i60.tinypic.com/2ebfn6b.jpg

keith williams

12th Mar 2014, 08:50

The problem here appears to be one of terminology.

If you go to page 221 of the Airbus document you will find a third limit called the "Contact Load Limitation".

a) Contact load limitation

• Definition: The contact load limitation is the maximum load acceptable in direct contact with the aircraft floor per surface unit. This limitation is used to prevent the load in direct contact with the floor from exceeding the capability of the horizontal floor panels (metal sheet, honey comb sandwich panels).
Note:

The Contact Load Limitation is called “Local load” limitation in the Airbus Weight and Balance manuals.

• Unit: kg/m2 or lb/ft2

S = Contact Area
W = Contact Load

Contact Load = Weight of the piece / Contact Area

The contact area is the surface in direct contact with the floor.

The JAR/EASA ATPL Mass & Balance syllabus includes only two limitations. These are:

1. Static Load Limit. This is the equivalent of the Contact load Limit in the Airbus document.

2. Running Load Limit. This is as described as the Running Linear Load Limit in the Airbus Document.

underfire

12th Mar 2014, 09:14

Yes, there is certainly terminology in place, which is why I ref the entire document.

There is also the Load Factor on page 226 that is for the 'apparent weight' of a heavy load, ie the centroid of the mass which would include the height of the object.

Reference back to the OP diagram and the question. That is a uniform distributed load per the document.

keith williams

12th Mar 2014, 10:09

That is all true.

The OP posted the same question in another website (BRISTOL GROUND SCHOOL). This suggests that he/she is doing the ATPL course or has recently completed it (probably with BGS). In the context of EASA ATPL ground school the term Static Load means:

Static Load Load / Contact Area

Using this definition and the data in the question posed the answer would be 667.

Reference back to the OP diagram and the question. That is a uniform distributed load per the document.

It is also a situation in which the Contact load Limit must be applied.

Although you gave the title of the Airbus document, you actually quoted only the page that supported your argument. You did not quote the page that supported the view of JT and some others in this thread. That may help you to win arguments but it is not a good way to enhance mutual understanding.

john_tullamarine

12th Mar 2014, 21:48

More importantly, a generic document is not the appropriate reference to determine the OEM's views, rather the particular aircraft's Loading Manual (by whatever name titled).

While it is rational to follow the OEM's prescribed Type practices, in the absence of such, pragmatism requires that the reasonably most conservative interpretation should apply.

underfire

12th Mar 2014, 21:57

not at all...Note in the page ref, example #2 where running load is exceeded.

I also explained several posts back as to when you start looking at concentrated loads, and when you look at the height of the load.

The contact load portion you ref is not applicable to the OP. Contact loading assumes there are no spreaders, it is the direct contact point of the object. The OP specifically ref spreaders.

and speaking quoting only one page...

You quote page 221, note the next page, 222. Contact load with spreaders and without. Note the surface area used with the spreaders.

http://i58.tinypic.com/2637uxs.jpg

john_tullamarine

12th Mar 2014, 22:33

I also explained .. when you look at the height of the load.

Perhaps you might expand on this ? Considerations of restraint we understand well, but for general loading considerations involving floor limitations ?

keith williams

12th Mar 2014, 23:16

underfire

You state:

The contact load portion you ref is not applicable to the OP. Contact loading assumes there are no spreaders, it is the direct contact point of the object. The OP specifically ref spreaders.

And:

You quote page 221, note the next page, 222. Contact load with spreaders and without. Note the surface area used with the spreaders.

But in the same post you have included a calculation for contact load, which includes spreaders:

Contact Load = (500 + 20 ) / 0.48 = 1084 kg/m squared.

If we apply exactly the same process to the OP we get

Contact Load = 2000 / 3 = 667 kg/m squared.

underfire

13th Mar 2014, 07:17

KW, That does not compute:

In the example shown in post, the area load with floor spreaders is 1.2x0.5,

In the OP example, the floor spreader load is 1.5x3=4.5, 2000/4.5=444.44

Note that area calc for a contact load with spreaders is the same as UDL.

JT,

With heavier loads, especially irregular shapes, you must look at the CG of the load. The resultant force, when the aircraft rotates or banks is very important. Restraints, depending on the angle of restraint, size, shape, or where the object is located, may not be able adequately distribute the load to the floor uniformly. The spreaders must be designed not to flex and create a punching shear or excessive bending moment in the floor. Max angle on a strap is 30 degrees, so a high CG of the load is going make things difficult. Straps are not infinitely rigid, so any flex redistributes the load to others.
Bank the ac 25 degrees, and that offset cg load will distribute unevenly to the spreaders, and any flex in the spreader will create a concentrated load.

In general, this is addressed in the Load Factors and Apparent Weight calcs, and these are certainly oversimplified with a perfect center of mass. It is the unusual loads, especially the ones that are near floor limits, that should be addressed.

john_tullamarine

13th Mar 2014, 21:25

underfire,

Perhaps you could PM me with a synopsis of your background so that we don't waste too much time going around in circles ? I do find some of your posts rather perplexing .. as does my good colleague, Mr Williams, I suspect ...

you must look at the CG of the load.

fair comment. However, other than for very asymmetric loads, I don't think folks generally worry too much in practice .. although the design folk certainly run this consideration into restraint system calculations.

Restraints .. may not be able adequately distribute the load to the floor uniformly.

an innovative thought ... the restraints are more likely to do just as well in turning flight as for straight and level. As to whether that be good or bad is the province of the design folks.

The spreaders must be designed not to flex and create a punching shear or excessive bending moment in the floor.

as suggested previously in several spots.

Max angle on a strap is 30 degrees

why ever might that be ? The restraint designer will use whatever angles suit the circumstances, considering geometric limitations, practical workability, and the boss's preferences ....

so a high CG of the load is going make things difficult.

why ? All taken care of in the back room head scratching.

Bank the ac 25 degrees, and that offset cg load will distribute unevenly to the spreaders, and any flex in the spreader will create a concentrated load.

again, unless we are looking at wildly unbalanced flight - and we ought not need to worry too much about that - how does the bank angle cause this to happen in a manner different to straight and level flight ?

For any real life half sensible spreader, the second statement is invalid.

In general, this is addressed in the Load Factors and Apparent Weight calcs

matters for the design PE folks to concern themselves with .. the pilots etc., only have to follow the load system requirements which should make the exercise very straightforward. Unless one is looking, say, at a basic pilot weight and balance theory course in which such things might be introduced for background understanding ... ?

It is the unusual loads, especially the ones that are near floor limits, that should be addressed.

likewise. The pilots etc. should just follow the system specified and the back room folks should have made sure that it all works according to Hoyle.

I look forward to your continuing responses ...

Alex Whittingham

14th Mar 2014, 13:30

I agree with JT's analysis of the physics and the ultimate authority of the OEM.

Reference the terminology and exam questions, there is no apparent dispute about what 'running load' means so I will disregard that. It seems we have two Airbus definitions which are 'Contact load' and 'Area Load'. There is only one definition in the JAA learning objectives which is related to this and that is:

Area Load or Floor Load
The load (or Mass) distributed over a defined area. Units of measurement used:
SI: N/m2, kg/m2 ; Non-SI: psi, lb/ft2

That definition could match either the Airbus 'Contact load' or the 'Area load', depending on the defined area.

KW, 'static load' is not an LO term, and I do not see, given the imprecise definition above, that you can deduce that 'area load'/'floor load' as defined in the LOs = the Airbus term 'contact load'. Do you have any more info on this?

keith williams

14th Mar 2014, 16:02

Alex, I must confess that I had not spotted the fact that the old “Static Load” had been replaced by “Area Load” in the learning objectives.

You are quite correct in saying that the definition now used in the LO’s, could match either the Airbus 'Contact load' or the 'Area load', depending on the defined area.

The thing that caught my eye in all of this was the fact that both of the numbers given in the original question (667 and 444) could be correct. 667 is correct for the AIRBUS Contact Load and 444 is correct for the AIRBUS Area Load.

As you say

I do not see, given the imprecise definition above, that you can deduce that 'area load'/'floor load' as defined in the LOs = the Airbus term 'contact load'

But your instructor appears to have done exactly that. When this same question appeared in your company website, he selected 667 (which matches the AIRBUS Contact Load) as being the correct answer.