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Erik[_5_] Erik[_5_] is offline
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Default How to straighten a 5x10ft, 3/4" steel plate

In article ,
Ignoramus8008 wrote:

Thanks, guys. I am having second thoughts about the heat method, as I
do believe that it will make the surface wavy and odd shaped. I do
know for sure that weight applied with a forklift and jacks/chains,
will flatten the surface acceptably. Given this knowledge, I will just
pull the surface from below using a welded on allthread and a
brace/nut, pulling the plate down using a C channel as a brace. It
should work. I have a 3/4" allthread available.


Thats a healthy chunk os steel you have... lets see, 60" X 120" X .75 is
5,400 cubic inches.

5,400 cubic inches divided 1,728 is 3.13 cubic feet.

Machinery's handbook claims steel weight of 491 lbs per cubic foot. So
3.13 cubic feet X 491 lbs comes to 1,536.83 lbs! (or .77 ton)

Loads of surface area to boot... it'd take a bunch of extra 'manly man'
torches to significantly heat it.

Lets see for the surface area... 60 X 120 inches is 7,200 square inches,
X 2 sides is 14,400 square inches. The edges add... 120 inches X 2
totaling 240... plus 2 X 60 inches for 120 inches for a circumference
of 360 inches, times .75 inches for a total edge area of 270 square
inches. (Another 1.88 square feet.

14,400 (total side) square inches + 270 (total edge) square inches gives
an overall total surface area of 14,670 square inches. Divided by 144
comes to 101.88 square feet.

101.88 square feet radiates boat loads of heat...

Erik (with the 'k', not a 'c')