bar bending schedule (BBS) | Basics of Bar Bending Schedule
bar bending schedule (BBS) |Basics of Bar Bending Schedule
Introduction Bar bending schedule-:
Bar Bending Schedule (BBS) indicates the bend shapes & cut lengths of bars according to the structural drawings issued by structural Engineer. Because bars are bent in different shapes according on the curvature of the member, each member will have its own BBS nature.
Basics of Bar Bending Schedule (BBS)
In this Post, I will discuss the basics of Bar Bending Schedule(BBS) & various sizes of reinforcement bars used.
STEPS USED IN BAR BENDING SCHEDULE:
- Calculate the length of bar
- Calculate the number of bars
Calculate the total length of bars
The total length of Bars = length of bar x number of bars
Check the diameter of bar used
Weight per meter of bar is calculated by using the formula;
Weight per meter = D2/162 Kg/m
Calculate the total weight of bars;
Total weight = Total length of steel calculated x weight per meter.
Calculate the number of bars
Number of bars = opposite length/spacing + 1
In case we need to calculate the number of stirrups;
Number of stirrups = (actual length of beam or column / spacing) + 1
For example
if we have a slab of dimensions Lx X Ly
Number of bars along Lx = Ly/spacing + 1
Similarly
Number of bars along Ly = Lx/spacing + 1
Stiruups
Number of stirrups is calculated by using the formula;
Number of stirrups = (Actual length of column or beam/spacing) +1
We will also see about some basics like
- Why we provide hooks?
- What is the function of stiruups?
- why we provide overlap?
- why we provide crank?
Diameters of reinforcement bars:
- 6mm,
- 8mm,
- 10mm,
- 12mm,
- 16mm,
- 20mm,
- 25mm,
- 32mm,
- 40mm dia
The standard length of a steel bar is usually kept as 12m (40 feet) for convenient transportation
Weight of steel bar:
The weight of a steel bar is calculated by formula
D2/162 Kg/m ( D is the diameter of the bar)
For example
if we have a bar of size 8mm
its weight in Kg/m will be equal to 8x8/162 = 0.39Kg/m.
Weight per bar = D2/162 x 12(12 is the standard length of steel ).
Example; weight of 8 mm steel bar = 82/162 x 12 = 4.74Kg/bar.
Similarly,
- 8 mm 0.395 kg/m
- 10 mm 0.610 kg/m
- 12 mm 0.89 kg/m
- 16 mm 1.58 kg/m
- 20 mm 2.46 kg/m
- 25 mm 3.87 kg/m
- 32 mm 6.32 kg/m
- 40 mm 9.87 kg/m
Why do we provide hooks & how to calculate hook length for bar bending schedule?
The function of hooks in stirrups are
The main function of a hook is to develop anchorage.
- To prevent seismic movements
- To prevent the splitting of concrete.
- To keep steel from slipping off the concrete.
- To hold the steel in position and keep the longitudinal bars in place.
Why is there a requirement for anchorage?
The tension bars should remain in position, otherwise, their purpose will not be fulfilled. So, to develop a bond strength between the steel bars & the concrete, development length requirements must be satisfied. Now there may be chances, where the required length is unavailable to meet the requirements of development length. In those cases, anchorages may be provided in the form of hooks.
Length of hook = 9d, while d is the diameter of the bar used
For example
if we have a bar of length L & the bar is provided with 2 no of hooks on either end.
The cutting length of this bar will be equal to the length of bar(L) + hook length (9d + 9d).
I.e. Cutting length = L + 9d + 9d
Bend calculation in bar bending schedule:
From the above image we have noted that, this steel bar is bent upon either end. These shapes are usually used in footings.
The length of this bend is taken from 10d to 16d in which d is the diameter of the bar.
The cutting length of this type of bar will be equal to straight length (L) + bends (10d to 12d)
i.e, Cutting length = L + 12d + 12d
OVERLAPPING (LAP LENGTH):
The standard length of reinforcement bars is 12m. If we add the straight bars directly from end to end, then there may be chances of slippage of steel which is prevented by overlapping the bars.
The aim of overlapping is to keep the steel bars connected so that the load can be safely transferred from one steel bar to the next steel.
What should be the overlapping length?
- For compression members (columns) the lap length should be taken as 40d (d diameter of bar).
- The reinforcement lapping zone is decided on the basis of critical zone in column.
- For tension members (beams) overlapping should be taken as 50d. (d diameter of bar)
- But practically we can take 50d for both columns & beams.
- If we need to overlap the reinforcement from the neck Column to footing, the lap length should be taken as 16d.
CRANK BARS (BENT UP BARS):
Cranked bars are those bars that can resist negative bending moments or simply hogging moments which is generally developed at support in top fiber. Bent up bars are provided in slabs as well as in beams since these members are subjected to flexural tensile stresses.
let us consider this example
As you can clearly observe from this iamge that, The Inclined length is known as crank length & is equal to 0.42d in which,
d = Slab or beam thickness – top and bottom cover
Formula for cutting length of crank bar in BBS:
Cutting length of cranked bar = L + 0.42d + 0.42d
RESIDENTIAL PLAN
#1 (40' x 80') - Click here
#2 (30' x 50') - Click here
#3 (30' x 24') - Click here
#4 (30' x 60') - Click here
#5 (30' x 70') - Click here
#6 (25' x 60') - Click here
#7 (20' x 60') - Click here
#8 (30' x 40') - Click here
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