Competency Assessment: Material Estimation & Costing

Subject: Civil Construction & Project Management
Level: Advanced Vocational Application

Scenario Description

You are the site supervisor for a rural infrastructure project. You have been tasked with constructing a cylindrical concrete water tank. Your objective is to calculate the precise volume of wet concrete required, determine the dry volume of materials needed based on standard shrinkage factors, and generate a final Cost Estimate (Bill of Quantities) for the client.

1. Technical Specifications

Parameter	Value
Inner Radius of Tank ($r$)	2.0 meters
Inner Height of Tank ($h$)	3.5 meters
Thickness of Walls ($t_w$)	0.15 meters (15 cm)
Thickness of Base Slab ($t_b$)	0.20 meters (20 cm)
Concrete Mix Ratio (Cement : Sand : Aggregate)	1 : 1.5 : 3 (M20 Grade equivalent)
Dry Volume Coefficient	1.54 (Standard safety factor for voids/shrinkage)

2. Market Rates (Current Fiscal Quarter)

Cement: $8.00 per bag (50kg bag = 0.0347 cubic meters)
Sand: $45.00 per cubic meter
Aggregate (Gravel): $40.00 per cubic meter
Labor Charge: $120.00 per cubic meter of concrete poured
Contingency Tax: 5% of total material + labor cost

Detailed Solution & Methodology

Step 1: Calculate Total Concrete Volume ($V_{total}$)

We must calculate the volume of the base slab and the volume of the hollow cylindrical walls separately.

A. Base Slab Volume ($V_{base}$)
The base extends to the outer edge of the walls. Therefore, the Outer Radius ($R$) is the Inner Radius ($r$) + Wall Thickness ($t_w$).

$$R = 2.0m + 0.15m = 2.15m$$ $$V_{base} = \pi \times R^2 \times t_b$$ $$V_{base} = 3.1416 \times (2.15)^2 \times 0.20$$ $$V_{base} \approx 2.904 \, m^3$$

B. Wall Volume ($V_{walls}$)
The walls are a hollow cylinder. We calculate the area of the ring annulus and multiply by height.

$$V_{walls} = \pi \times (R^2 - r^2) \times h$$ $$V_{walls} = 3.1416 \times (2.15^2 - 2.0^2) \times 3.5$$ $$V_{walls} = 3.1416 \times (4.6225 - 4.0) \times 3.5$$ $$V_{walls} = 3.1416 \times 0.6225 \times 3.5$$ $$V_{walls} \approx 6.845 \, m^3$$

C. Total Wet Volume

$$V_{total} = 2.904 + 6.845 = \mathbf{9.749 \, m^3}$$

Step 2: Calculate Dry Volume of Materials

Wet concrete shrinks when it dries. To find the quantity of raw materials to purchase, we multiply the wet volume by the Dry Volume Coefficient (1.54).

$$V_{dry} = 9.749 \times 1.54 = \mathbf{15.013 \, m^3}$$

Step 3: Material Breakdown (Ratio Analysis)

The mix ratio is 1 : 1.5 : 3. The sum of the ratio parts is:

$$Sum = 1 + 1.5 + 3 = 5.5$$

A. Cement Quantity

$$Volume_{cement} = \frac{1}{5.5} \times 15.013 = 2.729 \, m^3$$

To convert to bags (1 bag $\approx$ 0.0347 $m^3$):

$$Bags = \frac{2.729}{0.0347} \approx 78.6$$

Note: Always round up for procurement. Total: 79 Bags.

B. Sand Quantity

$$Volume_{sand} = \frac{1.5}{5.5} \times 15.013 = 4.094 \, m^3$$

C. Aggregate Quantity

$$Volume_{agg} = \frac{3}{5.5} \times 15.013 = 8.189 \, m^3$$

Step 4: Cost Estimation (Bill of Quantities)

Item	Quantity	Rate	Total Cost ($)
Cement	79 Bags	$8.00 / bag	$632.00
Sand	4.094 $m^3$	$45.00 / $m^3$	$184.23
Aggregate	8.189 $m^3$	$40.00 / $m^3$	$327.56
Material Subtotal			$1,143.79
Labor	9.749 $m^3$ (wet)	$120.00 / $m^3$	$1,169.88
Gross Total			$2,313.67

Final Calculation with Contingency Tax (5%)

$$Final Cost = 2313.67 \times 1.05 = \mathbf{\$2,429.35}$$

To the extent possible under law, the author has waived all copyright and related or neighboring rights to this educational work. This work is published from: India.