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General
Module 1. Perspective on Soil and Water Conservation
Module 2. Prerequisites for Soil and Water Conse...
Module 3. Design of Permanent Gully Control Struct...
Module 4. Water Storage Structures
Module 5. Trenching and Diversion Structures
Module 6. Cost Estimation
Lesson 32. Cost Estimation Examples
Computation for Material Requirement – Some Examples
Example 1: Calculate dry material required for 1 m^{3} cement concrete (CC) having density of 2.3 ton/m^{3} with 1:3:6 ratio.
Solution:
Ratio: 1:3:6 (cement: sand: 20 mm aggregate)
Sum = 1+3+6 = 10
From Table 31.1 (lesson 31), dry matter required for 1.0 m^{3} CC mix is 1.54 m^{3}
Cement requirement = = 0.15 m^{3}= 216 kg (using Table 32.1 below)= 4.3 bags of 50 kg each
Sand requirement = = 0.46 m^{3} = 1.22 ton
Aggregates = = 0.92 m^{3} = 1.56 ton (20 mm aggregates)
Table 32.1. Bulk density of different construction material
Material 
Bulk Density (kg/m^{3}) 

1440 

1600 

1700 

1950 

2700 

2650 

2160 
Example 2: Find dry matter for 1.0 m^{3} brick masonry with cement and sand mortar of 1:6 ratio.
Solution:
For 1.0 m^{3} brick work, 500 bricks are needed (from Table 31.1 described in lesson 31).
Cement: sand ratio = 1(cement):5(sand)
Sum = 1+5 = 6
For 1 m^{3} brick masonry, 0.3m^{3} cement mortar is required (see material requirement Table in lesson 31.1).Cement requirement = 0.05 m^{3} and
Sand requirement = 0.25 m^{3}
Some Example of Rate Analysis
Example 3: Determine unit cost for Gabion works
Solution:
The gabion work involves collection of stones (225 mm) and their arrangement in wire mesh accordingly. The material requires are GI wire mesh and stones. Labours (unskilled and semiskilled) worker are required for netting the wire mesh. The unit cost is computed as per the following analysis.
Particulars 
Quantity 
Rate (Rs/unit) 
Amount (Rs) 
Material 




18.2 kg 
60 
1092 

3.75 m^{3} 
200 
750 
Total material cost 
1842 

Labour 




no 
240 
120 

no 
220 
275 

180 
405 

Total labour cost 
800 

Total material and labour cost 
2642 

Add 3% contingency 
79.26(say 80) 

Grand total for 3 m^{3} 
2722 

Cost per m^{3} 
907.33(say 910) 
Example 4: Determine unit cost for logwood crib structures filled with stones.
Solution:
The logwood crib structures involve placement of stone (250 mm or more) in between the wooden pole to provide a barrier to flowing water. The required materials are wooden logs, nails, oils for painting, stones whereas labour requirement include skilled labour such as Carpenter, Mason, Painter and unskilled labour. The unit cost is computed as per the following analysis of rate for 15 m^{3} crib for span of 10 m.
Particulars 
Quantity 
Rate (Rs./unit) 
Amount (Rs.) 
Material 








22 poles 
120 
2,640 

30 poles 
140 
4,200 

20 kg 
60 
1,200 

5 lit 
40 
200 





5 lit 
30 
150 

15m^{3} 
200 
3,000 

Lumpsum 

1,000 
Total material cost 
12,390 

Labour 




10 nos. 
240 
2,400 

5 nos. 
240 
1,200 

2 nos. 
240 
480 

20 nos. 
180 
3,600 
Total labour cost 
7.680 

Total labour and material cost 
20,070 

Add 3% contingency 
602 

Grand total for 15 m^{3} 
20672 

Cost per m^{3} 
1378 (say 1380) 
Example 5: Determine unit cost for RCC work
Solution:
RCC work includes steel reinforcement of cement concrete. The most adopted cement concrete is of 1:2:4 mix of cement, sand and aggregates (volume basis). The density of concrete depends on the nature of aggregates used. Most frequently, 20mm stone ballast are used as aggregates. Reinforcement is done using steel bars. In watershed structures 12 mm steel bars are used. The unit cost can be computed as per the following analysis of rates.
Dry mortar required for 1m^{3} cement concrete = 1.54 m^{3}.
Thus for 10 m^{3} cement concrete, dry mortar volume will be 15.4 m^{3}
Cement requirement = 15.4/7 = 2.2 m^{3} = 64 bags of 50kg each
Sand requirement = 4.4 m^{3} and
Aggregates requirement = 8.8 m^{3}
Assuming density of concrete as 2300 kg/m^{3}, the weight of 10 m^{3} such concrete will be = 23 ton
Particulars 
Quantity 
Rate (Rs/unit) 
Amount(Rs) 
Material 



Cement grade 53 
64 bags 
280 
17,920 
Coarse sand (12 mm) 
4.4 m^{3} 
2000 
8,800 
Stone ballast 20mm 
8.8 m^{3} 
1200 
10,560 
Mild steel bar @1% reinforcement 
0.23 ton 
60000 
13,800 
Binding wires (1 mm) 
2 kg 
65 
130 
Total material cost 
51,210 

Labour 



Head mason (Raj mistri) 
1 
270 
270 
Mason (mistri) 
3 
240 
720 
Unskilled labour (beldar) 
12 
180 
2160 
Bhisti 
6 
220 
1320 
Sundries, T&P etc 
Lumpsum 

1000 
Total cost of labour 
5,470 

Centering and shuttering 



Timber planks and post 
On hire 

1,000 
Carpenter 
6 nos 
240 
1,440 
Unskilled labour (beldar) 
1 


Nails 
Lumpsum 

500 
T&P 
Lumpsum 

1,000 
SubTotal 
3,940 

Total of material and labour cost 
60,620 

Add 3% contingency 
1820 

Total cost for 10 m^{3} RCC 
62,440 

Cost of per m^{3} RCC 
6,244 (say 6,250) 
Example 6: Determine unit cost for Iclass brick work with 1:6 cement mortar
Solution:
First estimate cost for 10 m^{3} brick work in order to rationalize the labour requirement and mortar mix.
Dry matter requirement for mortar in brick work = 0.3m^{3} per m^{3} of brickwork. Thus 3.0m^{3} dry matter will be required for 10m^{3} brick work.
Cement requirement = (1/7)*3 = 0.43 m^{3} or 12.47 (say 13) bags of 50 kg each.
Sand requirement = (6/7)*3 = 2.58 m^{3}
The unit cost can be computed as per the following analysis of rates.
Particulars 
Quantity 
Rate (Rs/unit) 
Amount(Rs) 
Material 



Cement grade 53 
13 bags 
280 
3,640 
Coarse sand (12 mm) 
2.58 m^{3} 
2000 
5,160 
ClassI brick @500 per m^{3} 
5000 
4.50 
22,500 
Total material cost 
31,300 

Labour 



Head mason (Raj mistri) 
1 
270 
270 
Mason (mistri) 
10 
240 
2,400 
Unskilled labour (beldar) 
7 
180 
1,260 
Bhisti 
2 
220 
440 
Sundries, T&P etc. 
Lumpsum 

1,000 
Total cost of labour 
5,370 

Total of material and labour cost 
36,670 

Add 3% contingency 
1,100 

Total cost for 10 m^{3} RCC 
37,770 

Cost of per m^{3} RCC 
3,777 (say 3,800) 
Estimating and costing of some watershed management work  Examples
Example 7:
Estimate the cost to construct CCT in hard soil in 20 ha area. The distance between two row of CCT is kept as 25 m and width and depth of CCT is 50 cm each.
Solution:
The construction of CCT involves excavation only and so only labour is required with some T&P.
Length of CCT per ha can be calculated using following equation
Earthwork volume = 0.5*0.5*400 = 100 m^{3}
Particulars 
Quantity 
Rate (Rs/unit) 
Amount(Rs) 
Labour 



Unskilled labour @1.5 m^{3} per manday 
67 mandays 
180 
12,060 
Skilled labour 
2 mandays 
240 
480 
T&P 
Lumpsum 
 
500 
Total labour cost 
13,040 

Per m^{3} earth work 
130 
Example 8:
Gabion structures are proposed to stabilized 240 m long and 8 m wide gully. The average slope of the gully bed is 10% and the vertical interval may be taken as 2 meter. The proposed design is presented in Fig. 32.1. Estimate the project cost.
Example 2: Continuous Contour Trenches (CCT) in Heterogeneous Material
Department of Agriculture in Gaya District, Bihar proposes to make CCT in a catchment area of 40 hectares. The top soft soil stratum has a depth of 30 cm followed by hard soil for about 1 metre. The horizontal spacing between two successive rows CCT is 25 m. The width and depth of the CCT is 50 cm x 50 cm. Calculate the cost of construction of CCT.
Solution:
Fig. 32.1 Proposed design of gabion
Solution: Since the vertical interval is 2 meter and 10% slope, means gabion should be 20 m apart horizontally. Thus total number of gabion for the gully stabilization would be 240/20 = 12. The average gully width is 8 meter and 1 meter wide side wall is to be provided, the total width for earth work would be 8+2 =10m. The estimate involves computation of material and abstract of cost.
 Details of measurement and quantities
Sl. No. 
Particulars of work/items 
No 
Length (m) 
Width (m) 
Height (m) 
Quantity (m^{3}) 
1 
Excavation in foundation 
12 
10 
2 
0.6 
144 
2 
Gabion box filled with boulders of >200mm 







12 
10 
2 
1 
240 


12 
10 
1 
1 
120 


12 
4x2 (two sides per gabion) = 8 
1 
1 
96 
Total gabion work 
456 
 Abstract of cost
Sl. No. 
Particular of work/ items 
Quantity 
Unit 
Rate (Rs/Unit) 
Amount (Rs) 
1 
Excavation in foundation 
144 
m^{3} 
130(see example 7) 
18,720 
2 
Gabion work 
456 
m^{3} 
910(see example 3) 
414,960 
Total cost 
433,680 
References
Dutta, B.N. (1966). Estimating and Costing for Civil Engineering, Lucknow
Khanna, P.N. (1996). Indian Practical Civil Engineers’ Handbook
New Delhi: Engineers’ Publishers
Watershed Works Manual. (2007). Ministry of Rural Development, Government of India, pp.226273.
Mohan, S. C. et al. (2007). Training Manual on Soil Conservation & Watershed Management, pp. 380393
Sharda, V.N., Juyal, G.P., Prakash Chandra and Joshi, B.P. (2007). Training Manual – Soil Conservation and Watershed Management (Vol II Soil and Water Conservation Engineering). CSWCRTI, 218, Kaulagarh Road, Dehradun.