Today I will not discuss about the calculation of the need for the work of reinforced concrete columns or reinforced concrete beams because most of the work is almost the same as the calculation of reinforced concrete sloof, the fundamental difference in the three items is limited to the cross-sectional area, the number and diameter of the iron, type planned concrete, and on formwork work.
The work that gets my attention today is the work of a reinforced concrete floor plate, this floor plate is on the 2nd floor onwards as well as the roof that uses a reinforced concrete plate.
The calculation of labor and material requirements on the work of reinforced concrete floor plates is just an example and these calculations can be used if needed by anyone, hopefully useful.
This reinforced concrete floor plate has a function as a floor on a building more than one floor, but in many-storey buildings such as buildings that have more than five floors no longer use methods and materials such as reinforced concrete floor plates usually on two- or more-storey houses
In traditional house building, reinforced concrete floor plate has not been used considering the abundance of wood used as building material. In accordance with the development of the age of the use of wood materials ultimately considered expensive and complicated treatment so that reinforced concrete becomes a cheaper option.
We will plot a reinforced concrete floor plate with dimensions of 4.00 m x 2.00 m x 0.12 m and use a double layer Ø12 - 150 steel reinforcement.
In the work of reinforced concrete floor plate there are several items that we must count
- Concrete work (K-225)
- Occupation work
- Works formwork
The picture below is a drawing plan that we will count
) use autoCAD application
) use the autoCAD application
)use autoCAD applicationa. Concrete work (K-225)
In accordance with the volume calculation then the formula to solve it is
Length x width x height = m3
8.00 m x 2.00 m x 0.12 m = 1.92 m3
b. Occupation work
Is known :
| Items | Information |
|---|---|
| Plain coefficient of iron | 0.00616225 x Iron diameter ^ 2 |
| Iron diameter | Ø 12 |
| Count of iron / m | 0.00616225 x Ø 12 ^ 2 = 0.887364 Kg / m |
| Length of horizontal iron iron | (0,08 + 7,94 + 0,08) x2 = 16,24 m |
| Iron vertical iron length | (0.08 + 1.94 + 0.08) x2 = 4.24 m |
| Number of horizontal iron | ((8,00 / 0,15) + 1 = 54,33 Units |
| Vertical Iron Count | ((2.00 / 0.15) + 1 = 44.33 Units |
| The total length of iron | 943,1467 m |
| Total Weight total Iron | 0.887364 Kg / m x 943,1467 m = 836,9144 Kg |
c. Works Formwork
At formwork work we will calculate the bottom formwork and per side of concrete
| Items | Information |
|---|---|
| Bottom area 8.00 x 2.00 | 16.00 m2 |
| Concrete side 0.14 x 20.00 | 2.80 m2 |
| Total Amount | 18.80 m2 |
I will use the Indonesian National Standard Analysis to complete the calculation of reinforced concrete floor plate work.
a. Concrete work (K-225)
To complete 1 m3 of K-225 concrete work is required:
| Items | Information |
|---|---|
| Worker | 1,650 People Per Day |
| Stonemason | 0.275 People Per Day |
| Foreman | 0,028 People Per Day |
| Overseer | 0.083 Person Per Day |
| Portland Cement | 371,000 Kg |
| Concrete Sand | 698,000 Kg |
| Pebbles Net | 1,047,000 Kg |
| Water | 215 Liter |
In accordance with the calculation of the volume above then we will find out how many needs are needed with a volume of 1.920 m3
| Items | Information | Amount |
|---|---|---|
| Workers | 1,650 Person Per Day x 1,920 m3 | 3,168 Person |
| Stonmason | 0.275 Person Per Day x 1,920 m3 | 0.528 Person |
| Foreman | 0,028 Person Per Day x 1,920 m3 | 0.054 Person |
| Overseer | 0,083 Person Per Day x 1,920 m3 | 0,159 Person |
| Portland Cement | 371,000 Kg x 1,920 m3 | 712,320 Kg |
| Concrete Sand | 698,000 Kg x 1,920 m3 | 1,340,160 Kg |
| Pebbles Net | 1,047,000 Kg x 1,920 m3 | 2,010,240 Kg |
| Water Supply | 215 Liter x 1,920 m3 | 412,800 Liter |
b. Occupation work
To complete 1 Kg of plain concrete iron work is required:
| Items | Information |
|---|---|
| Worker | 0,0070 People Per Day |
| Blacksmith | 0,0070 People Per Day |
| Foreman | 0.0007 People Per Day |
| Overseer | 0.0004 People Per Day |
| Iron Concrete Plain | 1.0500 Kg |
| Concrete Wire | 0.0150 Kg |
In accordance with the calculation of the volume above then we will find out how many needs are needed with the volume 836.9144 Kg
| Items | Information | Amount |
|---|---|---|
| Worker | 0,0070 People Per Day x 836,9144 | 5,8584 People |
| Blacksmith | 0,0070 People Per Day x 836,9144 | 5,8584 People |
| Foreman | 0.0007 People Per Day x 836,9144 | 0,5858 People |
| Overseer | 0.0004 Person Per Day x 836,9144 | 0,3348 People |
| Iron Concrete Plain | 1.0500 Kg x 836,9144 | 878,7601 Kg |
| Concrete Wire | 0.0150 Kg x 836,9144 | 12,5537 Kg |
c. Works Formwork
To complete 1 m2 of floor formwork work is required:
| Items | Information |
|---|---|
| Worker | 0.660 Person Per Day |
| Carpenter | 0.330 Person Per Day |
| Foreman | 0.033 Person Per Day |
| Overseer | 0,033 Person Per Day |
| Wood Klass III | 0.040 m3 |
| Ordinary nail | 0.400 Kg |
| Oil Formwork | 0,200 m3 |
| Wood Klass II | 0.150 m3 |
| Multiplex 9 mm | 0,350 Sheet |
| Dolken Wood Ø 8 - 10, length 4,00 m | 6,000 Trunk |
In accordance with the above calculation then we will find out how many needs are needed on the work of floor plate formwork
| Items | Information | Amount |
|---|---|---|
| Worker | 0.660 Person Per Day x 18.80 m2 | 12,408 Person |
| Carpenter | 0.330 Person Per Day x 18.80 m2 | 6,204 Person |
| Foreman | 0,033 Person Per Day x 18.80 m2 | 0,620 Person |
| Overseer | 0,033 Person Per Day x 18,80 m2 | 0,630 Person |
| Wood Klass III | 0.040 m3 x 18.80 m2 | 0.752 m3 |
| Ordinary nails | 0.400 Kg x 18.80 m2 | 7,520 Kg |
| Oil Formwork | 0,200 Liter x 18,80 m2 | 3,760 Liter |
| Wood Klass II | 0.150 m3 x 18.80 m2 | 0.282 m3 |
| Multiplex 9 mm | 0,350 Sheet x 18,80 m2 | 6,580 Sheet |
| Dolken Wood Ø 8 - 10, length 4,00 m | 6,000 Trunk x 18,80 m2 | 112,800 Trunk |
In accordance with technical calculations using the analysis of Indonesian National Standard then to complete the work of reinforced concrete floor plate in accordance with the above mentioned specifications it is required
| Items | Amount |
|---|---|
| Workers | 21,434 Person |
| Stonmason,Blacksmith & Carpenter | 12,590 Person |
| Foreman | 1,260 Person |
| Overseer | 1,115 Person |
| Portland Cement | 712,320 Kg |
| Concrete Sand | 1.340,160 Kg |
| Gravel | 2,010,240 Kg |
| Water | 412,800 Liter |
| Iron Concrete Plain | 878,760 Kg |
| Concrete Wire | 12,554 Kg |
| Wood Klass III | 0.752 m3 |
| Ordinary nail | 7,520 Kg |
| Oil Formwork | 3,760 |
| Wood Klass II | 0.282 m3 |
| Multiplex 9 mm | 6,580 Sheet |
| Wooden Dolken | 112,800 Trunk |
This calculation is only a case example for young engineers and lay people to understand easily, these forms of calculation can be used properly by anyone and can continue to grow in accordance with the needs of the times.
This is all I can write on this occasion, for all the lack of writing as a servant of God I apologize.
Thank you to friends who have visited this blog and to all the actors of science who are willing to sacrifice energy, mind and time for the advancement of human civilization.
source:
en.wikipedia.org Technical analysis
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