**Introduction**

Construction work involves a lot of measurements. The materials used in construction are sold in the market based on unit quantities and these unit quantities have unit prices. Accurate estimation of construction cost involves accurate measurement of construction quantities. In this article, I will show how to accurately estimate the unit quantity of five common construction materials. With these unit quantities, one can get the total quantity required, append unit price, and get the total overall project sum. This article would be of assistance to beginners in quantity surveying and civil engineering as well as practicing engineers. It is not to replace their task. It would only show unit quantities. How to get the overall quantity, determine and append unit prices cannot be done without expertise in these fields.

The quantities to be estimated are:

- Blockwork
- Mortar for Block Laying and Plastering
- Concrete Work
- Reinforcement

**Determination of the Number of Blocks**

In Nigeria, there are about six (6) classes of a block made from usually (1:6) mix ratio of cement: sand. The usual block sizes are as shown below in the order of Length x Width x Breadth. These blocks can be hollow or solid depending on demand.

- 450 mm (18 inches) x 225 mm (9 inches) x 225 mm (9 inches)
- 450 x 200 x 225
- 450 x 150 x 225
- 450 x 125 x 225
- 450 x 100 x 225
- 450 x 75 x 225

Among these lists, 450 x 225 x 225 and 450 x 150 x 225 whether hollow or solid are commonest in building construction in Nigeria. Blocks are joined together by a mortar which usually has a thickness of 25 mm (1 inch) but this thickness can vary and the variation should be considered in determining mortar volume for work. In determining the area of the block within an area of the wall, the mortar thickness is necessarily considered.

Ideally, the area of a 450 x 225 mm block embedded in 25 mm mortar joints on all sides should be:

(25/2 + 450 + 25/2 ) x (25/2 + 225 + 25/2) = (12.5 + 450 + 12.5) x (12.5 + 225 + 12.5) = 475 x 250 = 118750 mm^{2} (0.11875 m^{2}) instead of 450 x 225 (101250 mm^{2} or 0.101250 m^{2}).

To determine number of blocks in 1 m^{2} of wall, then = 8.42 blocks

Adding 10% as waste, we have 8.42 + 0.842 = 9.262, say 10 blocks. Thus, we have 10 blocks in 1 m^{2} of wall

**Determination of the Quantity of Mortar for Laying and Bedding Blocks**

The mortar for laying and bedding block is measured in m^{3}/m^{2}. The volume of mortar required for the solid block is different from the volume required for a hollow block.

Generally, the volume of mortar required for a solid block = volume of the block (including mortar) – the volume of the block only

Gross volume of block (including mortar) = 475 x 225 x 250 = 26718750 mm^{3} = 0.02672 m^{3}

Volume of block only = 450 x 225 x 225 = 22781250 mm^{3} = 0.02278 m^{3}

Volume of mortar for solid block = 0.02672 – 0.02278 = 0.00394 m^{3}

For hollow blocks, the hollow of the block should be subtracted from the volume of mortar for the solid blocks to get the volume of mortar for the hollow blocks.

Table 1 below presents the block sizes, types, joint thickness, and mortar volume:

*Table 1: Block sizes and corresponding mortar required per block/per square meter of wall*

Assuming the block size specified above and the joint thickness are applicable to one work, one can read off the mortar volume required per block from Table 1 and multiply by the total number of blocks to get the total volume of mortar required for the block work.

Mortar usually comprises cement, fine aggregate (sand), and coarse aggregate (stone). Having determined the mortar volume required to lay bricks or blocks, the quantity of each of these materials can be determined from Table 2 based on some nominal mix ratio for 1 m^{3} of mortar to enable appending of prices to them.

*Table 2: Nominal mortar mix ratio and required constituent materials*

The same rule applies to mortar for laying and bedding blocks apply to mortar for plastering. In the latter case, one needs to determine the volume of mortar (m^{3}) required. Once the volume of mortar required is known, Table 2 can be used to get the quantity of each of the materials required within the volume. Remember when calculating these quantities to include allowances for wastes. In a follow-up article titled **Allowable Wastage of Construction Materials,** I have shown allowances for wastes that should be made for some materials in estimation. In a later article, I would also show the allowance to be made for shrinkage because some construction materials like cement, sand, and aggregate undergo shrinkage when they react with water and this shrinkage affects the volume of the plastic mass: mortar or concrete.

**Determination of Concrete Quantity**

Concrete is one of the key construction materials. Concrete consists of a mixture of cement, fine aggregate, and coarse aggregate and it is usually measured in cubic metres (m^{3}). To determine the quantity of concrete required for an item of work, compute the volume of that item of work by finding the product of its length, width, and depth (or height). This volume is the minimum volume. Put allowance for shrinkage and waste to get the true and most probable volume of the concrete. The constituent materials can be determined by design (design mix) or if a nominal mix would be used, Table 3 below shows the different constants one can use to determine the quantity of cement, sand, and stone. Let me give a demonstration of how these constants are obtained using a particular mix ratio.

Assuming we want to determine the number of materials required to obtain 1m^{3} of concrete using a nominal mix ratio of 1:2:4.

Sum the ratio as 1+2+4 = 7 of the total quantity of materials

Cement quantity = (1/7) x 1 m^{3} = 0.142857142 m^{3}

Allow 10% for waste and 35% for shrinkage, thus 0.142857142 x 1.1 x 1.35 = 0.212142857

Assuming the density of cement = 1440 kg

Mass of cement = density x volume = 0.212142857 x 1440 = 305.4857143 kg

Number of 50 kg bags required = 305.4857143/50 = 6.109714286 bags, say 6.1 bags

Sand quantity = (2/7) x 1 m^{3} = 0.285714285 m^{3}

Allow 10% for waste and 35% for shrinkage, thus 0.285714285 x 1.1 x 1.35 = 0.424285714 m^{3}, say 0.424 m^{3}

Stone quantity = (4/7) x 1 m^{3} = 0.571428571 m^{3}

Allow 10% for waste and 35% for shrinkage, thus 0.571428571 x 1.1 x 1.35 = 0.848571428 m^{3}, say 0.849 m^{3}

Consult Table 2 below for other nominal mix ration constants. The constants below are provided for 1 m^{3} of concrete. Thus when you get the volume of concrete required, multiply by the values below to get an equivalent for the particular volume of concrete concerned.

*Table 3: Nominal concrete mix ratio and required constituents*

**Determination of the Tonnage of Reinforcement**

Reinforcement is an important quantity in reinforced concrete. It is usually measured in lengths (m) or weight (kg and ton). It is the duty of the structural Engineer to provide the quantity in length of structural steel required in the building. The quantity estimator needs to be in contact with the structural engineer in this aspect to get the bar bending schedule (BBS) that species this. When the total length is cumulated, allowance is also made for waste, usually 5% when the bars are cut, bent, and fixed at the site, otherwise allow 1% if the bars are delivered to the site cut and bent. Table 4 below provides the classes of bars available in Nigeria together with their weight per metre length and the number of such lengths in 1000 kg (1 ton) of the rod.

*Table 4: Sizes of rods, weight per metre, and quantity*

Reinforcement bars are tied in position using tying wires often called binding wires. Table 5 below provides the tying wire requirement (in kg) per 1000 kg of each class of rod. In Nigeria, one roll (bundle) of binding wire weighs 20 kg.

*Table 5: Tying wire requirement per 1000 kg of rod*

In some light construction works such as drainage or ground floor slabs that do not bear much suspended or lateral loads, British Reinforcement Company (BRC) wire mesh or welded steel wire mesh is used to provide some tensile restraint. Table 6 below provides the fabric reinforcement to BS 4483.

*Table 6: Fabric reinforcement sizes*

Thanks for visiting mycivillinks today (mycivillinks; your civil links).