All about civil construction knowledge- PARAM VISIONS

Bar bending schedule (BBS) of a storm water drain. / Calculating the cutting length & weight of reinforcement in a storm water drain.

 Let us make BBS of the stormwater drain as shown below.




Given data:

RCC pardi depth = 450mm.

RCC pardi width = 150mm.

Drain bed width = 850mm.

Drain bed thickness = 150mm.

Stormdrain length = 8m.

Clear cover = 50mm.


Calculation:

A. Cutting length:

Given data:

Drain pardi vertical reinforcement     = T10 @ 200c/c

Drain pardi horizontal reinforcement = T8 @ 175c/c.

Drain bed reinforcement-1= T10 @ 150c/c.

Drain bed reinforcement-2 = T8 @ 175c/c.



1. Drain pardi vertical reinforcement:

a. Cutting length of outer vertical reinforcement

= [ {a } + {b} + {c } + {d } - (2nos.× 90° bend)]



= [{pardi width -(2nos.× cover)} + {(pardi depth - cover) + (bed thickness -cover- bed reinforcements )}+ {pardi width  -cover} + {175mm.} - ( 2nos. × 2d)]

= [{150mm. -(2nos.× 50mm.)} + {(450mm. - 50mm.) + (150mm. - 50mm.-10mm.-8mm. ) }+ {150mm. -50mm.} + {175mm.} - ( 2nos. × 2 × 10mm. )]

= [{50mm.} + {482mm.} + {100mm.} + {175mm.} - (40mm.)]

= 767mm.

= 0.767m.


b.  Inner vertical reinforcement 

 Cutting length of inner vertical reinforcement

=  [ {a } + {b} + {c } + {d } - {2nos.× 90° bend}]



= [{pardi width -(2nos.× cover)} + {(pardi depth - cover) + (cover + top bed reinforcements )}+ {cover} + {175mm.} - {2nos. × 2d}]

= [{150mm. -(2nos.× 50mm.)} + {(450mm - 50mm.) + (50mm + 10mm. +8mm. )}+ {50mm.} + {175mm.} - {2nos. × 2 × 10mm.}]

= [{50mm.} + {468mm.} + {50mm.} + { 175mm.} - {40mm.}]

= 703mm.

0.703m.


2. Drain pardi horizontal reinforcement:

 Cutting length of horizontal reinforcement

= [ drain length - (2nos. × cover)]

= [ 8000mm. - ( 2nos. × 50mm.)]

= 7900mm.

=7.9m.


3. Drain bed reinforcement -1:

Cutting length of bed reinforcement-1 

= horizontal reinforcement of pardi = 7900mm.= 7.9m.


4. Drain bed reinforcement -2:

Cutting length of bed reinforcement -2

= [{2nos. × a} + {b} - {2nos. × 90° bend}]



= [{ 2nos.×( bed thickness - 2nos. × cover)} + {bed width - 2nos. × cover} -{2nos. × 2d}]

= [{ 2nos.× ( 150mm. - 2nos. × 50mm.)} + {850mm. - 2nos. × 50mm.} -{2nos. × 2 × 8mm.}]

= [ { 100mm.} + {750mm.} - {32mm.}]

= 818mm. = 0.818m.


Note: Cutting length of top & bottom reinforcement -2 is equal.


B. No. of bars:

1. No of drain pardi vertical bar

= [(drain length - 2nos. × cover) ÷ c/c distance] +1

= [8000mm. - 2nos. × 50mm.) ÷ 200mm.] +1

= 39.5 +1

= 40.5

By rounding off = 41nos.


2. No of drain pardi horizontal bar

= [(pardi depth - cover) ÷ c/c distance] +1 

=  [(450mm. - 50mm.) ÷ 175mm.] +1 

= 2.28 +1

= 3.28nos.

By rounding off = 4nos.


3. No of drain bed bar-1

 =  [(bed width - 2nos. × cover) ÷ c/c distance] +1 

 = [(850mm. - 2nos. × 50mm.) ÷ 150] +1 

= 5+1

= 6nos.


4. No of drain bed bar-2

 =  [(drain length - 2nos. × cover) ÷ c/c distance] +1 

 = [(8000mm. - 2nos. × 50mm.) ÷ 175] +1 

= 45.14+1

= 46.14

By rounding off = 47nos.

Total no. of bed bar-2 for top & bottom

= [2sides × 47nos.]

= 94nos.


Now, let us prepare a BBS table for the stormwater drain.

Sl. No.

Type of Bar

Dia.

in mm.

Nos.

Length

in m.

Total length

in m.

Weight

in Kg/m.

Total

 bar wt.

in kg.

1.

Drain pardi vertical bar.

  

a.

Outer bar.

10

41

0.767

31.447

0.612

  19.24

b.

Inner bar.

10

41

0.703

 28.823

0.612

  17.64

2.

Drain pardi

horizontal bar

8

4

7.90

31.60

0.395

  12.48  

3.

Drain bed

Bar-1

10

6

7.90

47.4

0.612

  29.00

4.

Drain bed

Bar-2

8

94

0.818

76.89

0.395

  30.37

5.

       Total weight of bars =

  108.73

6.

         Add 3% wastage =

      3.26 

7.

The grand total wt. of rebar's =

 112.00

 


Note:

1. The depth of stormwater drain varies according to the road level & slope. But to understand the calculation, a constant value is taken. 

Thank you for going through these calculation steps. Have a good day πŸ˜„.

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Bar bending schedule (BBS) of a Chajja (sunshade). / Calculating the cutting length & weight of reinforcement in a window projection (sunshade).

 Let us make BBS of the chajja or RCC sunshade as shown below.




Given data:

Length of sunshade = 1500mm.

Width of sunshade = 900mm.

Sunshade thickness (front) = 75mm.

Lintel beam width = 230mm.

Lintel beam depth = 350nmm.

Main bar = 10∅ @ 150c/c.

Distribution bar = 8∅ @ 150 c/c.

Sunshade clear cover =20mm.

Lintel beam cover = 25mm.


Calculation:

A. Cutting length:

1. Cutting length of main bar

= [{sunshade thickness - (2nos. × cover)} + (sunshade width - cover) + {2nos.×  (lintel width - cover)} + 100mm. + (100mm. - lintel cover) - (3nos. × 90° bend.) ]

=  [{75mm. - (2nos. × 20mm.)} + (900mm. - 20mm.) + {2nos.×  (230mm. - 20mm.)} +100mm.+ ( 100mm.- 25mm.) - (3nos. × 2 × 10mm.) ]

=  [{75mm. - (40mm.)} + (880mm.) + {2nos.×  (210mm.)} +100mm.+ ( 75mm.) - (60mm.) ]

= [ 35mm. + 880mm. + 420mm. + 115mm.]

= 1450mm.

= 1.45m.


2. Cutting length of distribution bar

= [ sunshade length - (2nos.× sunshade cover)] 

= [ 1500mm. - ( 2nos. × 20mm.)]

= 1460mm.

= 1.46m.


B. No. of bars:

1. No of main bars

= [{sunshade length - (2nos. × cover)} ÷ c/c distance ] +1

= [{1500mm. - (2nos. × 20mm.)} ÷ 150mm. ] +1

= [ {1460mm.}  ÷ 150mm. ] +1

= 9.73 +1

= 10.73

By rounding off = 11 nos.


2.  No of distribution bars

= [{sunshade width - cover - main bar dia} ÷ c/c distance ] +1

= [{900mm. - 20mm. - 10mm} ÷ 150mm. ] +1

= [ {870mm.}  ÷ 150mm. ] +1

= 5.8 +1

= 6.8

By rounding off = 7 nos.


If you want to know how to prepare BBS of lintel beam, Go through the articleπŸ‘‡

πŸ’ ° How to calculate the cutting length & make BBS of the Lintel bar.


Now, let us prepare a BBS table for the Chajja or sunshade.

Sl. No.

Type of Bar

Dia.

in mm.

Nos.

Length

in m.

Total length

in m.

Weight

in Kg/m.

Total

 bar wt.

in kg.

1.

Main bar.

10

11

1.45

15.95

0.612

   9.76

2.

Distribution bar.

8

7

1.46

10.22

0.395

   4.04 

3.

       Total weight of steel bars =

 13.80

4.

         Add 3% wastage =

   0.414 

5.

The grand total wt. of rebar's in sunshade =

14.21

 

Thank you for going through these calculation steps. Have a good day πŸ˜„.

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How to calculate the volume of the dome slab? / Calculating concrete volume of the semi-spherical slab.

Let us calculate the volume of the dome slab as shown below.



Now, let us observe a section of a dome slab having a semi-spherical shape as shown below.



Given data:

Slab thickness = 0.15m. = 150mm.

Outer diameter of dome slab = D = 6.3m.

Height of dome slab = Inner radius = 3m.

Inner radius of slab = r = 3m.


The volume of the semi-spherical slab is calculated by the formula

= [2/3 × Ο€ × { ( R )3( r )3 }]

Where

R = outer radius

      = D/2 = [6.3m ÷ 2] = 3.15m.


The volume of the semi-spherical slab

= [2/3 × 3.142 ×  { ( 3.15m. )3( 3m )3 }]

= [2/3 × 3.142 × { 31.256 - 27}]

= [ 2.094 × 4.256 ]

= 8.912 cum.

= 314.72 cu ft.


The outer surface area of the semi-spherical dome slab

=  [2 × Ο€ × ( R )2]

=  [2 × 3.142 × ( 3.15 )2]

= 62.35 sq m.

= 671.16 sq ft.


Thank you for going through these calculation steps. Have a good day πŸ˜„.
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What should be the minimum distance between reinforcement bars?/ Minimum spacing between individual reinforcements.

 According to the IS-456:2000, the minimum spacing between the individual reinforcement bars is determined as follows.




A. Minimum horizontal spacing:




As per IS-456:2000, the minimum horizontal distance between the individual bars shall be the maximum of the following three conditions.


1. Equal diameter bars:



As shown in the above drawing, if all the bars spaced horizontally are of equal diameter, i.e. 20mm., the spacing as per condition-1 is 20mm.


2. Unequal diameter bars:



As shown in the above drawing, if all the bars spaced horizontally are of unequal diameter, i.e. 25mm. & 16mm, the spacing as per condition-2 is 25mm.

3. Aggregate size:

 Suppose if we use a coarse aggregate of 20mm. size, the minimum spacing as per condition - 3 is

= [nominal size of coarse aggregate + 5mm.]

= [20mm. + 5mm.] = 25mm.

Here, the maximum value in all three conditions is 25mm.

 So, we have to consider 25mm. as the minimum horizontal spacing.


If we use a coarse aggregate of  25mm. size., the minimum spacing will be

= [25mm.+ 5mm.] = 30mm.


 Here, we have to consider 30mm. (greater value)  as the minimum horizontal spacing.



B. Minimum vertical spacing:



As per IS-456:2000, the minimum vertical distance between the individual bars shall be the maximum of the following three conditions.


1. Minimum requirement:



 In any situation ( condition-1), the minimum vertical bar spacing shall not be less than 15mm.

If the diameter of the bars is  ≤12mm., the minimum vertical distance shall be 15mm. In this case, we cannot consider 12mm. ( bar dia.) as the minimum spacing.


2. Aggregate size:

 Suppose we use a coarse aggregate of 20mm. size, the minimum vertical spacing as per condition - 2 is

= [ 2/3 × nominal size of coarse aggregate ]

= [2/3 × 20mm.] = 13.33mm.



3. Bar diameter:





As you can observe in the above drawing, if the diameter of the vertical layer of bars is 20mm., the minimum vertical spacing shall be equal to the bar diameter, i.e. 20mm.

Out of all these three conditions, the maximum value is 20mm.

 Here, we have to consider 20mm. as the minimum vertical spacing.


Note:

1. Whatever might be the permutation & combination of conditions, in all cases you have to choose the greater value as a minimum spacing.

2. Minimum spacing is the clear span ( not c/c distance) between the individual bars. 


Thank you for going through this article. Have a good day πŸ˜„.

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How to calculate the angle of staircase?/ Calculating slope of staircase.

 Let us now find out the angle or slope of the staircase as shown below.

Type-1:

If riser & tread are known.




Given data:

Riser = 160mm.

Tread = 250mm.


Slope 

     = [riser ÷ tread]

     = [ 160mm.÷ 250mm.]

     = 0.64

Here, slope = tanΞΈ

So, 

tanΞΈ = 0.64

ΞΈ = tan-1 0.64

  = 32.62°


2. Type-2

If the height & horizontal distance of the staircase are known.



Given data:

Height of landing = 1300mm.

Horizontal distance = 2000mm.


tanΞΈ = [opposite ÷ adjacent]

        = [height ÷ horizontal distance]

         = [ 1300mm.÷ 2000mm.]

     = 0.65

ΞΈ = tan-1 0.65

  = 33.02°


Thank you for going through these calculation steps. Have a good day πŸ˜„.
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Bar bending schedule (BBS) of a pile cap.(Part-2) / Calculating the cutting length & weight of reinforcement in a pile cap.

  Part -1   πŸ‘ˆ Back     


5. Side face or roundup bar:

Theoretically, the side face bars are the stirrups to hold the pile cap reinforcements as shown in the drawing.




But practically, if the size of the pile cap is large, the side face bars need lapping as the length exceeds 12m. So, in such cases, the side face bars are provided in two equal U-shape parts as shown below.



The cutting length of the side face bar

= [{2nos. × ( a +b )} +( 2nos. × hook length) - (3nos. × 90° bend) - (2nos. × 135° bend)]

Where,

a = [ pile width - ( 2 × cover) - (2nos. ×1/2 × dia. of bar)]

   = [ 2500mm. - ( 2 × 50mm. ) - ( 2nos. × 1/2 × 12mm.)]

 a  = 2388mm.


b= [ pile length - ( 2 × cover) - (2nos. ×1/2 × dia. of bar)]

   = [ 3200mm. - ( 2 × 50mm. ) - ( 2nos. × 1/2 × 12mm.)]

  b = 3088mm.


The cutting length of the side face bar

= [{2nos. × ( 2388mm. + 3088mm. )} +( 2nos. × 10d) - (3nos. × 2d) - (2nos. × 3d)]

= [{2nos. × ( 5476mm. )} + ( 2nos. × 10 × 12mm.) - (3nos. × 2 × 12mm.) - (2nos. × 3 × 12mm.)]

[{10958mm.} + ( 240mm.) - (72mm.) - (72mm.)]

= 11054mm.

= 11.054m. < 12m.

Hence no lapping is required.


B. No of rebars: 

1. No. of bottom bar layer -1 required

  = [{ length of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1

 =  [{ 3200mm. - (2nos. × 50mm.)} ÷ 125mm. ] + 1

=   [{ 3100mm.} ÷ 125mm. ] + 1

= 24.8 +1

= 25.8.

By rounding off = 26 nos.


2. No. of Top bar layer -1 required

  = [{ length of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1

 =  [{ 3200mm. - (2nos. × 50mm.)} ÷ 150mm. ] + 1

=   [{ 3100mm.} ÷ 150mm. ] + 1

= 20.66 +1

= 21.66

By rounding off = 22 nos.


3. No. of bottom bar layer -2 required

  = [{ width of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1

 =  [{ 2500mm. - (2nos. × 50mm.)} ÷ 125mm. ] + 1

=   [{ 2400mm.} ÷ 125mm. ] + 1

= 19.2 +1

= 20.2

By rounding off = 21 nos.


4. No. of top bar layer -2 required

  = [{ width of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1

 =  [{ 2500mm. - (2nos. × 50mm.)} ÷ 150mm. ] + 1

=   [{ 2400mm.} ÷ 150mm. ] + 1

= 16 +1

= 17 nos.



Now, let us prepare a BBS table for the pile cap.

Sl. No.

Type of Bar

Dia.

in mm.

Nos.

Length

in m.

Total length

in m.

Weight

in Kg/m.

Total

 bar wt.

in kg.

1.

Bottom bar

Layer-1

20

26

3.806

98.956

2.467

 244.12

2.

Bottom bar

Layer-2

20

21

4.466

93.786

2.467

 231.37  

3.

Top bar

Layer-1

12

22

4.004

88.089

0.887

   78.13

4.

Top bar

Layer-2

12

17

4.68

79.56

0.887

   70.57

5.

Side face bar

12

5

11.054

 55.27

0.887

   49.02

6.

       Total weight of steel bars =

  673.21

7.

         Add 3% wastage =

    20.20

8.

The grand total wt. of rebar's in pile cap =

 693.41

 

Thank you for going through these calculation steps. Have a good day πŸ˜„.

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Bar bending schedule (BBS) of a pile cap.(Part-1) / How to calculate the cutting length & weight of reinforcement in a pile cap?

  Let us make BBS of the pile cap as shown below.





The pile cap section is shown below mentioning different terms used while doing BBS calculation.




Given data:

Length of the pile cap = 3200mm.

Width of the pile cap = 2500mm.

Depth of the pile cap = 1500mm.

Clear cover = 50mm.

Pile cutoff length = 75mm.

Bottom bar layer-1 =20∅ @ 125 c/c.

Bottom bar layer-2 =20∅ @ 125 c/c.

Top bar layer-1 =12∅ @ 150 c/c.

Top bar layer-2 =12∅ @ 150 c/c.

Side face or roundup bar = 12∅-5 nos.


Calculation:

A. Cutting length:

1. Bottom bar layer-1:




Cutting length of the bottom bar layer-1

= [pile width + (2nos.× a ) - {(2nos. × clear cover ) + (2nos. × side face bar dia.) + (2nos. × 1/2 × bar dia.) + (2nos.× 90° bend)]




Where

    = [(1/2 × pile cap depth) + (1/2 × 300mm.) - pile cutoff - clear cover -  (1/2 × bar dia)]

   = [ (1/2 × 1500mm.) + ( 1/2 × 300mm.) - 75mm. - 50mm. - ( 1/2 × 20mm.)]

   = [ 750mm. + 150mm. - 75mm. - 50mm. - 10mm.]

  a = 765mm.

90° bend = 2d.


Cutting length of the bottom bar layer-1

= [2500mm. + (2nos.× 765mm. ) - {(2nos. × 50mm.) + (2nos. × 12mm.) + (2nos. × 1/2 × 20mm.) + (2nos.× 2 × 20mm.)]

= [ 2500mm. + 1530mm. - { 100mm. + 24mm. + 20mm. + 80mm.}]

= [ 4030mm. - 224mm.]

= 3806mm.

= 3.806m.


2. Top bar layer -1 :

Cutting length of the top bar layer-1

= [pile cap width + (2nos.× b ) - {(2nos. × clear cover ) + (2nos. × side face bar dia.) + (2nos. × 1/2 × bar dia.) + (2nos.× 90° bend)]

Where b

    = [(1/2 × pile cap depth) + (1/2 × 300mm.) - clear cover -  (1/2 × bar dia)]

   = [ (1/2 × 1500mm.) + ( 1/2 × 300mm.) - 50mm. - ( 1/2 × 12mm.)]

   = [ 750mm. + 150mm. - 50mm. - 6mm.]

  b = 844mm.

90° bend = 2d.


Cutting length of the top bar layer-1

= [2500mm. + (2nos.× 844mm. ) - {(2nos. × 50mm.) + (2nos. × 12mm.) + (2nos. × 1/2 × 12mm.) + (2nos.× 2 × 12mm.)]

= [ 2500mm. + 1688mm. - { 100mm. + 24mm. + 12mm. + 48mm.}]

= [ 4188mm. - 184mm.]

= 4004mm.

= 4.004m.


3. Bottom bar layer-2:



Cutting length of the bottom bar layer-2

= [pile cap length + (2nos.× c ) - {(2nos. × clear cover ) + (2nos. × side face bar dia.) + (2nos. × 1/2 × bar dia.) + (2nos.× 90° bend)]

Where c 

    = [(1/2 × pile cap depth) + (1/2 × 300mm.) - pile cutoff - clear cover - bottom bar layer-1 dia - (1/2 × bar dia)]

   = [ (1/2 × 1500mm.) + ( 1/2 × 300mm.) - 75mm. - 50mm. - 20mm. - ( 1/2 × 20mm.)]

   = [ 750mm. + 150mm. - 75mm. - 50mm. -20mm. - 10mm.]

   c = 745mm.

90° bend = 2d.


Cutting length of the bottom bar layer-2

= [3200mm. + (2nos.× 745mm. ) - {(2nos. × 50mm.) + (2nos. × 12mm.) + (2nos. × 1/2 × 20mm.) + (2nos.× 2 × 20mm.)]

= [ 3200mm. + 1490mm. - { 100mm. + 24mm. + 20mm. + 80mm.}]

= [ 4690mm. - 224mm.]

= 4466mm.

= 4.466m.


4. Top bar layer -2 :

Cutting length of the top bar layer-2

= [pile cap length + (2nos.× d ) - {(2nos. × clear cover ) + (2nos. × side face bar dia.) + (2nos. × 1/2 × bar dia.) + (2nos.× 90° bend)]

Where d

    = [(1/2 × pile cap depth) + (1/2 × 300mm.) - clear cover - top bar layer-1 dia. - (1/2 × bar dia)]

   = [ (1/2 × 1500mm.) + ( 1/2 × 300mm.) - 50mm. - 12mm. - ( 1/2 × 12mm.)]

   = [ 750mm. + 150mm. - 50mm.- 12mm. - 6mm.]

   d = 832mm.

90° bend = 2d.


Cutting length of the top bar layer-2

= [3200mm. + (2nos.× 832mm. ) - {(2nos. × 50mm.) + (2nos. × 12mm.) + (2nos. × 1/2 × 12mm.) + (2nos.× 2 × 12mm.)]

= [ 3200mm. + 1664mm. - { 100mm. + 24mm. + 12mm. + 48mm.}]

= [ 4864mm. - 184mm.]

= 4680mm.

= 4.68m.


Continued πŸ‘‰ Part-2

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