We Already Discussed How To Prepare Bar
bending Schedule For Rectangular And Square Shape Slab. Now We Are Going To
Learn How To Calculate Reinforcement For Circular Slab.

In Circular Slab Reinforcement Length Is
Changed Due To Spacing Of The Reinforcement.

For This Cause Using Pythagoras Theorem To
Find Out The Length Of Each Reinforcement.

L= Diameter Of Slab – Nominal Cover

For Pythagoras Theorem L1= (√R

^{2 }– H_{1}^{2})2 H_{1,}H_{2,}H_{3 }=Spacing Of Reinforcement
L2=
(√R

^{2 }– H_{2}^{2})2 R = Radius Of Slab ( Excluding Nominal Slab )
Same Steps For Other Reinforcements

Example For Circular Slab Reinforcement
Calculation

Find The Reinforcement Details Having The
Diameter Of 3 m Providing Main
Reinforcement 10 mm Dia And Distributor 8 mm Dia At Spacing 200 mm And Nominal
Cover Is 25 mm

✦ L = 3 – 2 ( 0.05 )

L =2.95 m

✦ L1 = (√R

^{2 }– H_{1}^{2}) X 2
L1 =
(√1.475

^{2 }– 0.2^{2})2
L1= 2.92 M

✦ L2= (√R

^{2 }– H_{2}^{2})2
L2= (√1.475

^{2 }– 0.4^{2})2
L2= 2.84 M

L3= (√R

^{2 }– H_{3}^{2}) X 2
L3= (√1.475

^{2 }– 0.6^{2}) X 2
L3= 2.70 M

✦ L4 = (√R

^{2 }– H_{4}^{2}) X 2
L4 = (√1.475

^{2 }– 0.8^{2}) X 2
L4 = 2.48 M

✦ L5 = (√1.475

^{2 }– 1^{2}) X 2
L5 = 2.16 M

✦ L6 = (√R

^{2 }– H_{6}^{2}) X 2
L6 = (√1.475

^{2 }– 1.2^{2}) X 2
L6 = 1.72 M

✦ L7 = (√R

^{2 }– H_{7}^{2}) X 2
L7 = (√1.475

^{2 }– 1.4^{2}) X 2
L7 = 0.93 M

Total Length Of Main And Distribution
Reinforcements

= L + 2(L1)+ 2(L2)+ 2(L3)+ 2(L4)+ 2(L5)+ 2(L6)+
2(L7)

= 2.95 + 2(2.92) + 2(2.84)+ 2(2.70)+ 2(2.48)+
2(2.16)+ 2(1.72)+ 2(0.93)

= 34.45 M

Total Weight Of 10 Mm Dia Main
Reinforcement

Weight Of 10 Mm Bar Per Metre Length =
0.617 Kg /M

Weight = 34.45 X 0.617

= 21.26 Kg

Total Weight Of 8 Mm Dia Main Reinforcement

Weight Of 8 Mm Bar Per Metre Length = 0.395
Kg /M

Weight = 34.45 X 0.395

= 13.60 Kg

Total Weight Of Reinforcement For This Slab
= 21.26 + 13.60

=34.88 Kg

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