# Bar Bending Scheduled Of One Way Slab (BBS)

## Bar Bending Scheduled Of One Way Slab

Bar Bending Scheduled Of One Way Slab (BBS):- Today We are going to discuss the Bar Bending Schedule of One way Slab.

BBS is an important thing in building construction or civil engineering.

Bar Bending Scheduled Of One Way Slab

Given,

Diameter of Main Bar              = 12 mm.

Diameter of distribution Bar  = 10 mm.

Spacing                                   = 150 mm.

Top and Bottom Clear Cover   = 25 mm.

Slab Thickness                       = 150 mm.

Development length (Consider) = 40 d

first, We will find out the number of the main bars.

Calculation

No. Of Main bars formula =  Length Of Slab / Spacing + 1

Step 1

1).

No. Of Main bars  =    L y  / Spacing + 1

No. Of Main bars  =    6000 / 150 + 1

= 41 Nos.

No. Of Crank bars  = 21 Nos.

No. Of Straight Bar = 20 Nos.

2).

No. Of Distribution bars  =    L x  / Spacing + 1

No. Of Distribution bars  =    2500 / 150 + 1

= 17 Nos.

Step 2

Cutting length

(i). Cutting Length Of Main Bar (Crank) = Clear span of Slab (L x) + (2 x development Length) + ( 2 x inclined Length) – ( 45* Bend x 4 ) – ( 90* Bend x 2)

Cutting Length Of Main Bar (Crank) = 2500+(2 x 40 d) + ( 2 x 0.42 d ) – ( 1 d x 4 ) – ( 2 d x 2 ).

Inclined Length = 0.42 D

For 45*        = 1 d

For 90*       = 2 d

### To understand in details watch the video

Cutting Length Of Main Bar (Crank) = 2500+(2 x 40 d) + (2 x 0.42 d) – (1 d x 4) – (2 d x 2).

= 2500 + (2 x 40 x 12) + (2 x 0.42 d) – (1 x 12 x 4) –  (2 x 12 x 2).

= 2500 + 960 + ( 2 x 0.42 d) – 48 – 48

What is D,

D = Slab Thickness – 2 Sides Clear Cover – Diameter of Bars.

= 150 – 50 – 12

= 88 mm.

D = 88 mm.

Cutting Length = 2500 + 960 + ( 2 x 0.42 x 88) – 96

=  3460 + 73.92 – 96

– 3437.92 mm.  or

= 3.437 meter

Cutting Length Of Main ( Crank ) Bar = 3.437 meter

For 21 no. of bars = 21 x 3.437 = 72.177 meter

Weight of steel     =  D x D/162 x L

= 12 x 12 / 162 x (72.177)

–   0.89 x 72.177

=     64.23 Kg.

### 1(ii). Cutting Length Of Main Bar (straight) = Clear span of slab (Lx) + ( 2 x development length )

= 2500 + ( 2 x 40 d )

– 2500 + ( 2 x 40 x 12 )

= 2500 + 960

– 3460 mm or 3.46 meter

For 20 No. Of Bars = 20 x 3.46    = 69.2 meter

Weight of steel      = 0.89 x 69.2 = 61.58 Kg

Bar Bending Schedule Basics: Especially for Freshers Civil Engineers

### 1(ii). Cutting Length Of Distribution Bar

Length of the slab (L y) – ( Cover on both sides )

= 6000 – 50 = 5950 mm or 5.95 meter

For 17 no. of bars    = 17 x 5.95 = 101.15 meter.

Weight of steel        =  10 x 10 /162 x (101.15)

= 61.70 Kg.

## 3).  Extra Top Bar

( L x / 4 ) / Spacing + 1

=      2500 / 4 x 150 +1

=  5 Nos.

For two Sides    =      2 x 5    = 10 Nos.

Length Of Extra Top Bar = Length of Same As Distribution Bar.

=  5.95 meter

For 10 No. Of Bars = 10 x 5.95 = 59.5 meter

Weight Of Steel      = 0.89 x 59.5 = 52.95 Kg.

Total Wt. Of Steel   = 64.23 + 61.58 + 61.70 + 52.95

Total Weight Of  Steel    = 240.46 Kg.

I hope this information can be useful for you guys.

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