Wood framed floor strong enough for a car?

[aczlan]

When I worked for a company that built sheds, we'd double-layer the plywood floor for people who wanted to store cars in them. Floors were usually framed with 2x6 on 12" centers.

My neighbor has such a building, its 12x20ish. He parks his lawnmower and a project car in it with no problems.

Aaron Z

 

[s219]

This is not correct. A few ways of seeing that without doing any analysis would be to ask what the load would be if the tires went flat while the car was parked there (the load would not change even though the tire pressure went to zero). Or, from a different perspective, I have a lawnmower and an SUV that have the same tire pressure, but one weighs about 400# and the other about 4000#.
There is a complex relationship between tire pressure, size of the tire, the rigidity of the tire material, and the resulting contact area of the tire. But the contact area is always relatively small, so the important thing is to know the weight distribution on each wheel, which comes from knowing the total weight and center of gravity of the vehicle.

I beg to differ -- people have been using that relationship between tire psi, tire load, and contact area to do design work for decades. The only requirement is that you are within the design range of tire inflation and that it's a standard type of tire. A flat tire, a rigid tire (such as run-flats with internal structure), and other scenarios violate the rule of thumb, so the rule won't work in those case (or may only be approximate).

The easiest way to verify this, for any doubters, is to park one front tire of a car on a piece of graph paper and gently wiggle the steering wheel. Then back off, and look at the imprint left on the paper. You can count up area of the contact patch, then do the math. You will see that tire pressure comes out to be the weight on the wheel divided by the contact area. It's a great science project topic for kids. The more basic the tread pattern, the easier it is to verify.

As to the comparison of a 400# mower to a 4000# vehicle with the same tire pressures, you will see that the vehicle will have a much larger contact patch than the mower (would have to be 10X larger). One could make the same argument between a bike tire at 120psi and a cement truck tire at 120psi, but again, there are vastly different contact areas.

 

[Gary Fowler]

If you are looking for cheap and short term lifespan, a portable building might work, but I wouldn't expect it to last more than about 10 years even with 2x6 or 2x8 runners and double plywood flooring.

If I were looking at cheap square footage, I would opt for a SeaCan cargo container for $1500-2000 each delivered. If you want larger area, get two space them about 24+ feet apart, put trusses across each one to span the entire width and put on a roof and even wall in the ends if you want. You could then have a 40 foot wide building with 24 feet of "garage" and 16 feet of waterproof/ratproof storage area on the sides. With or without concrete slab, they make good garages or shops.
IF you are looking at aesthetics and want something more permanent, I would follow advice by EddieWalker and others and just remove the stumps and back fill with compacted dirt then build on a concrete slab that will last several lifetimes.

 

[BeezFun]

The easiest way to verify this, for any doubters, is to park one front tire of a car on a piece of graph paper and gently wiggle the steering wheel. Then back off, and look at the imprint left on the paper. You can count up area of the contact patch, then do the math. You will see that tire pressure comes out to be the weight on the wheel divided by the contact area. It's a great science project topic for kids. The more basic the tread pattern, the easier it is to verify.

Well I love a good science project, so I went off and did this with my car. Corolla is 2400# with 30psi tires, contact patch is slightly elliptical, front is 6.5" long, rear is 5.5" long, which all works out to a contact pressure of 18psi. I was surprised it was this close, so I went off and found this article that uses data from the tire manufacturer Avon. There's a table that shows how contact pressure varies with loading for a range of tire pressures, below is the table for fixed tire pressure of 28psi. The right hand column should be equal to 28psi for all values of loading, but it's not. It varies by about a factor of 10 with load, and is never closer than a factor of two to the tire pressure. So this indicates contact pressure is a pretty bad predictor of tire load. The table's a little mangled from the cut and paste:

28 PSI
Tire load (lb) Loaded Radius (mm) Contact Patch Length (mm) Patch Area(sq-in) Patch Pressure(psi)
66.14 317.0 98.46 37.39 1.77
119.05 315.7 113.95 43.27 2.75
174.17 314.6 125.53 47.67 3.65
227.08 314.3 128.50 48.80 4.65
282.19 312.7 143.27 54.41 5.19
335.10 311.9 150.08 56.99 5.88
390.22 310.9 158.16 60.06 6.50
445.33 310.1 164.32 62.40 7.14
498.24 309.4 169.52 64.38 7.74
553.36 308.3 177.36 67.35 8.22
608.48 307.5 182.83 69.43 8.76
663.59 306.5 189.42 71.93 9.23
718.71 305.8 193.90 73.63 9.76
773.82 305.2 197.64 75.05 10.31
828.94 304.4 202.52 76.91 10.78
884.05 303.8 206.09 78.26 11.30
936.96 303.1 210.17 79.81 11.74
994.28 302.3 214.73 81.54 12.19
1047.20 301.7 218.08 82.82 12.64
1102.31 301.1 221.37 84.07 13.11
1157.43 300.3 225.68 85.70 13.51
1212.54 299.6 229.37 87.10 13.92
1267.66 299.0 232.48 88.28 14.36
1322.77 298.3 236.05 89.64 14.76

I looked up the formula for contact patch length:

L = 0.7 * a * r[SUB]f[/SUB] * ( d / r[SUB]f[/SUB] + 2.25 * ( d / r[SUB]f[/SUB])[SUP]1/2[/SUP] )

where:
a has a value 1.0 for a static tire (this equation can be used for tires in motion)
r is unloaded tire radius
d is ratio of load to vertical stiffness

which shows that the contact length is non-linear with respect to tire pressure. That agrees with the empirical data in the table above. I'm still puzzled why my experiment produced a result that was only off by 40%, I would have expected it to be off by at least a factor of two based on the better data.

Anyway, that was interesting. I think it is a good science project but probably not a good way to design structures.

 

[vtsnowedin]

Well I love a good science project, so I went off and did this with my car. Corolla is 2400# with 30psi tires, contact patch is slightly elliptical, front is 6.5" long, rear is 5.5" long, which all works out to a contact pressure of 18psi. I was surprised it was this close, so I went off and found this article that uses data from the tire manufacturer Avon. There's a table that shows how contact pressure varies with loading for a range of tire pressures, below is the table for fixed tire pressure of 28psi. The right hand column should be equal to 28psi for all values of loading, but it's not. It varies by about a factor of 10 with load, and is never closer than a factor of two to the tire pressure. So this indicates contact pressure is a pretty bad predictor of tire load. The table's a little mangled from the cut and paste:

28 PSI
Tire load (lb) Loaded Radius (mm) Contact Patch Length (mm) Patch Area(sq-in) Patch Pressure(psi)
66.14 317.0 98.46 37.39 1.77
119.05 315.7 113.95 43.27 2.75
174.17 314.6 125.53 47.67 3.65
227.08 314.3 128.50 48.80 4.65
282.19 312.7 143.27 54.41 5.19
335.10 311.9 150.08 56.99 5.88
390.22 310.9 158.16 60.06 6.50
445.33 310.1 164.32 62.40 7.14
498.24 309.4 169.52 64.38 7.74
553.36 308.3 177.36 67.35 8.22
608.48 307.5 182.83 69.43 8.76
663.59 306.5 189.42 71.93 9.23
718.71 305.8 193.90 73.63 9.76
773.82 305.2 197.64 75.05 10.31
828.94 304.4 202.52 76.91 10.78
884.05 303.8 206.09 78.26 11.30
936.96 303.1 210.17 79.81 11.74
994.28 302.3 214.73 81.54 12.19
1047.20 301.7 218.08 82.82 12.64
1102.31 301.1 221.37 84.07 13.11
1157.43 300.3 225.68 85.70 13.51
1212.54 299.6 229.37 87.10 13.92
1267.66 299.0 232.48 88.28 14.36
1322.77 298.3 236.05 89.64 14.76

I looked up the formula for contact patch length:

L = 0.7 * a * r[SUB]f[/SUB] * ( d / r[SUB]f[/SUB] + 2.25 * ( d / r[SUB]f[/SUB])[SUP]1/2[/SUP] )

where:
a has a value 1.0 for a static tire (this equation can be used for tires in motion)
r is unloaded tire radius
d is ratio of load to vertical stiffness

which shows that the contact length is non-linear with respect to tire pressure. That agrees with the empirical data in the table above. I'm still puzzled why my experiment produced a result that was only off by 40%, I would have expected it to be off by at least a factor of two based on the better data.

Anyway, that was interesting. I think it is a good science project but probably not a good way to design structures.

Regardless of the tire pressure the whole weight of the vehicle has to be supported by the floor under it carried out to what ever point carries that load. Say you have a 4000 pound vehicle measuring seven feet by eighteen feet. That comes out to 31 pounds per square foot which isn't much but consider the stringer under the front axle if it hangs sideways under both front tires. Then you have a load imposed at two points six feet apart at about 1500 pounds each that might be carried over to a sill plate five feet away on each side. I'd want an 8X8 clear timber or gluelamed to do that but you guys have fun with your 2x4s.

 

[KennyG]

Say you have a 4000 pound vehicle measuring seven feet by eighteen feet. That comes out to 31 pounds per square foot which isn't much but consider the stringer under the front axle if it hangs sideways under both front tires. Then you have a load imposed at two points six feet apart at about 1500 pounds each that might be carried over to a sill plate five feet away on each side. I'd want an 8X8 clear timber or gluelamed to do that but you guys have fun with your 2x4s.

As I mentioned above, this misses the diaphragm effect of tying the joists together with the flooring, plywood or planks. This is very effective in distributing the load over a length of the joists and over several joists at one time. As has also been noted, if you use 2x4s you are limited in spans which is why the 4x4 base is used, but for a short span, the 2x4 is fine.

 

[s219]

Well I love a good science project, so I went off and did this with my car. Corolla is 2400# with 30psi tires, contact patch is slightly elliptical, front is 6.5" long, rear is 5.5" long, which all works out to a contact pressure of 18psi. I was surprised it was this close, so I went off and found this article that uses data from the tire manufacturer Avon. There's a table that shows how contact pressure varies with loading for a range of tire pressures, below is the table for fixed tire pressure of 28psi. The right hand column should be equal to 28psi for all values of loading, but it's not. It varies by about a factor of 10 with load, and is never closer than a factor of two to the tire pressure. So this indicates contact pressure is a pretty bad predictor of tire load. The table's a little mangled from the cut and paste:

28 PSI
Tire load (lb) Loaded Radius (mm) Contact Patch Length (mm) Patch Area(sq-in) Patch Pressure(psi)
66.14 317.0 98.46 37.39 1.77
119.05 315.7 113.95 43.27 2.75
174.17 314.6 125.53 47.67 3.65
227.08 314.3 128.50 48.80 4.65
282.19 312.7 143.27 54.41 5.19
335.10 311.9 150.08 56.99 5.88
390.22 310.9 158.16 60.06 6.50
445.33 310.1 164.32 62.40 7.14
498.24 309.4 169.52 64.38 7.74
553.36 308.3 177.36 67.35 8.22
608.48 307.5 182.83 69.43 8.76
663.59 306.5 189.42 71.93 9.23
718.71 305.8 193.90 73.63 9.76
773.82 305.2 197.64 75.05 10.31
828.94 304.4 202.52 76.91 10.78
884.05 303.8 206.09 78.26 11.30
936.96 303.1 210.17 79.81 11.74
994.28 302.3 214.73 81.54 12.19
1047.20 301.7 218.08 82.82 12.64
1102.31 301.1 221.37 84.07 13.11
1157.43 300.3 225.68 85.70 13.51
1212.54 299.6 229.37 87.10 13.92
1267.66 299.0 232.48 88.28 14.36
1322.77 298.3 236.05 89.64 14.76

I looked up the formula for contact patch length:

L = 0.7 * a * r[SUB]f[/SUB] * ( d / r[SUB]f[/SUB] + 2.25 * ( d / r[SUB]f[/SUB])[SUP]1/2[/SUP] )

where:
a has a value 1.0 for a static tire (this equation can be used for tires in motion)
r is unloaded tire radius
d is ratio of load to vertical stiffness

which shows that the contact length is non-linear with respect to tire pressure. That agrees with the empirical data in the table above. I'm still puzzled why my experiment produced a result that was only off by 40%, I would have expected it to be off by at least a factor of two based on the better data.

Anyway, that was interesting. I think it is a good science project but probably not a good way to design structures.

Phew, that was a long article to read, but I suspect the tires they tested were carrying some load in the sidewalls. They do mention the ideal case of a balloon though, and that is a good analogy. If you do a free body diagram of the forces on a tire, what you see is that pressure cancels itself out in all directions except vertical where it acts on the contact patch (so that net force is pressure times area, p*A, where p is a gauge pressure relative to ambient). In the vertical direction, you then have a weight (W) pressing down, p*A pushing up, and then some possible contribution of the sidewalls pushing up (I'll call that cumulative sidewall force S):

W = p*A + S

For the rule of thumb to work, S has to be much much smaller than p*A such that it can be neglected and W ~ p*A. That does not happen in all tires. Some have stiff sidewalls, and others act like an inflatable structure such that S is only small over a certain range of pressures. The rule won't scale linearly to extremes, such as a flat tire, or very high pressure (tire would blow up before the contact patch goes to zero).

The rule works exceptionally well for aircraft landing gear tires, which is where I first became familiar with it many years ago. I've since tested it on a few trailer and car tires and also seen good results. But it has some limitations.

I don't know how you did the math on the Corolla, or tallied up the area (I wouldn't use a formula or linear dimension -- I'd tally up squares on graph paper. It's probably safe to assume 55-60% weight on the front and 40-45% on the rear as a first cut.

If I can find some of my graph paper imprints, I will post a picture. I am pretty sure I saved some. The ones I did for a Honda S2000 came out right on the money (that car was near 50/50 weight distribution, so it was easy to figure out).

 

[BeezFun]

They do mention the ideal case of a balloon though, and that is a good analogy.

Yes, the actual behavior of a tire is somewhere between a balloon and the steel rim with no tire mounted.

If you do a free body diagram of the forces on a tire, ...

What you've described is the free body diagram for the contact patch, not for the tire. The free body diagram for the tire includes the vertical force of the rim acting along the bead surface of the tire pressing down, the out of plane force of the rim along the bead keeping the tire from blowing off the rim, and the force of the ground pushing up on the contact patch area. That diagram then includes the complexity of accounting for the deformation of the tire tread surface, where you have to bend steel and fiber cords in order to create a flat spot on the tire surface, as well as the vertical forces transmitted through the sidewall up into the rim which you mentioned.

The rule works exceptionally well for aircraft landing gear tires, which is where I first became familiar with it many years ago. I've since tested it on a few trailer and car tires and also seen good results. But it has some limitations.

I believe this, since an aircraft tire is designed primarily to absorb the shock of landing, so you want it to act as much as possible like a balloon without exploding.

I don't know how you did the math on the Corolla, or tallied up the area (I wouldn't use a formula or linear dimension -- I'd tally up squares on graph paper. It's probably safe to assume 55-60% weight on the front and 40-45% on the rear as a first cut.

I looked up requirements for the tread void area. For passenger tires it has to be a minimum of 10%. For a tire to carry an M+S snow rating it has to have a minimum 25% void area. So I used 80% tread pattern area since I have plain old passenger tires. The length of the contact area showed the relative distribution in my case, the fronts were 6.5", the backs were 5.5", which agrees with your numbers.

If I can find some of my graph paper imprints, I will post a picture. I am pretty sure I saved some. The ones I did for a Honda S2000 came out right on the money (that car was near 50/50 weight distribution, so it was easy to figure out).

That would be interesting. I was thinking about the science experiment, it would be instructive to use a bicycle instead of a car, and do the test 3 ways. One with only the tube mounted on the rim, which should correspond almost exactly. A second with only the steel rim to show what happens to an equation when the denominator goes to zero, and finally with the tire mounted to see how much different it is from the tube only answer.

 

[vtsnowedin]

Here is something useful.
http://www.awc.org/Publications/update/WFCM2001FullPage.pdf

 

[Jim Timber]

I would opt for a SeaCan cargo container for $1500-2000 each

Up here they're $2500 for a 20'er at auction. 40's run about $3500.

Access to the build site is also an issue with connex containers. You need sufficient room and turf to get a truck in and out from under it to unload.

I've looked into these as I like their rodent free nature and inherent waterproofness - but delivery to my site is $500, and I'd need a helicopter to take it from the road to where I intend to put my shed. I hate to think what that 600' journey would cost.