How to Calculate the Cutting Length of Spiral Bar or Helix Bar for Pile Foundation | Bar Bending Schedule:

How to Calculate the Cutting Length of Spiral Bar or Helix Bar | BBS Bar Bending Schedule, commonly referred to as “BBS” is a comprehensive list that describes the location, mark, type, size, length and number, and bending details of each bar or fabric in a Reinforcement Drawing of a Structure.

This process of listing the location, type and size, number of and all other details is called “Scheduling”. In context of Reinforcement bars, it is called bar scheduling. In short, Bar Bending Schedule is a way of organizing rebars for each structural unit, giving detailed reinforcement requirements.

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Cutting Length of Spiral Bar or Helix Bar:

To calculate the quantity of steel spirals ( or spiral ties) in steel cage (or sometimes rebar cage) for piles ( reinforced concrete or bored pile), you have to have the following information ( check the details in the drawings):

a) diameter of pile
b) length of pile
c) pitch of spirals (spacing of spirals)
d) size of spiral (diameter of bar)

Example 1:

See More : How to Make Bar Bending Schedule for Pile Foundation Reinforcement | BBS for Pile(Opens in a new browser tab)

Example 2:

For example, you are given a 1.5 m diameter bored pile with a length of 20 meters where spiral pitch is 100 mm. Use a 12 mm diameter spiral ties.
To solve the quantity in kilograms, follow the steps below :

a) Compute the net diameter of bored pile, i.e. diameter of bored pile less clear concrete cover, say 50 mm. Hence, net diameter = 1.5 m – 2*0.05 = 1.4 m.
b) Calculate the circumference of the spiral, consider it as a circle or ring, so C = pi*d, where pi=3.1416 and d = 1.4 m, therefore C= 4.40 m
c) Determine the no. of spirals by dividing the length of pile by the given pitch, so number of spirals = 20/0.1 +1 =201
d) Determine the total length of spirals by multiplying its number by its circumference, hence, Total length = 201 x 4.4 =884.4 m
e) Unit weight of a 12 mm diameter steel = 0.888 kg/m
f) Finally, total quantity of spirals expressed in kgs = total length x unit weight = 785.35 kgs.

Conclusion:

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