Billion Dollar Photo? Civil Engineer Opinion Wanted (Vanity)

Street Tracks Gold Shares ^ | Dec 9, 2005 | Travis McGee

[2oakes]

http://www.cwc.org/plastic/pl_htm/PL-00-1rpt.htm

A little bit tangential, but these plastic pallets are rated to take over 200,000 lbs of static load.

[Mr. Lucky]

Don't I feel stupid.

The point is that it's photoshopped. No owner of that amount of gold would permit it to be photographed.

[river rat]

I think what may be confusing some folks - is the fact they can't clearly see there is a CENTER support on each pallet, and a sheet of material bridging the THREE pallet riser beams.

The end of the center support is covered in part by the I.D. label..

This center support would make it possible to stack the shown number of gold bars in the manner shown -- assuming wooden beams and bridging sheets of suitable material and condition.

Semper Fi

[mr_hammer]

I regularly see steel ingots being prepared for a melt furnace stacked that way and there is nothing unusual about the wood skids holding up under the forces you can see in the photo.

I have stacked over 6000 lbs of steel on a wooden skid for storage with little problems.

P.S. many heavy equipment riggers load a 20,000 pound piece of equipment on a few oak 4x4 and ship the machine over the road that way.

[Baraonda]

"We need to know the exact weight of the gold above"

You can't w/o seeing all the actual bars. How many of them are there? That is the missing term.

[PAR35]

On plain wooden pallets!

I don't accept that proposition. From the pictures, the decking of the pallets appears to be 2x4s, and the risers appear to be 4x8x. Much sturdier than an ordinary pallet.

[Baraonda]

Moroever the warehousing of the gold doesn't seem approriate.

[Travis McGee]

I see the three risers, but still, the load does not extend out over the side risers. It's a sheer strength problem, not only compression. The sheer strength of those horizontal wooden floors is only a fraction of their compressive strength.

[Travis McGee]

I have no problem with your numbers, when you are talking about a load (steel beams, heavy equipment on a full length steel skid) bearing down on the risers in compression. No problem at all.

But if you look carefully at the bottom corner of that bottom pallet on the right, you can see it's not lined up in compression, as they would be if the ingots bridged between the risers.

The ingots do not extend over the risers. There is no bridging effect, or very little. MOst of the load is in SHEER where the horizontal floors meet the vertical risers.

As you know, the horizontal wood's sheer strength is only a fraction of its compressive strength.

[MosesKnows]

the ingots in the photos look longer than 7 inches long to me.

The bars in the photo appear longer than 7 inches, or, less than 3 5/8 wide. The ratio of length to width for the gold bars stored at Ft. Knox is approximately 2:1. The bars in your photo appear to have a ratio closer to 3:1.

[annalex]

Look at the thickness of the pallet beams and the fact that the pallet floor is solid -- no gaps between floor boards. I googled images for "wood pallet" and on the first 10 pages I did not see a pallet quite like that. These are special order pallets.

Also, I do not count 80 per pallet, I see 3 x 5 = 15 each corner and 5 x 2 stacked inside, to the total of 70, and that is for the most heavily loaded pallets. Most stacks of pallets has 3-high and 4-high loads per pallet.

[river rat]

NOW - I see your point...
The entire stack height weight WOULD be better distributed if the bars were loaded to put more weight directly over the side risers.....avoiding or minimized exposure to shear forces being applied to any of the bridging planks.

I suspect the bars are being INTENTIONALY stacked inside the perimeter to prevent damage/loss of gold due to impacts. Another possibility is the clown who designed the pallet or wrote the loading procedure, screwed up!

Interesting problem --
If I had that much gold to stack -- I'm certain I would want a more sophisticated stacking technique!

Semper Fi

[Steve Van Doorn]

I would say no. You have a fake pic.

lets say each bar weighs in at 30 pounds each pallet has 84 ingots (7x3x4) and then stacked 6 high this comes to around 15000 pounds exerted on the bottom pallet.

A standard heavy duty wood pallet can take around 3500 pounds (standard metal ones can take 6000 pounds.) It looks like you have 15000-pound load on the bottom pallets you are way over the limits of a standard heavy-duty pallet. Granted these seem to be even stronger pallets then what I am referring to.

It is way over their weight limits in my opinion. Don’t forget to take into account the drop weight when the forklift places the top pallet on the stack.

[Tokra]

Why not have it sent to me via UPS and I'll try to determine if it's real.

[Pete'sWife]

I want to know how you got into my basement to take the photo.

[Travis McGee]

I'd just build these custom pallets with 5 or even 7 web risers, then it's a moot point.

I mean, how much money did they save on lumber, given the value of the load?

[Travis McGee]

There are a lot of unknown variables, but it sure seems like a cheesy and unstable way to stack millions of dollars in gold.

[Hermann the Cherusker]

My only question involves the pallets at the bottom of the stacks . . . How do they hold up the entire load above them without breaking

The load is largely transmitted to the floor through the beams on the sides and center of the pallets.

Ummm ... look at the picture.

Its quite clear the side beams are floating in space on every stack and are not directly supported (they are cantilevered out by the bottom planks, which appear to be 1x8's). The gold is stacked on the top plate of wood of each pallet up to the side beams, but not over them.

The gold is a simple distributed load, but more of the load will be acting on the center support than the ends.

g = the load of gold per foot
M = force on the middle beam
E = force on the outside beam
x = length of gold stack
b = width of beam beyond the edge of gold stack

Solving the normal equations of force and Moment at M:

gx = M + 2*E
E*(x+b) = (g*x/2)*(x/4)

Gives:

M = g*x + (g*x^2)/[4*(x+b)]
E = (g*x^2)/[8*(x+b)]

If x = 2.5 ft, b = 0.25 ft, and g = 900 lb/ft. (2250 lb/2.5 ft.), then:

M = 1739 lb and E = 256 lb

The load from the middle beam is going to be imposed directly down the middle of the stack, with slight distribution to the sides of the middle on each of the stacks below, while the load on the edge beams will act as a point load on the edge of the gold stack and should cause bending and a rotational deflection in the end of the 1x8 plank that lifts it slightly off the stack below, due to a lack of direct support under the end beam.

Most of the load is going straight down the middle to the floor, and should end up imposing a roughly 50 psi pressure on the floor in the 2/3 sq. ft. of the two planks on the bottom pallet that is centered on either end of the middle beam.

The only real question is whether there is a type of wood that can support the roughly 10,500 lbs load in shear being imposed on it in the bottom pallet. I suspect there is, since these loads are near to the type of loads imposed on railway ties - 40,000 lbs on an 8"x14" section of wood. This is 10,500 lbs on a 3"x8".

[RightWhale]

Don’t forget to take into account the drop weight when the forklift places the top pallet on the stack.

The fork lift operator is a seniority position. That is, that is not an inexperienced young operator.

[Lekker 1]

"I'm just guessing, but the ingots in the photos look longer than 7 inches long to me."

My wife also has a hard time judging 7 inches.