Estimation is one aspect of cost engineering that often precedes bidding. Other components of cost engineering include cost control, value engineering, planning scheduling, construction management, investment appraisal, risk analysis, conceptual studies, and profitability analysis of engineering projects and processes. Cost engineering is the duty of a cost engineer who is trained by education and professional experience to apply the principles of engineering cost management. Estimation of engineering works is usually done according to the principles of the Bill of Engineering Measurement and Evaluation (BEME) in order to get the estimated cost (the probable cost of construction). This basically applies to engineering works. Quantity Surveyors who are trained in cost estimation usually employ the principles in Bill of Quantities (BOQ). What is really the difference between these two?

A wiki online commenter said that BOQ is a factored estimate of all the materials, equipment, and workmanship required for all items of work in a building project while BEME is accurate measurements of materials, a considerate judgment of workmanship involved, and exact assessment of the equipment required to carry out every item of work involved in buildings, culverts, drains, bridges, rail lines, canals, tunnels, and other civil-structural facilities. Another commenter says that the difference only lies in the fact that BOQ is used for building works while BEME is used for civil engineering works. I did not subscribe fully to their opinion and I do not also have a clearer opinion. However, the reader is free to provide a more acceptable difference maybe as a comment to this post if one has such

Preparation of BOQ/BEME for buildings and roads requires first the understanding of the units of different items involved in each project. This is important because, without the application of accurate units of measurement, an accurate estimation cannot be done. Tables 1 and 2 show the units of measurement of items in building and road works respectively.

* Table 1: Items in building and units of measuremen*t

**Table 2:** Items in road works and units of measurement

The determination of the different quantities of each item of work proceeds through a process known as taking-off (though this is not the subject of this article). The taking-off sheet may not necessarily form part of the bill document presented for bidding but it does help the engineer for accurate estimation of quantities.

**SOME REQUIREMENT FOR ESTIMATION OF COST OF BUILDING PROJECT**

It is ideal to start the estimation for building works from the foundation. Before estimating for each item, one needs to consider the unit as provided above for guidance. For instance, the estimation of excavation of the foundation is measured in m^{3}. One needs to get the product of the proposed length of excavation and the width and depth in order to get the volume. Generally, costs are appended per unit of measurement of the item. This cost includes both materials and labour costs. For example, the cost of producing one (1) cubic metre of concrete should include the cost of all constituent materials comprising cement, sand, and gravel required for the one cubic metre of concrete, the cost of hauling those materials to the production site, the cost of manpower that would mix the materials and delivery of the material into formwork/mould.

**Example**

**Determine the cost of production of 1 cubic metre (1m ^{3}) of concrete using a 1:2:4 mix ratio**

Assuming,

Density of concrete = 2500 kg/m^{3}

Density of cement = 1440 kg/m^{3}

Density of fine aggregate = 1600 kg/m^{3}

Density of coarse aggregate = 1650 kg/m^{3}

**Sum of ratio = 1+2+4 = 7**

Considering 2% of entrained air

2% of 1 m^{3} = (2/10) x 1 = 0.02 m^{3}

Thus to avoid shortage, take 1 m^{3} to be 1.02 m^{3}

**Calculation of volume of materials from the mix ratio**

Volume of cement = (1/7) x 1.02 = 0.1457 m^{3}

Volume of sand = (2/7) x 1.02 = 0.2914 m^{3}

Volume of coarse aggregate = (4/7) x 1.02 = 0.5829 m^{3}

**Determination of mass of quantities from the volume (N.B: Mass = volume x density)**

Mass of cement = 0.1457 x 1440 = 209.808 kg

Mass of sand = 0.2914 x 1600 = 466.24 kg

Mass of coarse aggregate = 0.5829 x 1650 = 961.785 kg

**Determination of water-cement ratio using grade 42.5 cement**

Assuming the target mean strength, k = 20 N/mm^{2}

Water-cement ratio (W/C) = (83 – k)/84 = (83 – 20)/84 = 0.75

Thus, the weight of water = 209.808 x 0.75 = 157.36 kg

The water content = 157.36 x 2.6 = 409.125 litres

**Rate of production of 1 m ^{3} of concrete**

Assuming,

1 bag of cement = 50 kg = ~~N~~ 2600 (This is assuming value. Always use current value in analysis)

In 1m^{3} of concrete, we have 209.808 kg = 209.808/50 = 4.196 bags, say 5 bags

Cost of 5 bags of cement = 5 x 2600 = ~~N~~ 13,000

1 trip of sharp sand = 5000 kg = ~~N~~ 30,000 (assumed)

In 1 m^{3} of concrete, we have 466.24 kg

Cost of 466.24 kg = (466.24/5000) x 30,000 = ~~N~~ 2,796

1 trip of chippings = 10,000 kg = ~~N~~ 105,000 (assumed)

In 1m^{3} of concrete, we have 961.785 kg

Cost of 961.785 kg = (961.785/10000) x 105,000 = ~~N~~ 10,098.7

10 litres of water = ~~N~~ 5 (assumed)

409.125 litres = (409.125 x 5)/10 = ~~N~~ 210

Labour work (mixing and casting of the concrete) = ~~N~~ 1000 for 1 bag of cement (the value can change when using a concrete mixer).

~~N~~ 5000 would be labour charge for the 5 bags of cement

Total cost = 13000 + 2796 + 10,098.7 + 210 + 5000 = ~~N~~ 31,104.7

Take the cost of 1m^{3} of concrete as ~~N ~~33,000

**Determination of rates for 1kg of reinforcement**

The weights in kg of 1m length of the following rods are as follows:

**Y8 = 0.395 kg**

**Y10 = 0.616 kg**

**Y12 = 0.888 kg**

**Y16 = 1.578 kg**

**Y20 = 2.466 kg**

Assuming one ton (1,000kg) of 12 mm of TMT rod is ~~N~~ 410, 000, 1kg would cost (410,000/1,000) = ~~N~~ 410

Assuming the cost of labour and transport for 1kg of rod = ~~N~~ 100

Total cost for 1kg of the rod = 410 + 100 = ~~N~~ 510

**Determination of** **Rates for Formwork**

Assuming the cost of 3.0 x 0.3m well sawn timber of 25mm thickness = ~~N~~ 1,400 /m^{2}

Assuming cost of transportation = ~~N~~ 100 /m^{2}

Assuming cost of carpentry work = ~~N~~ 500 /m^{2}

Total cost for 1m^{2} of formwork = 1400 + 100 + 500 = ~~N~~ 2000

Cost of bracing and other accessories = 20% of cost of total cost above = (20/100) x 2000 = ~~N~~ 400

Rate of formwork per square metre = 2000 + 400 = ~~N~~ 2,400

**SOME REQUIREMENTS FOR ESTIMATION OF COST OF ROAD PROJECT**

For the estimation of road works just like the concrete works, to get the rate per unit of the item, it is necessary to factor in the cost of hiring equipment if that is required, the cost of diesel consumption for the equipment, the cost of the operator and the number of days the equipment would work. With all these, one can determine the cost of the item per unit of time. Table 3 shows the typical cost of using different equipment in road construction works and their rates. These rates are subject to variation.

**Table 3: **Rates of common equipment in road construction work

**Example**

**Cost of producing and laying 1m ^{2} of asphalt**

Assuming we are required to estimate the cost to produce and lay asphalt on a square metre of road

Estimated thickness of road = 5 cm (0.05 m) or (usually 3cm (0.03)) for individual works

Area of asphalt (m) = 1000 x 8 (excluding shoulders) = 8000 m^{2}

Tonnage of asphalt = 2.35 ton/m^{3}

Total tonnage = 8000 x 0.05 x 2.35 = 940 tons

Assuming asphalt cost is ~~N~~ 40, 000 per ton (subject to variation)

Asphalt cost = 940 x 40, 000 = ~~N~~ 37, 600,000

Cost of hiring of Paver = cost per day x estimated number of days = 100,000 x 5 days = ~~N~~ 500, 000

Diesel consumption for Paver = diesel consumption per day (litres) x cost of a litre of diesel x number of days = 50 x 260 x 5 = ~~N~~ 65, 000

Cost of operation of Paver = cost of operation per day x estimated number of days = 30, 000 x 5 = ~~N~~ 150, 000

Cost of hiring of roller (Smoothwheel and Pneumatic) = cost of operation per day x estimated number of days = 70,000 x 5 x 2 Rollers = ~~N~~ 700, 000

Diesel consumption for Roller = 60 x 260 x 5 x 2 = ~~N~~ 156, 000

Cost of operation of Roller = 30,000 x 5 x 2 = ~~N~~ 300, 000

Cost of Rakers = 8000 x 4 persons x 5days = ~~N~~ 160, 000

**Sub-total = N 39, 631, 000**

Add 20% profit = ~~N~~ 7, 926, 200

**Grand total = N 47, 557, 200**

Cost per square metre to lay the asphalt = 47,557,200/8000 = ~~N~~ 5, 944.65/m^{2}, say ~~N~~ 6, 000

**Cost of Preparing 1 m ^{3} of Earthwork**

Earthwork is usually measured in cubic metre (m^{3}).

Assuming depth of compacted earthwork = 0.2 m, volume = 1000 x 8 x 0.2 = 1600 m^{3}

(Assuming using a 6 m^{3 }truck to haul earthwork, 5 m^{3} may be used in computation because of voids and losses on transit).

No of trucks required = 1600/5 = 320 trucks

If cost of one truck hauling soil = ~~N~~ 15, 000, total cost = 320 x 15, 000 = ~~N~~ 4, 800, 000

Assuming the grader can spread 62.5m length of road (width inclusive) in one day, the number of days required = 1000/62.5 = 16 days

Cost of hiring Grader for 16 days = 150,000 x 16 = ~~N~~ 2, 400, 000

Diesel consumption for Grader = 150 x 260 x 16 = ~~N~~ 624, 000

Cost of operation = 30,000 x 16 = ~~N~~ 480, 000

Cost of hiring Roller (Smooth-wheeled roller and Sheepsfoot roller) = 90,000 x 16 x 2 = ~~N~~ 2, 880, 000

Diesel consumption = 60 x 260 x 16 x 2 = ~~N~~ 499, 200

Cost of operation = 30,000 x 16 x 2 = ~~N~~ 960,000

Cost of Lowbed = 140,000 x 2 = ~~N~~ 280, 000

**Sub-total = N 8, 123, 200**

Add 20% profit = ~~N~~ 1, 624, 640

**Total = N 9,747, 840**

Cost per cubic metre of earthwork = 9,747,840/1600 = ~~N~~ 6, 092.4, say ~~N~~ 6, 100

**Thanks for visiting MYCIVILLINKS today. God bless you**

## 4 Comments

Nice information. How do you take care of profit in the production of the concrete, reinforcement and formwork rates

Thank you Abubakar. In formal project setting, profit is usually calculated as a percentage of total contract sum and the percentage depends on the size of the project and contract sum, however, 10% is usually common.

Very good work. Please can I have it ?. I’m in mamoudouibrahim5161@gmail.com

Wow! That’s a very nice article.