INTRODUCTION :-
To decide about tentative cost of project in advance, various items are required to be estimated. The approximate quantities of all the items are calculated as per the specifications. These quantities of items are then bifurcated to calculate the volume and cost of the material, labours, machineries, plants and equipments.
The volume of materials, labours and machineries required ,will help to decide whether the specifications and methods are workable or not, as per the local availability. The volume of materials estimated in advance, helps to decide about the required storage facilities. The volume of labours converted into mandays helps to decide about area required for transit camp of the workers.
All these related information helps to draw the primary layout of the site.
After confirming these primary requirements, the specifications, methods and procedures are finally decided. The rates of various materials, labours and machineries are analysed as per the final specifications. The rates of individual items are then analysed considering the consumption of material, labour, machinery and their market rates. The total cost is then derived with the help of initially surveyed quantity and the rate derived for each item.
The various overhead costs are then added for the consumable items (fuel, oil etc.), supervision charges, administrative expenses, contingencies and such costs which could add as per the importance and uniqueness of the project.
To judge the feasibility of a project, preliminary estimates are helpful. These preliminary estimates help to decide about, whether to start the project and how to start the project.
If the feasibility is proved, through the preliminary estimates, the detailed estimates can be worked out.
Procedure for preparing preliminary estimates
To ascertain the tentative cost of a project at a glance, in short time; the preliminary estimates are prepared. The preliminary estimates of ownership type buildings are prepared on the following guidelines.
The total cost of project = Direct cost + Indirect cost
WHERE
Direct cost includes,
- Cost of
- Cost of
- Cost of Labour.
Indirect cost includes:-
- Land development
- Legal
- Administrative
- Supervision
- Infastructural development
- Cost of Security
- Miscellaneous
- Any other costs as per the peculiarity of the
To fulfill the time essence of preparing preliminary estimates, certain rules are derived, based on practical past experience. These thumb rules help to decide the approximate project cost, quickly.
SOME OF THE THUMB RULES IN BUILDING CONSTRUCTION:-
To ascertain the preliminary cost of a residential building, certain assumptions are made for the rate of consumption of materials per square feet area of building. These assumptions are purely based on practical observations and records. These observations are converted into some thumb rules. These thumb rules will help to ascertain the preliminary cost of a building, easily.
SR.NO. | MATERIAL | QUANTITY PER SFT |
1. | Cement | 0.45 Bags |
2. | Steel | 2.2 to 3.5 Kg |
3. | Sand | 2.25 Cft. |
4. | Metal | 1.10 Cft |
5. | Ruble | 0.30 Cft. |
6. | Murrum | 075 Cft |
7. | Bricks 9″ x 4″ x 3″For plinth | 4 Nos. |
8. |
Bricks 6″ x 9″ x 4″ For external walls in superstructure |
4 Nos. |
9. |
Bricks 4″ x 9″ x 3″ For internal walls in superstructure |
4 Nos. |
10. | Sanala | 0.22 bags |
11. | Ceramic tiles | 0.80 Sft. |
12. | Ceramic skirting | 0.25 Sft |
13. | Bathroom Floorings | 0.10 Sft |
14. | Dado | 0.3 Sft |
15. | Kitchen platform | 0.012 Rft or minimum 8 feet per flat |
16. | Chequered tiles | 0.08 Sft |
17. | Staircase flooring | 0.10 Sft |
18. | Brickbat for waterproofing | 0.05 Cft |
19. | Door frames | 0.06 Nos. |
20. | Door shutters | 0.08 Sft |
21. | Staircase railing | 0.003 Rft |
22. | Aluminium window | 0.09 Sft |
23. | Light point including common areas | 0.022 Nos. |
24. | Power points | 0.02 Nos. |
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DETAILED ESTIMATE: –
To prepare a detailed estimate, quantities of all the items are worked out in detail, with the help of finalized specifications and detailed drawings. The total project cost is then calculated by taking into account the quantities worked out and considering the prevailing market rates. The rate analysis of the individual item is done with the help of market rates of labours, materials, stipulated task work, cost of machinery, tools & plants. The total process study of an item helps to decide the exact quantum of material, labour, task work and components of machinery in that item.
The detailed estimate gives us the total cost of a project and helps us to take the following decisions.
- To decide the exact date of starting the
- To decide the total project
- To plan the work with the help of suitable tools e.g. Bar
- To decide the required cash
- To decide about various itemwise contracts and most suitable type of
- To freeze the bill of
- To take the advantages of fluctuations in market
An exercise will be self-explanatory to understand the process of estimating and costing.
Consider a bedroom having clear dimensions of 4.27m x 3.35m in plan. The Floor to floor height is 3.05m.
The sizes of the openings: Door (D1): – 1m x 2.1m, Door (D2): – 0.75m x 2.1m, Window (W1): – 2m x 1.2m
Â
Columns
C-1 : 0.23 X 0.60
C-2 : 0.23 X 0.60
C-3 : 0.23 X 0.45
C-4 : 0.23 X 0.45 C1<
BeamsÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â C4
B-1 : 0.23 X 0.60
B-2 : 0.23 X 0.45
B-3 : 0.23 X 0.45
Â
Doors
D-1 : 1.0 X 2.1
D-2 : 0.75 X 2.1
Â
Window
B2 < C2 <
B3 C3
B1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â W1
Estimate the quantities of materials and labours for 6″ thick B.B.M.
Sr. No. |
Particulars | Nos. |
Length (L) mtr. |
Breadth (B) mtr. |
Height (H) mtr. |
Area in Sq.mtr. |
1. | Long walls | Â | Â | Â | Â | Â |
Â | L1 | 1 | 4.11 | – | 2.45 | 10.05 |
Â | L2 | 1 | 3.74 | – | 2.45 | 9.16 |
2. | Short walls | Â | Â | Â | Â | Â |
Â | S1 | 1 | 2.82 | – | 2.60 | 7.33 |
Â | S2 | 1 | 2.45 | – | 2.60 | 6.37 |
Â | Â | Total B.B.M. (A) | 32.92 | |||
Â 3. |
Â Deductions |
Â |
Â Â Â – – Â 2 |
Â | Â | Â |
Â | Doors | Â | Â | Â | Â | |
Â | D1 | 1 | 1 | 2.1 | 2.10 | |
Â | D2 | 1 | 0.75 | 2.1 | 1.57 | |
Â | Window | Â | Â | Â | Â | |
Â | W1 | 1 | – | 1.2 | 2.40 | |
Â | Total deduction for openings (B) | 6.07 | ||||
Net B.B.M. quantity (A-B) | 26.85 |
Â
Therefore the total quantity of B.B.M. for bed room = 242.5 square feet Material Analysis for 6″ thick B.B.M. in C.M. 1:6
- 6″ Bricks: –
bricks required for 100 sft B.B.M. work Volume of B.B.M.= 100 sft x 6″ = 50 cft. Consider 30% wet mortar.
Total quantity (Volume) of bricks= 50 x 0.7= 35 cft.
Volume of one brick= 9â€™â€™ x 6″ x 4″= 0.75 x 0.50 x 0.33 = 0.12 cft. No. of bricks required = volume of brick work / volume of one brick
= 35/0.12
= 291.66 Say 292 No.
Add 2% wastageÂ Â Â = 292 x 1.02 = 297.84Â Â Â Â Â Say 300 Nos. Nos. of bricks required for 100sftÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 300 Nos.
Nos. of bricks for 242.5 sftÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 242.5 x 300 =727.5 say 730 Nos.
- Cement: –
Quantity of mortar = 15 cft. For C.M. 1:6
Cement = 1/7 x 15=2.14 cft
Volume of one bag of cement = 1.25 cft.
Quantity of cement = 2.14/1.25 =1.712 say 1.75bags
Quantity of cement for 242.5 sft B.B.M.= 242.5/100 x 1.75 =4.243 Quantity of cement = 4.243 Say 4.25 bags
Quantity of sand = 15cft â€“2.14 cft = 12.86 cft.
Quantity of sand for 242.5 cft B.B.M.= 242.5/100 x12.86 =31.18 cft Quantity of sand = 31.18 cft Say 32 cft.
Water = As required for workable mix.
Â
Rate Analysis:-Â Â Â Â Â Â Â Â Â Â Â For 6″ B.B.M. in C.M. 1:6 for 100 sft.
Sr No. | Particulars | Quantity | Unit | Rate/unit | Amount |
1. |
Material a)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Cement b)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Sand c)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Bricks |
Â 1.75 12.86 300 |
Â Bags Cft Nos. |
Â 160/- 20/- 3.5/- |
Â 280=00 257=20 1050=00 |
Â | Total cost of material (M) | Â | Â | 1587=20 | |
2. |
Labour a)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â B.B.M. b)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Curing |
Â 100 – |
Â Sft – |
Â 4/- – |
Â 400=00 10=00 |
Â | Total cost of labour (L) | Â | Â | 410=00 | |
3. | Total material + labour (M+L) | Â | Â | 1997=20 | |
Â | Add 1% for supervision charges | Â | Â | 19=97 | |
Â | Total cost of B.B.M (100 sft) | Â | Â | 2017=17 | |
Â | Â | Say | Â | 2020=00 |
Therefore cost of B.B.M. for 100 Sft = Rs.2020/- From the example above,
Cost for B.B.M. for 242.5 sft is :-
242.5/100 x2020 =4898.5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Say Rs.4900/-
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