----- Original Message -----
Sent: Friday, April 20, 2001 12:55
AM
Subject: Re: Pier loads for bale
wall
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