spiral formula 2

[mwitt]

I am trying to calculate a spiral rebar cage that is 5' 3" in diameter, 55' in heigth, and has 6" between loops. i believe this is called pitch. i know the weight of the rebar per foot but cannot find a suitable formula for the spiral length.

 

[RWF7437]

It's NOT a spiral, although it is often called that. It is a helix.

I get: length of one loop = 3.14159 x 5.25 ft = 16.49 feet
Number of loops = 55 feet / 0.50 foot spacing = 110 loops
Total Length = 110 loops x 16.49 ft/loop = 1814 feet +/-
Weight of #4 Bar = 0.668 lb/ft
Total weight = 0.668 x 1814 = 1212 lb.

Actual will be a little more to allow for laps and splices.

good luck

 

[Taro]

In the U.S., spiral is the correct term for this type of reinforcing as defined in the ACI code.

 

[BarryEng]

Unwrap the circle into a flat plane. I will use metric. Your diameter is 1,600 mm, & I will assume a 20 dia bar. The effective (mean) diameter is 1,620 mm with one spiral advance of 150 mm. The bar length is calculated by the pythaggie theorem.

= sqrt(1620^^2 + 150^^2)
= 1,627 mm

That is the length of one loop. Rest of calcs as per RWF7437 above.

 

[RWF7437]

Can't agree with BarryEng.

1627 mm = 5.33 feet ( this is the Diameter of ONE loop )
The length ( circumference of that loop is Pi x Dia. =3.14x5.33= 16.74 feet )
The total length of the spiral, which is composed of 110 such loops would be 1841 feet. ( within 1.5% of my original calculation).

I don't know what the "pythaggie theorem" is or what it has to do with this question unless the column is triangular and this is some slang for the Pythagorian Theorum.

 

[BarryEng]

It was slang for Pythagorus (I was just trying to be a bit light hearted). In Wikepedia it is pythagoras (in dictionary) or Pythagorus (in encyclopedia) & it is theorem (in dictionary) or theorum (in encyclopedia).

Sorry for the error, I just was not concentrating.

Circum = pi x D
= pi x 1627
= 5111 mm

Spiral length (one hoop) = sqrt(5111^^2 + 150^^2)
= 5114 mm

 

[dicksewerrat]

Don't forget to add the bars you are using to maintain your 6 inch spacing.

Richard A. Cornelius, P.E.

http://WWW.amlinereast.com

 

[IFRs]

In this case, approximating the length of one 360 degree path of the re-bar with a circle is likely to be close enough. If the distance between bars was a major fraction of the diameter, it would be short by too much to ignore.

 

[bimr]

The length of the helix spiral is the hypotenuse of a triangle whose other sides are the circumference of the round part and the lead distance formed by the width of the spiral. Therefore, the formula would be: helix length = square root of [(PI x diameter)^2 + (width length)^2].

= square root of [(3.14 x 5'-3")^2 + ( 6"^2)]

= square root of [272.03 + .25] = 16.500938 feet

For 110 spirals, length = 1815.10 feet.

detailed explanation:

http://216.221.100.200/rmr_faq/Layout/Helix.html

Adjusting for the added diameter of a #4 bar, the length will be:

= square root of [(3.14 x 5'-3.5")^2 + ( 6"^2)]

= square root of [276.37.03 + .25] = 16.631779 feet

For 110 spirals, length = 1829.50 feet.