Calculating lbs/sq ft.

   / Calculating lbs/sq ft. #1  

raykos

Silver Member
Joined
May 22, 2007
Messages
132
Location
Central Pa.
Tractor
JD2305
Hi all,

If I wanted to build a floor out of 2x's, spaced 16" O.C., covered with 3/4" plywood, and park my tractor on it,(max 3000 lbs.), how do I figure out the lbs/sq. ft. the floor will need to support?

I'm thinking that I probably take the length & width of the wheel base as the sq. ft. "foot print", & then divide the weight by this "foot print"? But, I'm not positive?

Thanks,
Ray
 
   / Calculating lbs/sq ft. #2  
Hi all,

If I wanted to build a floor out of 2x's, spaced 16" O.C., covered with 3/4" plywood, and park my tractor on it,(max 3000 lbs.), how do I figure out the lbs/sq. ft. the floor will need to support?

I'm thinking that I probably take the length & width of the wheel base as the sq. ft. "foot print", & then divide the weight by this "foot print"? But, I'm not positive?

Thanks,
Ray

You really need to get the tire foot print and the load on each wheel to do what you want to do. That will give much higher loadings than what you propose.

If it were me I sprinkle some powder or flour on some pieces of plywood and drive the tractor on them, mark the foreward and rear edge of the footpint , then drive off. I'd measure the length and width of each "footprint" (it's liable not to be a rectangle but it will be close enough).
Because the cg of the tractor is unknown, you have to weigh each wheel to get the load to calculate the loading in psf.

I'd guess that 2/3 of the weight on the rear tires so that 1000lb per rear wheel for a 3000lb tractor. Divide that by the rear foot print area and you 'll have the loading for the rear tire. Multiply that 1.25 for a safety factor and use that for the rear tire loading. repeat the process for the front wheel at 500lb per front tire. that should get you in the ballpark.
 
   / Calculating lbs/sq ft. #3  
Hi all,

If I wanted to build a floor out of 2x's, spaced 16" O.C., covered with 3/4" plywood, and park my tractor on it,(max 3000 lbs.), how do I figure out the lbs/sq. ft. the floor will need to support?

I'm thinking that I probably take the length & width of the wheel base as the sq. ft. "foot print", & then divide the weight by this "foot print"? But, I'm not positive?

Thanks,
Ray

That is the way I have always done it.

Sure, if you just count the tire contact area you will get a much higher #/sq. ft. number, but you are ignoring that all the space where the tires are not in contact has no load at all on it.

You just have to be certain that there is enough load transfer between the floor joists to equalize the loads on them, or you may overload individual joists.
 
   / Calculating lbs/sq ft. #4  
No way would I put a tractor, or any other vehicle for that matter, on 2X4s. I would think maybe 2X8s, 12" OC, and a 1" deck. Better yet, concrete on grade.
 
   / Calculating lbs/sq ft. #5  
Just because a framed floor (joists and decking) is designed to support say a 40# live load (per square foot) does not mean it would support a tractor of which its entire weight is supported by the four points where the tires bear.

over time you would end up with a floor that has four big dips (birdbaths)....

it is all about distributing the weight over the entire structure
 
   / Calculating lbs/sq ft. #6  
Here's a reference that discusses floor joists and loads - this one is about residential loads so it dosen't directly answer your question but the basic concepts are applicable:
Understanding Loads and Using Span Tables - Publications - BM&WT - UMass Amherst

A real valuable source of documents I found through a post here on TN several years ago is the US Army' General Dennis J. Reimer Training and Doctrine Digital Library

Found here: https://rdl.train.army.mil/soldierPortal/soldier.portal

You can find all kinds of construction and tractor-related documents there (mostly search under "Engineer" for this kind of stuff)
- like this one:
https://rdl.train.army.mil/soldierPortal/atia/adlsc/view/altfmt/9736-1

That said - I don't think plain 2x lumber will support your tractor - for long...

Best of luck to you

WVBill
 
   / Calculating lbs/sq ft. #7  
Also keep in mind that wood however treated will over time abosorb moisture. I think concrete is a better choice. As a swag most likely reinforced at about 4 to 6 inches.
 
   / Calculating lbs/sq ft. #9  
That is the way I have always done it.

Sure, if you just count the tire contact area you will get a much higher #/sq. ft. number, but you are ignoring that all the space where the tires are not in contact has no load at all on it.

.

That's the point! you need to size floor structure for the maximum "point loads" which are where the the tire contact is made. To use the average load of the area under the tractor will lead to undersized structure. That will likely cause a failure unless you put some huge safety factor in to use for sizing.
 
   / Calculating lbs/sq ft. #10  
Study the links that Bill provided, good stuff there. Yes, you have to worry about the point loading but there is also the overall loading and how the joists are supported. This gets into spans and such. Don't forget to figure all the scenarios, not just the one where your tractor is parked in the middle. There could be higher loads during approach and such. Wood will work, given enough of it. They've been making wooden covered bridges to hold much more than your tractor for a long time now.
 

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