What size barn for square bales

In a 28'X72'X 13.5' 3 sided shed with 5 bents dumping from the side I can put 2000 bales (4 dumps in each shed).

In another barn, dumping long wise (36x60X13.5 on the low points) I net 2 dumps per 12'x12' bent (200 bales). Barn has 15 of these bents so it holds right at 3000.

I only net 99 bales per load. With 1 tie tier and the top tier set to 2 wide I usually dont have to get out and use stack poles.

You don't mention what size balewagon you are considering, but with a 105 bale wagon (and 14x18x40 bales) you're looking at 5*14"=70", and 3*40=120", for an area of 58 sq ft per stack. Figure 5' at each building wall, and 1' between each stack. Three stacks wide with area between and on each side would be 42' wide and and 6' front to back, for a total of 250 sq ft containing 312 bales (104 x 3). You will also require a "bulkhead" to support the start of the stack. This is made using bales from the top of the first load, and we'll figure 200 sq ft for it. So, to fit 5000 bales you'd need 312/250 = 5000/x, solving for x = 4006 sq ft, plus 200 sq ft for the bulkhead, is 4206 sq ft, so we'll assume 4250 sq ft, since the stacks wont be perfectly tight front to back (the bottom usually pulls out slightly when dumping).

So for a 105 bale wagon you're looking at a building 42' x 100', as long as one end is completely open. If you have a door to go through, then you need to account for the length of the tractor and bale wagon. If you are comfortable with getting closer to the walls you could probably get by with a 40' x 100' building.

For a 56 bale wagon if you went 3 stacks wide you'd be looking at a width of 32' (5' on each side wall, 1' between stacks, 80" wide stacks), and at least 4800 sq ft for 5000 bales. Again, this would assume that at least one end is open. A 32'x150' building would be 4800 sq ft.

Assuming 14 inches wide, 36 inches long, each bale will consume 3.48 square feet for a total of 17,400 if you don't stack your bales.

Obviously you will stack, so the question is how high? If you stack 10 bales high (10 bales times 18" = 15 foot in height), divide the 17,400 by 10 for total square footage needed of 1,740.

9 bales high would require 1,934 square feet and 7 bales high would require 2,486 square feet.

All square footage is assuming tight stacks with no gaps and no walkways.

Hope this helps,

Dug

Well said. Seems like I always plan to build just the right size, then all my stuff won't fit! Probably should do the math then multiply by two (or three).

Have a good day,

Dug