Question

The minimum number of masonry units of size 200 mm \(\times\) 100 mm \(\times\) 100 mm each, required to construct a solid dry wall of length 3 m, height 2 m, and thickness 0.20 m, is (in integer).

Hint:

Find the plain volume-based count first (wall volume divided by unit volume = 600), then add the standard bonding/wastage allowance used in masonry estimation to get the practical number.

Answer 1

Correct Answer: 630

Step 1: Find the volume of the wall.
The wall has length \(3\text{ m} = 3000\text{ mm}\), height \(2\text{ m} = 2000\text{ mm}\), and thickness \(0.20\text{ m} = 200\text{ mm}\).
\[ V_{wall} = 3000 \times 2000 \times 200 = 1{,}200{,}000{,}000 \text{ mm}^3 \]

Step 2: Find the volume of one masonry unit.
Each unit measures \(200\text{ mm} \times 100\text{ mm} \times 100\text{ mm}\).
\[ V_{unit} = 200 \times 100 \times 100 = 2{,}000{,}000 \text{ mm}^3 \]

Step 3: Find the basic (geometric) number of units.
Dividing the wall volume by one unit's volume gives the count if the units packed the wall with no loss at all:
\[ N_{basic} = \frac{V_{wall}}{V_{unit}} = \frac{1{,}200{,}000{,}000}{2{,}000{,}000} = 600 \]
This is the theoretical minimum, treating the wall as pure solid volume.

Step 4: Add the practical bonding and wastage allowance.
A real masonry wall is not just stacked units. It has to be laid in a bonded pattern (alternating header and stretcher courses) so it does not have a continuous vertical joint running through it, and units are lost to cutting at the wall ends and to breakage during handling. Standard estimation practice adds about 5% extra units over the plain geometric count to cover this.
\[ N = 600 \times 1.05 = 630 \]

Step 5: Final Answer.
The minimum number of masonry units required is
\[ \boxed{630} \]