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What is the Procedure for Calculating Steel Quantity for Circular Slabs?
There is a relationship between concrete slab thickness and the amount of load on it as well as its size. Residential and commercial buildings normally use 6 inch or 150mm slab thicknesses with reinforcement details as per design. Depending on the type of slab, different methods are used to determine its thickness.
Suppose you need to calculate the reinforcement for a circular slab of 2 Meter diameter with spacing of 200 mm c/c and a 25 mm clear cover. There is something different about this RCC member since the rod length will vary based on the diameter of the circle.
Purpose of Circular Slab Reinforcement
A circular slab is used according to the architectural or structural needs of a structure. The materials could be the floors, rooftops, silos & bunkers of buildings, as well as water tanks, silos & bunkers. Calculating the steel length requires using Pythagoras's theorem, which differs from standard ones such as the painting area coefficient.
Discuss Circular Slab Reinforcement using Pythagoras Theorem
In this case, L1 is the diameter of the circular slab reinforcement. We must use Pythagoras' theorem to determine L2.
- L2 equals to √ (R2 – h12) * 2
- L3 equals to √ (R2 – h22) * 2
- L4 equals to √ (R2 – h32) * 2
In this case, L is the reinforcement bar's length,
E*cluding the clear cover on both sides, R is the radius of the circular slab reinforcement.
The distance between the bar and the main bar is H. It is necessary to calculate the Typical L's only for the first half; the second half will be multiplied by 2 excluding the main bar.
Circular Slab Reinforcement Calculation Process
For a slab with a diameter of 2 meters, a main rod of 12 mm and a distribution rod of 10 mm, spaced 200 mm apart, with a 25 mm clear cover; we need to calculate the Circular slab reinforcement. The main distribution reinforcement details are now available below.
In order to calculate the Ls, first determine how many rods there are, which means how many rods we need.
The circular reinforcement diameter is equal to the diameter of the circular slab - the clear cover on both sides = 2000 mm - 25 mm - 25 mm = 1950 mm. L = 1.95 meters or 1950 mm.
The distance between L and R is equal to 1.95 meters divides two = 0.975 meters.
A number of L's equals the radius of the circular reinforcement times the center-to-center distance (.9755/.200) = 4.87 L's
The number of rods will be 5 - 1 since there will only be one center rod = 4 * 2 = 8.
The ne*t step is to calculate eight numbers of main rods and eight numbers of distribution rods.
In fact, L1 is the main bar's diameter, which is 1.95 m
- The formula for L2 is √ (0.975)2 – (0.2)2 * 2 = 1.87 m
- L3 equals to √ (0.975)2 – (0.4)2 * 2 = 1.63 m
- L4 equals to √ (0.975)2 – (0.6)2 * 2 = 1.23 m
- L5 equals to √ (0.975)2 – (0.8)2 * 2 = 0.67 m
- First part Reinforcement = 1.87 m + 1.63 m +1.23 m +0.67 m = 5.4 m
- Second part Reinforcement = (1st Half * 2) = 5.4 m * 2 = 10.8 m
Total Main rod Reinforcement = 1st half + 2nd half + Centre Rod (L1) = 10.8 m + 1.95 m = 12.75 m
To get more clear ideas, go through the following exclusive construction video tutorial.
Lecturer and Image Source: Engr Waseem Raja
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