Engineeringcivil: HOW TO CALCULATE STEEL QUANTITY FOR SLAB?-REINFORCEMENT CALCULATION :TWO WAY SLAB

A Vision To Clear Solutions.

SLAB REINFORCEMENT CALCULATION

One way Slab Reinforcement DetailingThe Biggest misconception is considering beam outer to outer as slab span.Please refer the below diagram.

Let's take an example of the below one way slab diagram.

Given

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

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

Top & Bottom clear cover is 25mm

Consider Development length as 40d

Thickness of slab -150mm

One way slab Bar bending Schedule Calculation

Step-1

First ,find number of rods required for main reinforcement and distribution
Number of Bars Formula=(Length of slab/spacing)+1
Number of Main Bars=(Ly/spacing)+1=(5000/150)+1=34 nos
Number of Distribution Bars=(Lx/spacing)+1=(2000/150)+1=14 nos

Step 2
Find cutting length of main bars and distribution bars

Cutting Length of Main Bar,
= Clear Span of Slab (Lx) + (2 X Developement Length) +(1xInclined length) – (45°bend x 2)
Crank Length = 0.42 D, We have already discussed this on Cutting Length of Main Bar post
= 2000 + (2 x 40 x 12) + (1 x 0.42 x D) – (1d x 2)
= 2000 + 960 + 0.42D – (1x12x2) = 2960 +0.42D – 24

D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 2960+(0.42 x 88) – 24 = 2973 mm or 2.97 m

Step 3

Find length of distribution bar
= Clear Span (Ly) + (2 x Development Length (Ld))
= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A
Number of top bars = (Lx/4) / spacing + 1 = (2000/4) / 150 +1 = 4 Nos x 2 sides = 8Nos
Length of Top Bar (L) = same as distribution bars = 5.64 m

Bar Bending Schedule for One Way Slab

Two way slab Reinforcement Detailing

Now Calculate Bar Bending Schedule for Two Way Slab.

Let’s take an example the below two-way slab diagram

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.

Top and Bottom Clear Cover is 25 mm

Development length – 40 d

Thickness of Slab – 150 mm

One Way Slab Bar Bending Schedule Calculation

Step 1

First, find number of rods required for main reinforcement and distribution

Number of Required Bars Formula = (Length of slab / spacing) + 1

Number of Main Bars = Ly / spacing + 1 = (4000/150) + 1 = 27 nos

Number of Distribution Bars = Lx / spacing + 1 = 3000/150 + 1 = 21 nos

Step 2
Find cutting length of main bars and distribution bars

Cutting Length of Main Bar,
= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)
Inclined Length = 0.42 D
= 3000 + (1 x 40 x 12) + (1 x 0.42 x D) – (1d x 2)
= 3000 + 480 + 0.42D – (1x12x2) = 3480 + 0.42 D -24
D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 3480+(0.42 x 88)-24 = 3492.9 mm or 3.49 m

Step 3

Find cutting length of distribution bar

= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)

Inclined Length = 0.42 D
= 4000 + (1 x 40 x 8) + (1 x 0.42 x D) – (1d x 2)
= 4000 + 320 + (0.42×88) – (1x8x2) = 4340.96mm or 4.34 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A

Number of Top bars on Ly side = (Lx/5) / spacing + 1 = (3000/5) / 150 +1 = 5 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 4000+(24012) = 4000+960 = 4960 mm or 4.96 m

Number of Top bars on Lx side = (Ly/5) / spacing + 1 = (4000/5) / 150 + 1 = 6 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 3000+(24012) = 3000+960 = 3960 mm or 3.96 m

Bar Bending Schedule for Two Way Slab

NOTE

Here we have taken development length as 40d. However, In practice, the structural drawing itself has this dimension or else you have to extend the rod up to the beam end minus clear cover.

Engineeringcivil

Sunday, July 8, 2018

SLAB REINFORCEMENT CALCULATION

One way Slab Reinforcement DetailingThe Biggest misconception is considering beam outer to outer as slab span.Please refer the below diagram.

Let's take an example of the below one way slab diagram.

Step-1

First ,find number of rods required for main reinforcement and distribution
Number of Bars Formula=(Length of slab/spacing)+1
Number of Main Bars=(Ly/spacing)+1=(5000/150)+1=34 nos
Number of Distribution Bars=(Lx/spacing)+1=(2000/150)+1=14 nos

Step 2
Find cutting length of main bars and distribution bars

D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 2960+(0.42 x 88) – 24 = 2973 mm or 2.97 m

Step 3

Find length of distribution bar
= Clear Span (Ly) + (2 x Development Length (Ld))
= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A
Number of top bars = (Lx/4) / spacing + 1 = (2000/4) / 150 +1 = 4 Nos x 2 sides = 8Nos
Length of Top Bar (L) = same as distribution bars = 5.64 m

Bar Bending Schedule for One Way Slab

Two way slab Reinforcement Detailing

Now Calculate Bar Bending Schedule for Two Way Slab.

Let’s take an example the below two-way slab diagram

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.

Top and Bottom Clear Cover is 25 mm

Development length – 40 d

Thickness of Slab – 150 mm

One Way Slab Bar Bending Schedule Calculation

Step 3

Find cutting length of distribution bar

= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)

Inclined Length = 0.42 D
= 4000 + (1 x 40 x 8) + (1 x 0.42 x D) – (1d x 2)
= 4000 + 320 + (0.42×88) – (1x8x2) = 4340.96mm or 4.34 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A

Number of Top bars on Ly side = (Lx/5) / spacing + 1 = (3000/5) / 150 +1 = 5 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 4000+(24012) = 4000+960 = 4960 mm or 4.96 m

Number of Top bars on Lx side = (Ly/5) / spacing + 1 = (4000/5) / 150 + 1 = 6 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 3000+(24012) = 3000+960 = 3960 mm or 3.96 m

Bar Bending Schedule for Two Way Slab

NOTE

Here we have taken development length as 40d. However, In practice, the structural drawing itself has this dimension or else you have to extend the rod up to the beam end minus clear cover.

Sunday, July 8, 2018

SLAB REINFORCEMENT CALCULATION

One way Slab Reinforcement DetailingThe Biggest misconception is considering beam outer to outer as slab span.Please refer the below diagram.

Let's take an example of the below one way slab diagram.

Step-1

First ,find number of rods required for main reinforcement and distributionNumber of Bars Formula=(Length of slab/spacing)+1Number of Main Bars=(Ly/spacing)+1=(5000/150)+1=34 nosNumber of Distribution Bars=(Lx/spacing)+1=(2000/150)+1=14 nos

Step 2Find cutting length of main bars and distribution bars

D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mmLength of Main Bar = 2960+(0.42 x 88) – 24 = 2973 mm or 2.97 m

Step 3

Find length of distribution bar= Clear Span (Ly) + (2 x Development Length (Ld))= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,Please refer the drawing section A-ANumber of top bars = (Lx/4) / spacing + 1 = (2000/4) / 150 +1 = 4 Nos x 2 sides = 8NosLength of Top Bar (L) = same as distribution bars = 5.64 m

Bar Bending Schedule for One Way Slab

Two way slab Reinforcement Detailing

Now Calculate Bar Bending Schedule for Two Way Slab.

Let’s take an example the below two-way slab diagram

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. Top and Bottom Clear Cover is 25 mm Development length – 40 d Thickness of Slab – 150 mm One Way Slab Bar Bending Schedule Calculation

Step 3

Find cutting length of distribution bar

= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)

Inclined Length = 0.42 D= 4000 + (1 x 40 x 8) + (1 x 0.42 x D) – (1d x 2)= 4000 + 320 + (0.42×88) – (1x8x2) = 4340.96mm or 4.34 m

Step 4

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,Please refer the drawing section A-A

Number of Top bars on Ly side = (Lx/5) / spacing + 1 = (3000/5) / 150 +1 = 5 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 4000+(2*40*12) = 4000+960 = 4960 mm or 4.96 m

Number of Top bars on Lx side = (Ly/5) / spacing + 1 = (4000/5) / 150 + 1 = 6 Nos

Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)

= 3000+(2*40*12) = 3000+960 = 3960 mm or 3.96 m

Bar Bending Schedule for Two Way Slab

NOTE Here we have taken development length as 40d. However, In practice, the structural drawing itself has this dimension or else you have to extend the rod up to the beam end minus clear cover.

First ,find number of rods required for main reinforcement and distribution
Number of Bars Formula=(Length of slab/spacing)+1
Number of Main Bars=(Ly/spacing)+1=(5000/150)+1=34 nos
Number of Distribution Bars=(Lx/spacing)+1=(2000/150)+1=14 nos

Step 2
Find cutting length of main bars and distribution bars

D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 2960+(0.42 x 88) – 24 = 2973 mm or 2.97 m

Find length of distribution bar
= Clear Span (Ly) + (2 x Development Length (Ld))
= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A
Number of top bars = (Lx/4) / spacing + 1 = (2000/4) / 150 +1 = 4 Nos x 2 sides = 8Nos
Length of Top Bar (L) = same as distribution bars = 5.64 m

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.

Top and Bottom Clear Cover is 25 mm

Development length – 40 d

Thickness of Slab – 150 mm

One Way Slab Bar Bending Schedule Calculation

Inclined Length = 0.42 D
= 4000 + (1 x 40 x 8) + (1 x 0.42 x D) – (1d x 2)
= 4000 + 320 + (0.42×88) – (1x8x2) = 4340.96mm or 4.34 m

Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A

= 4000+(24012) = 4000+960 = 4960 mm or 4.96 m

= 3000+(24012) = 3000+960 = 3960 mm or 3.96 m

NOTE

Here we have taken development length as 40d. However, In practice, the structural drawing itself has this dimension or else you have to extend the rod up to the beam end minus clear cover.