**HOW TO CALCULATE STEEL QUANTITY FOR SLAB? - BAR BENDING SCHEDULE**

Hii before going to calculate bar bending schedule for slab you should know to find your slab is whether

**one-way slab or two-way slab.**If not its**ok**we will guide through thisOne Way Slab | Ly/Lx > 2 |

Two Way Slab | Ly/Lx < 2 |

**In one-way slab,**Main bars are provided in shorter direction (Cranked bars) and distribution bars are provided in Longer direction (Straight bars).

**in Two-way slab**Main bars (cranked bars) are provided in both directions.

Q/A 79 |

**Distribution bars:**These bars are straight bars.

**Main bars:**These bars are cranked bars. Main bars are cranked at an angle of 45 Degree with the length of 0.42D

therefore,

D = Depth of Slab - Top cover – Bottom cover

**Extra Bars:**The extra bar is provided at the bottom of Cranked bars to maintain the framework of the slab. The length of Extra bar is L/4

**.**

**OK**

**We Think You Got A Clear Picture Of Slab Reinforcements**

**Now Take Look Of This Detailing**

**GIVEN DATA FOR FLOOR SLAB 1**

**SIZE OF FLOOR SLAB 1**

Length= 2950mm

Breadth= 6000mm

Beam size = 225x300mm

Main bars are 12 mm in diameter @ 150 mm centre to centre spacing

Distribution bars are 8 mm in diameter @ 150 mm centre to centre spacing. (Main Bar & Distribution Bar Difference)

Top and Bottom Clear Cover is 25 mm

Thickness of Slab – 150 mm

**Therefore, To Find Its One Way Or Two Way Slab**

Lx = 2950mm (shorter span)

Ly = 6000mm (Larger span)

Ly/Lx= 6000/2950 = 2.033

One Way Slab | Ly/Lx > 2 |

**Hence its one-way slab**

So, We need to find

·

**Main bar**for shorter span·

**Distribution bar**for longer span·

**Top bar**for main rod on both sides for critical structure**LCET**

**ED**

**TRIVIA**

**1.**Main reinforcement bar is normally used at the bottom of the slab.

Distribution bars are placed on the top of the main bar.

**2.**Main bar is used in shorter direction but distribution bar is used in longer span.

**3.**Higher dimension bar is used as main reinforcement bar.

Lower dimension bar is used as distribution bars.

**4.**Main reinforcement bar is used to transfer the bending moment to beams.

Distribution bars are used to resist the shear stress, and cracks developed at the top of the slab

**STEP 1**

Find Cutting Length Of

**Main Bars**For Floor Slab 1**Length Of Main Bars**= (depth of beam/2) + (width of beam/2) + (1 x inclined length) – (45° bend x 2) +

**length of shorter span**+ width of beam + (0.3 x intermediate slab length)

**Inclined length = 0.42D**

**D = Slab thickness – 2 side clear cover (top cover + bottom cover)**

D = 150 – (2 x 25) = 100mm

Inclined length = 0.42 x 100 = 42mm

**45° bend = 1d**

d = diameter of bar

45° bend = 1 x 12 = 12mm

**Intermediate slab**

L = 3100mm

**Therefore,**

**Length Of Main Bars**= (300/2) + (225/2) + (1 x 42) - (12 x 2) + 2950 + 300 + (0.3 x 3100) =

**4730.5**

**Cutting length Of Main Bars for floor slab 1= 4.8m**

**STEP 2**

No Of Bars Required For Cutting Main Bars

**No of bars = (opp length/spacing) + 1**

Opposite length of bar = 6000

Spacing = 150mm c/c as per given details

**= (6000/150) + 1**

**No of bars = 41 nos**

**STEP 3**

Total Length Of Bar Required For

**Main Bars**Of Floor Slab 1= no of bar x total length of bar

= 41 x 4.80 = 196.8m = 197m

**Total length of bar required for main bars of**

**Floor Slab 1 =**

**197m of 12mm rod**

**STEP 4**

Find Cutting Length Of Distribution Bars For Floor Slab 1

**Length of distribution bars =**

**(breadth of beam/2) + length of longer span + (breadth of beam/2)**

Length of distribution bars = (225/2) + 6000 +(225/2) = 6225mm

**Length of distribution bars = 6.225m**

**STEP 5**

No Of Bars Required For Cutting Distribution Bars

**No of bars = (opp length/spacing) + 1**

Opposite length of bar = 2950

Spacing = 150mm c/c as per given details

**= (2950/150) + 1**

**= 20.6666666667 = 21 nos**

**No of bars = 21nos**

**STEP 6**

Total Length Of Bar Required For

**Distribution Bars**Of Floor Slab 1= No Of Bar X Total Length Of Bar

= 21 x 6.225 = 130.725m

**Total length of bar required for distribution bars of Floor Slab 1 = 131m of 8mm rod**

**STEP 7**

**Top Bar (Extra):**Top Bars Are Provided At The Top Of Main Bar As Critical Length (L/4) Area

Number of top bars = (Lx/4) / spacing + 1

Number of top bars = ((2950/4) / 150) + 1 =5.91 = 6 nos of bars x 2 sides =

**12 nos****Number of top bars = 12 nos of bars**

**STEP 8**

Total Length Of Bar Required For

**Top Bars**Of Floor Slab 1**Length of top bars = Length of Distribution Bars**

Cutting Length Of top Bars =

**6.225m**

Total length of top bars =

**6.225**x 10 = 27.25m**Total length of top bars of Floor Slab 1= 63 m of 8mm rod**

**GIVEN DATA FOR FLOOR SLAB 2**

**SIZE OF FLOOR SLAB 1**

Length= 3100mm

Breadth= 6000mm

**Therefore,**

Lx shorter span = 3100mm

Ly larger span = 6000mm

Beam size = 225x300mm

Main bars are 12 mm in diameter @ 150 mm centre to centre spacing

Distribution bars are 8 mm in diameter @ 150 mm centre to centre spacing. (Main Bar & Distribution Bar Difference)

Top and Bottom Clear Cover is 25 mm

Consider Development length as 40 d

Thickness of Slab – 150 mm

**Therefore, To Find Its One Way Or Two Way Slab**

Lx = 3100mm (shorter span)

Ly = 6000mm (Larger span)

Ly/Lx= 6000/3100 = 1.93

Two Way Slab | Ly/Lx < 2 |

**Hence its Two-way slab**

So, We need to find

·

**Main bar**for shorter span·

**Main bar**for longer span·

**Top bar**for main rod on both sides for critical structure**STEP 1**

Find Cutting Length Of

**Main Bars**For Floor Slab 2 (A2-B2 direction)**Length Of Main Bars (A2-B2)**= (depth of beam/2) + (width of beam/2) + (1 x inclined length) – (45° bend x 2) + length of span + width of beam + (0.3 x intermediate slab length)

**Inclined length = 0.42D**

**D = Slab thickness – 2 side clear cover (top cover + bottom cover)**

D = 150 – (2 x 25) = 100mm

Inclined length = 0.42 x 100 = 42mm

**45° bend = 1d**

d = diameter of bar

45° bend = 1 x 12 = 12mm

**Intermediate slab**

L = 3000mm

**Therefore,**

**Length Of Main Bars**(A2-B2) = (300/2) + (225/2) + (1 x 42) - (12 x 2) + 3100 + 300 + (0.3 x 3000) = 4580.5

**Cutting length Of Main Bars for floor slab 2= 4.6m**

**STEP 2**

No Of Bars Required For Cutting Main Bars

**No of bars = (opp length/spacing) + 1**

Opposite length of bar = 6000

Spacing = 150mm c/c as per given details

**= (6000/150) + 1**

**No of bars = 41 nos**

**STEP 3**

Total Length Of Bar Required For

**Main Bars**Of Floor Slab 2= no of bar x total length of bar

= 41 x 4.6= 188.6m = 190m

**Total length of bar required for main bars of**

**Floor Slab 2**

**=**

**190m of 12mm rod**

**STEP 4**

Find Cutting Length Of

**Main Bars**For Floor Slab 2 (C2-D2 direction)**Length Of Main Bars (C2-D2)**= (depth of beam/2) + (width of beam/2) + (1 x inclined length) – (45° bend x 2) + length of span + (width of beam/2)

**Inclined length = 0.42D**

**D = Slab thickness – 2 side clear cover (top cover + bottom cover)**

D = 150 – (2 x 25) = 100mm

Inclined length = 0.42 x 100 = 42mm

**45° bend = 1d**

d = diameter of bar

45° bend = 1 x 12 = 12mm

**Intermediate slab**

L = 3000mm

**Therefore,**

**Length Of Main Bars**(C2-D2) = (300/2) + (225/2) + (1 x 42) - (12 x 2) + 6000 + (225/2) = 6393m

**Cutting length Of Main Bars for floor slab 2= 6.4m**

**STEP 5**

No Of Bars Required For Cutting Main Bars

**No of bars = (opp length/spacing) + 1**

Opposite length of bar = 3100

Spacing = 150mm c/c as per given details

**= (3100/150) + 1**

**No of bars = 21.6666666667 = 22**

**STEP 6**

Total Length Of Bar Required For

**Main Bars**Of Floor Slab 2= no of bar x total length of bar

= 22 x 6.4= 140.8m = 141m

**Total Length Of Bar Required For Main Bars Of**

**Floor Slab 2**

**=**

**141m Of 12mm Rod**

**STEP 7**

**Top Bar (Extra)**(A2-B2): Top Bars Are Provided At The Top Of Main Bar As Critical Length (L/4) Area

Number of top bars = (Ly/4) / spacing + 1

Number of top bars = ((6000/4) / 150) + 1 = 11 nos of bars x 2 sides = 22 nos

**Number of top bars = 22 nos of bars**

**Top Bar (Extra)**(C2-D2): Top Bars Are Provided At The Top Of Main Bar As Critical Length (L/4) Area

Number of top bars = (Lx/4) / spacing + 1

Number of top bars = ((3100/4) / 150) + 1 =6.16 = 7 nos of bars x 2 sides = 14 nos

**Number of top bars = 14 nos of bars**

**.**

**STEP 8**

Total Length Of Bar Required For

**Top Bars**Of Floor Slab 2 (A2-B2)**Length of top bars = Length of Opposite Bars**(C2-D2)

Cutting Length Of top Bars =

**3.325m**

Total length of top bars =

**3.325**x 14 = 46.55m**Total length of top bars of Floor Slab 2= 47 m of 8mm rod**

Total Length Of Bar Required For

**Top Bars**Of Floor Slab 2 (C2-D2)**Length of top bars = Length of Opposite Bars**(A2-B2)

Cutting Length Of top Bars =

**6.225m**

Total length of top bars =

**6.225**x 14 = 87.15m**Total length of top bars of Floor Slab 1= 88m of 8mm rod**

**RESULT,**

**FLOOR SLAB 1,**

Main Bars = 185m Of 12mm Rod

Distribution Bars = 131m Of 8mm Rod

Top Bars = 63m Of 8mm Rod

**FLOOR SLAB 2,**

Main Bars = 190m of 12mm rod

Main Bars = 141m Of 12mm Rod

Top Bars = 88 m of 8mm rod (A2-B2)

Top Bars = 47m of 8mm rod (C2-D2)

**TOTAL**

12mm rod required = 516m

8mm rod required = 329m

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