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I was calculating loads for a pier foundation and came across the
figure of 25# / sqft / course of bales. So for a 36' X 10' wall (360
sq ft) we are looking at a load of 63,000#. 360x25x7=63000#. Is this
the correct way to calculate this? If so thats makes an even more tremendous
load than I suspected. The number of peirs needed to safely hold this, an
possibly to add a second floor, is very large. Am I calculating this
correctly? Thanks Jeff
Hi Jeff- This is one reason those code people always want you to hire
an engineer :-). It seems to me quite a bit more information would be
needed to give you a real answer here, such as type of piers you are thinking
about, type of foundation, wind load, snow load, roof slope, roof material,
siesmic zone.... It seems the enignoid types are all busy, along with the rest
of the list these (spring) days ;-).
Now, yesterday I couldn't spell carpinter, and today I are one...
but it seems to me your 25 pound/sq. ft/ course figure would be for a
square foot of floor space, not vertical wall space. You're multiplying by 10'
high, *and* multiplying by 7 courses high. Are these two string bales, or
3 string? I think the 25#/sq ft is high, too, if thats for a square foot
of floor area, per course. 2 string bales around here are about 50 pounds,
and 40"x18" (hieght is factored in by the number of courses, the way
you're doing it). That's around 50#/5 sq. ft, or 10#/sq. ft.
36'x18"wide, divided by 12"/ft, gives 54 sq ft under the bale
wall. 54sq ft x 10#/sq ft/course x 7courses (7 courses??)=3780 pounds of
bales. That sounds more in the ballpark. 7 courses of 2 string bales
would be more like 8.2 feet high, actually. I'm not sure how many courses
you were planning on there.
Is this sitting on a raised wood floor attached to a pole frame? Or
sitting on a concrete slab, with posts holding up a second story and roof?
Questions, questions...
Cheers- WasatchBill
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