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Materials for 10 m³ brickwork (1:6) numerical

Short answer

For 10 m³ brickwork, wet mortar volume is 2.485 m³, dry volume 3.156 m³ (×1.27). Using 1:6 mix, cement = (1/7)×3.156 = 0.451 m³ (13 bags of 50 kg) and sand = (6/7)×3.156 = 2.705 m³. Bricks: 4777 nos.

Estimate the quantities of materials for 10 m³ brick work where size of brick is 57mm × 115mm × 240mm, thickness of mortar is 10 mm and mortar of cement sand ratio is 1:6.

Given: brickwork volume =10 m3= 10\ \text{m}^3; brick size 57 mm x 115 mm x 240 mm (actual size, without mortar); mortar joint =10= 10 mm; mortar 1:6 (cement : sand).

Step 1: Size of brick with mortar

(0.25×0.125×0.067)=0.002094 m3(0.25 \times 0.125 \times 0.067) = 0.002094\ \text{m}^3

Step 2: Number of bricks

N=100.002094=4776.1≈4777 bricksN = \frac{10}{0.002094} = 4776.1 \approx 4777\ \text{bricks}

Step 3: Volume of bricks without mortar

4777×0.24×0.115×0.057=7.515 m34777 \times 0.24 \times 0.115 \times 0.057 = 7.515\ \text{m}^3

Step 4: Wet volume of mortar

10−7.515=2.485 m310 - 7.515 = 2.485\ \text{m}^3

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

2.485×1.27=3.156 m32.485 \times 1.27 = 3.156\ \text{m}^3

Step 6: Cement and sand for 1:6 (sum of ratio =7= 7)

Cement=17×3.156=0.451 m3=649 kg≈13 bags (50 kg, 0.0347 m3 each)\text{Cement} = \frac{1}{7} \times 3.156 = 0.451\ \text{m}^3 = 649\ \text{kg} \approx 13\ \text{bags (50 kg, 0.0347 m}^3\text{ each)} Sand=67×3.156=2.705 m3\text{Sand} = \frac{6}{7} \times 3.156 = 2.705\ \text{m}^3

Answer: Bricks = 4777 nos.; cement = 0.451 m³ (about 13 bags); sand = 2.7 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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