Mortar for 1.5 m³ brickwork numerical

Faces of a brick

A brick (standard 240 x 115 x 57 mm) has the following faces and parts:

• Stretcher (side) face: the longest vertical face, 240 x 57 mm; it shows in a stretcher course.
• Header (end) face: the small end face, 115 x 57 mm; it shows in a header course.
• Bed (top / bottom) face: the horizontal face, 240 x 115 mm, on which mortar is spread.
• Frog: a depression (about 10 mm deep) in the top bed face which holds mortar.
• Arris: the edge where two faces meet.

           240 (length)
      +----------------+
     /|               /|
    / |  top (bed)    / |  57
   +----------------+  |  (height)
   |  | header face |  |
   |  +-------------|--+
   | /  stretcher   | /  115 (width)
   +----------------+

Mortar for 1.5 m³ brickwork in 1:4

Given: brickwork volume $= 1.5\ \text{m}^3$; brick size 57 mm x 115 mm x 240 mm (actual size, without mortar); mortar joint $= 10$ mm; mortar 1:4 (cement : sand).

Step 1: Size of brick with mortar

$$(0.25 \times 0.125 \times 0.067) = 0.002094\ \text{m}^3$$

Step 2: Number of bricks

$$N = \frac{1.5}{0.002094} = 716.4 \approx 717\ \text{bricks}$$

Step 3: Volume of bricks without mortar

$$717 \times 0.24 \times 0.115 \times 0.057 = 1.128\ \text{m}^3$$

Step 4: Wet volume of mortar

$$1.5 - 1.128 = 0.372\ \text{m}^3$$

Step 5: Dry volume of mortar (add 27% for dry bulk, voids in sand, wastage and shrinkage)

$$0.372 \times 1.27 = 0.472\ \text{m}^3$$

Step 6: Cement and sand for 1:4 (sum of ratio $= 5$)

$$\text{Cement} = \frac{1}{5} \times 0.472 = 0.094\ \text{m}^3 = 136\ \text{kg} \approx 2.7\ \text{bags (50 kg, 0.0347 m}^3\text{ each)}$$ $$\text{Sand} = \frac{4}{5} \times 0.472 = 0.378\ \text{m}^3$$

Answer: Mortar = 0.372 m³ (wet), 0.472 m³ (dry); cement 0.094 m³ and sand 0.378 m³; bricks = 717 nos.