Question

1. A grade C7 concrete with nominal mix of 1:3:8 by volume is to be used for floor in dairy and beef cattle house. 2 bags of Ordinary Portland Cement (OPC) and moist aggregates are used. 50 kg of cement have a volume of 37 litres. Given that the moisture contents of the sand and stones are 4% and 1.5% respectively, the bulk densities of the sand and stones are 1200 kg/m³ and 1800 kg/m³ respectively and the cost of the sand and stones are N1500:00 per ton and N2500:00 per ton respectively. Calculate i. The volume of aggregates in the mix. ii. Calculate the required total volume of the ingredients assuming 30 % for decrease in volume when mixed and 5 % for waste iii. The weight of the aggregates. iv. The amount of water contained in the aggregates. v. The costs of the aggregates. vi. The water/cement ratio.​

Answer 1

i. The volume of aggregates in the mix:

The total volume of the mix can be calculated by adding the volumes of cement, sand, and stone.

Given that 2 bags of OPC have a volume of 37 litres, the volume of cement in the mix is:

2 bags * 37 litres/bag = 74 litres

The volume of sand in the mix can be calculated as follows:

Volume of sand = 1/3 * Total volume of mix - Volume of cement

= 1/3 * (74 + V) - 74

where V is the volume of aggregates (sand and stone) in the mix.

Similarly, the volume of stone in the mix can be calculated as follows:

Volume of stone = 8/3 * Total volume of mix - Volume of cement - Volume of sand

= 8/3 * (74 + V) - 74 - (1/3 * (74 + V))

= 8/3 * (74 + V) - 98

Now, we can substitute the given values to obtain the equation:

V/1000 * (1 - 0.04) * 1200 + V/1000 * (1 - 0.015) * 1800 = Volume of sand + Volume of stone

Simplifying and solving for V, we get:

V = 251.8 litres

Therefore, the volume of aggregates in the mix is 251.8 litres.

ii. The required total volume of the ingredients:

The required total volume of the ingredients can be calculated as follows:

Total volume of ingredients = Total volume of mix / (1 - Decrease in volume - Waste)

= (74 + 251.8) / (1 - 0.30 - 0.05)

= 400.5 litres

Therefore, the required total volume of the ingredients is 400.5 litres.

iii. The weight of the aggregates:

The weight of the sand can be calculated as follows:

Weight of sand = Volume of sand * Bulk density of sand * (1 - Moisture content of sand)

= 251.8/1000 * 1200 * (1 - 0.04)

= 286.77 kg

The weight of the stone can be calculated as follows:

Weight of stone = Volume of stone * Bulk density of stone * (1 - Moisture content of stone)

= (74 + 251.8)/1000 * 1800 * (1 - 0.015)

= 626.31 kg

Therefore, the weight of the aggregates is 286.77 + 626.31 = 913.08 kg.

iv. The amount of water contained in the aggregates:

The amount of water contained in the sand can be calculated as follows:

Water in sand = Volume of sand * Moisture content of sand

= 251.8/1000 * 0.04

= 0.01 m³

The amount of water contained in the stone can be calculated as follows:

Water in stone = Volume of stone * Moisture content of stone

= (74 + 251.8)/1000 * 0.015

= 0.0079 m³

Therefore, the total amount of water contained in the aggregates is 0.01 + 0.0079 = 0.0179 m³.

v. The costs of the aggregates:

The cost of the sand can be calculated as follows:

Cost of sand = Weight of sand * Cost per tonne / 1000

= 286.77 * 1500 / 1000

= N430.16

The cost of the stone can be calculated as follows:

Cost of stone = Weight of stone *