There has been a lot of discussion lately on the strength of concrete. It
was mentioned in at least one messages the problem with digging the size
hole specified for the base. For example, the Tri-Ex LM354HD tower, that is
being held for me until I move, calls out for a 3'-6" square by 7'-6" deep
base. There is no way to dig this hole and just pour concrete in. You
would have to dig a much large hole and build a form for the concrete. Then
after a few days, remove the form, and back fill the hole with compacted
earth. Another way is to have the hole drilled which might make my question
mote.
I have been involved in building a number of commercial towers, the largest
was a 450' guyed tower in Florida. While I am not a structural engineer, I
know from this experience that a base in undisturbed earth seems to requires
less concrete than one with backfilled earth. By example, a 110" three
legged self supporting microwave tower had a 3' diameter base about 8' deep
poured in a drilled hole for each leg. This is about the same size required
for my Tri-Ex.
Another point, soil density is not the same everywhere. The coastal Florida
locations where I have experience had very sandy soil with a high water
table. The base design was done only after soil samples and density tests
were completed by a soil testing company. It should be noted that the
drilled hole used a special concrete pouring and drilling method using a
drillers mud or slurry and the concrete was pumped from the bottom up. The
drillers mud held the wet sand in place until the concrete was poured and
was forced out by the heavier concrete. This is not something the average
Ham would undertake. I believe that Tri-Ex and the other tower manufactures
of Ham towers design a base that will work anywhere.
So we get to my question. The 3'-6" by 7'-6" base would require about 3.5
cubic yards of concrete which would weight 13,500 pounds or almost 7 tons.
If the base were modified for ease of digging to say 5' square by 4' deep or
3.7 cubic yards it would weight 14,400 pounds. What are the opinions about
this modified base being equal to the specified one?
73
Ted W4ZE
w4ze@arrl.net
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