Lesson 32. Cost Estimation Examples

Computation for Material Requirement – Some Examples

Example 1: Calculate dry material required for 1 m3 cement concrete (CC) having density of 2.3 ton/m3 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 m3 CC mix is 1.54 m3

Cement requirement =Example 1 = 0.15 m3= 216 kg (using Table 32.1 below)= 4.3 bags of 50 kg each

Sand requirement =  = 0.46 m3 = 1.22 ton

Aggregates = = 0.92 m3 = 1.56 ton (20 mm aggregates)

 Table 32.1. Bulk density of different construction material

Material

Bulk Density (kg/m3)

  1. Cement

1440

  1. Aggregates (10 mm)

1600

  1. Aggregates (20 mm)

1700

  1. Aggregates (40 mm)

1950

  1. Solid rock

2700

  1. Sand

2650

  1. Cement mortar

2160

 

Example 2: Find dry matter for 1.0 m3 brick masonry with cement and sand mortar of 1:6 ratio.

 Solution:

For 1.0 m3 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 m3 brick masonry, 0.3m3 cement mortar is required (see material requirement Table in lesson 31.1).Cement requirement = 0.05 m3 and

Sand requirement = example_20.25 m3

 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 semi-skilled) 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

 

 

 

  1. GI wire 10 gauge with 10cm X 10cm spaced in 3mX1m size (standard size available in the market). For 3 m3 gabion, 14 m2 surface area of GI wire is needed. Thus weight of the GI wire @1.3 kg/m2 including wastage

18.2 kg

60

1092

  1. Stones of size > 225 mm including wastage (25%)  and transportation at site

3.75 m3

200

750

Total material cost

1842

Labour

 

 

 

  1. Skilled labour for wire netting

no

240

120

  1. Semi skilledlabour for placing stones in the net

no

220

275

  1. Unskilled labour

180

405

Total labour cost

800

Total material and labour cost

2642

Add 3% contingency

79.26(say 80)

Grand total for 3 m3

2722

Cost per m3

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 m3 crib for span of 10 m.

 

Particulars

Quantity

Rate (Rs./unit)

Amount (Rs.)

Material

 

 

 

  1. Wooden logs

 

 

 

  1. Vertical post in two rows 1 meter Centre to Centre and span of 10 meters; 2.15 m long and 100-120 mm in diameter

22 poles

120

2,640

  1. Horizontal post of 3.0 meter long and 100-120 mm in diameter placed 50 cm apart

30 poles

140

4,200

  1. 200 to 250 mm long iron nails for joining logs at 88 joints (say 100); weight of 100 nails @200 gm per nails

20 kg

60

1,200

  1. Oil Painting

5 lit

40

200

  1. Creosote oil

 

 

 

  1. Turpentine oil

5 lit

30

150

  1. Stones at site (> 250mm)

15m3

200

3,000

  1. T&P (Drills, hacksaw, Hammer etc.)

Lumpsum

 

1,000

Total material cost

12,390

Labour

 

 

 

  1. Carpenter

10 nos.

240

2,400

  1. Mason

5 nos.

240

1,200

  1. Painter

2 nos.

240

480

  1. Unskilled

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 m3

20672

Cost per m3

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 1m3 cement concrete = 1.54 m3.

Thus for 10 m3 cement concrete, dry mortar volume will be 15.4 m3

Cement requirement = 15.4/7 = 2.2 m3 = 64 bags of 50kg each

Sand requirement = 4.4 m3 and

Aggregates requirement = 8.8 m3

 Assuming density of concrete as 2300 kg/m3, the weight of 10 m3 such concrete will be = 23 ton

 

Particulars

Quantity

Rate (Rs/unit)

Amount(Rs)

Material

 

 

 

Cement grade 53

64 bags

280

17,920

Coarse sand (1-2 mm)

4.4 m3

2000

8,800

Stone ballast 20mm

8.8 m3

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

Sub-Total

3,940

Total of material and labour cost

60,620

Add 3% contingency

1820

Total cost for 10 m3 RCC

62,440

Cost of per m3 RCC

6,244

(say 6,250)

Example 6: Determine unit cost for I-class brick work with 1:6 cement mortar

Solution:

First estimate cost for 10 m3 brick work in order to rationalize the labour requirement and mortar mix.

 Dry matter requirement for mortar in brick work = 0.3m3 per m3 of brickwork. Thus 3.0m3 dry matter will be required for 10m3 brick work.

 Cement requirement = (1/7)*3 = 0.43 m3 or 12.47 (say 13) bags of 50 kg each.

Sand requirement = (6/7)*3 = 2.58 m3

 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 (1-2 mm)

2.58 m3

2000

5,160

Class-I brick @500 per m3

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 m3 RCC

37,770

Cost of per m3 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

 

Example 7

Earthwork volume = 0.5*0.5*400 = 100 m3

Particulars

Quantity

Rate (Rs/unit)

Amount(Rs)

Labour

 

 

 

Unskilled labour @1.5 m3 per manday

67 mandays

180

12,060

Skilled labour

2 mandays

240

480

T&P

Lumpsum

-

500

Total labour cost

13,040

Per m3 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

image_1

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 (m3)

1

Excavation in foundation

12

10

2

0.6

144

2

Gabion box filled with boulders of >200mm

 

 

 

 

 

 

  1. Bottom

12

10

2

1

240

 

  1. top

12

10

1

1

120

 

  1. side

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

m3

130(see example 7)

18,720

2

Gabion work

456

m3

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.226-273.

  • Mohan, S. C. et al. (2007). Training Manual on Soil Conservation & Watershed Management, pp. 380-393

  • 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.

Last modified: Friday, 14 March 2014, 9:00 AM