### WHAT IS BAR BENDING SCHEDULE (BBS) AND HOW TO PREPARE BAR BENDING SCHEDULE FOR ONE WAY ROOF SLAB WITH DESIGN EXAMPLE.

WHAT IS BAR BENDING SCHEDULE AND ITS REQUIREMENT

Bar Bending Schedule Is Nothing But The Process Of Calculation Of Steel Bar Length And Total Weight Of Steel Bars In Reinforced Cement Concrete Element.
The Bar Bending Schedule Provide The Detailed Quantity Of Steel Bars In Rcc.
Each Type Of Rcc Elements Have Steel In Different Shapes .The Shapes Include Band Up Bars Cranked Hooks And Over Lap.
Different Shapes Have Different Formulas To Find Out Its Length .
Proper Drawing And Rc Design Details Are Required To Prepare Bar Bending Schedule.
In Case The Drawing Are Not Available The Quantity Of Steel Is Calculated Approximately In Percentage  Basis.
The Percentage Of Steel Varies From Structure To Structures And Design To Design.
The Absence Of Detailed Design And Drawing The Percentage Of Reinforcements Are Assume For The Follows.

Lintel And Beam – 1  To 2 %
Columns               – 1  To 5 %
Foundation          – 0.5 %  To  0.8 %
Roof Slab             – 0.7 To 0.8 %

The Above Percentage Calculations Is Just  For Approximate Calculation Its May Differ From Actual Requirement Of Steel.
DIFFERENT TYPES OF BAR BENDING AND ITS LENGTH.

The Following Table Provide The Length  Of Bars In Different Shapes.

HOW TO PREPARE BAR BENDING SCHEDULE IN ONE WAY SLAB.

To Prepare The Bar Bending Schedule For The Given Drawing And Rcc Design Data

Room Size  =  6.10m X 3.05 M ( Inner Dimension )
Wall Thick  =  0.23 M
Main Bars  =  10mm Dia At 265 mm C/C
Distributor  =  8 Mm Dia 330 mm C/C
Extra Reinforcement In Bend Up Portion = 8 mm Dia

SOLUTION

Min Reinforcement 10mm Dia At 265 mm C/C Alternative Rods Bends Up 1/5 Clear Span Of Both Ends.
Distribution Reinforcement 8 Mm Dia At 33 mm C/C
Distribution Extra Reinforcements At Bends At Bend Up Portion 8 mm Dia Bars 2 Nos Each Side.
Cover Bottom And Top  = 15 mm
Covers At Sides               = 40 mm

LENGTH OF BAR

(a)    FOR SHORT SPAN

Length Of Main Steel = Clear Span + 2 Bearing – 2 Covers +Length Of Bend Up + 2 Hooks
D = Depth – 2 Covers – Dia Of Bar
= 125 – 2(15) – 10
= 85 mm
= 0.085 m
= 3.05 + 2 ( 0.23 ) - 2(0.04)+ 0.45 (0.085)+ 18 (0.01)
= 3.65 m

(B)FOR LENGTH SPAN

Length Of Distributor ( 8 Mm Dia ) = Clear Span + 2 Bearings – 2 Covers + 9 Dia
= 6.10 + 2( 0.23 ) – 2 (0.04) + 18 ( 0.08)
= 6.624 M

NO OF RODS

No Of Main Reinforcement = ( Long Span – (2 X Covers )+2 Bearings)/Spacing)+1
= (( 6.10 + 2(0.23 ) - 2 (0.04))/0.265
= 25 Nos + 1
= 26 Nos
No Of Distributors  = ( Short Span  – (2 X Covers )+2 Bearings)/Spacing)+1
= (( 3.05 + 2(0.23 ) - 2 (0.04))/0.33
= 11 Nos + 1 No
= 12 Nos
8 Mm Distribution Rods 4 Nos Top At Bend Up Portion (2 Nos For Each Side )
Total Distributors = 11 + 4
=15 Nos + 1
= 16 Nos

BAR BENDING SCHEDULE OF ONE WAY SIMPLY SUPPORTED SLAB
 S.NO DESCRIPTION OF BARS DIA IN MM NO.S LENGTH(M) TOTAL LENGTH WT PER METER IN KG TOTAL WT IN KG 1 Main Reinforcement 10 Mm 26 3.65 94.9 0.62 58.84 2 Distributors 8 Mm 16 6.624 105.98 0.395 41.86 Total Weight 100.70 Kg

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