# Math solution is offer for beginners

### Math solution is offer for beginners

Problem:  There is a footing plan in the figure. The size of the footing is:

8′-6 “× 8′-6″ 2′-0 ”

One layer of brick soling should be given on the floor of the footing,

CC should be cast 3 “thick on the soling.

The 20 mm diameter rod will fit 4 “C / C on both sides of the footing.

Based on all this information, the necessary estimate of the footing has to be made.

## Solution: –

1) Matikata calculation should be done at the beginning.

Since the shutter has to be built. Therefore,

in addition to the size of the footing, the soil should be dug for

more than 6 “or 15 cm on both sides.

### Earth Work

1 × 9′-6 “9′-6″ 2′-6 ”

= 225.625 Cft

In the case of formwork, the calculation will

come only by multiplying the excavated height with

the actual range around the footing. In this case,

solving and CC have to be done inside the framework.

The amount of soling and CC is correct.

### Formworks = 1×2(8′-6″+8′-6″)×2′-6″

= 85.00 Sft.

3) When determining the soling of a brick,

the length and width should  multipli

according to the actual size of the footing.

### BFS Works = 1 × 8′-6 “× 8′-6” = 72.25 Sft.

4) The calculation of CC will be the same as soling, just multiply the thickness with it.

CC Works (1: 3: 6)

= 1 × 8′-6 “× 8′-6″ 0′-3 ”

= 18.06 Cft.

5) The calculation of RCC will be related to the actual size of the footing.

However, the ratio must written.

### RCC Works (1: 2: 4)

= 1 × 8′-6 “× 8′-6” × 2′-0 “= 144.50 Cft.

6) Calculate the rod. From the actual measurement of the footing,

cover 3 “on both sides of the covering and subtract the number by spacing.

Multiply the length of the rod by covering the number without covering.

The total length of the rod can found. Multiply the weight of the rod.

Since the size of the footing is call square.

So if we find the number in any one and multiply by 2,

we get all the numbers. Notice again in the figure,

one rod has a 3 “covering, the next rod has a 9” covering.

In such cases, a rod would be larger. The next rod will be smaller.

In this way, it has arrange one after the other.

That means the rod will save.

Now the number of rods

= [{8′-6 “- (2 × 3”) ÷ 4 “] + 1

= 25 Nos.

If you put rods 4 “in a row on one side,

it will take 25 rods.

Half of these 25 rods will be big and half will be small.

That means 13 will be bigger, which will be the size

= 8′-6 “- (2 × 3″) = 8′-0 ”

And 12 will be smaller than that (covering 9 “)

= 8′-6″ – (2×9″) = 7′-0”

### Estimate for Reinforcement:-

20 mm Dia Rod @ 4″C/C Both way.

= 2×13×8′-0″ = 208.00 Rft.

= 2×12×7′-0″ = 168.00 Rft.

Total = 376.00 Rft.

We know  20mm the rod will have a single weight= (20×20)÷162.2 = 2.46 Kg/m

= 2.46÷3.281 = 0.75 Kg/Rft

Total Rod = 376.00×0.75 = 282.00 Kg.