How are the bricks for 10 m3 of brickwork calculated (brick 230x110x55, 10 mm joints)?

Divide 1 m3 by the volume of one brick with mortar (0.24 x 0.12 x 0.065) = 534.19 per m3, then multiply by 10

Divide 1 m3 by the volume of the brick without mortar

Divide 10 m3 by 0.25 x 0.12 x 0.07 only for the first course

Multiply the wall area by the number of courses only

How is the dry mortar for the 1:6 brickwork worked out?

Wet mortar = 1 - brick-only volume, x10; dry = 1.27 x wet = 3.26 m3; cement is 1/7 of it

Dry mortar = 1.54 x wet; cement is 1/6 of it

Dry mortar = wet mortar; cement is 1/6 of it

Dry mortar = 1.27 x wet; cement is 1/6 and sand 6/6

For 10 m3 of PCC 1:3:6, what is the dry volume and cement quantity?

Dry volume 15.40 m3; cement 1/10 of it = 1.54 m3 = 44.4 bags

Dry volume 12.7 m3; cement 1.27 m3

Dry volume 15.40 m3; cement 1/3 of it

Dry volume 10 m3; cement 1.0 m3 = 28.8 bags

Asked 2/22
2067 Asar (old course) · 4×2 marks

Materials: 10 m³ brick 1:6, 10 m³ PCC 1:3:6

Short answer

i) 10 m³ brickwork 1:6 (brick 230x110x55 mm, 10 mm joint): 5,342 bricks, 13.4 bags cement, 2.79 m³ sand. ii) 10 m³ PCC 1:3:6: dry factor 1.54, cement 44.4 bags, sand 4.62 m³, aggregate 9.24 m³.

Calculate the quantities of materials required for the following works: i) 10 m³ brick masonry in 1:6 cement mortar ii) 10 m³ PCC (1:3:6) in foundation.

Assumptions: dry volume 1.54 × wet for concrete and 1.27 × wet for mortar; 1 m³ cement = 28.8 bags. The brick size is not given, so 230 × 110 × 55 mm with 10 mm joints is taken.

(i) 10 m³ brickwork in 1:6 mortar

Brick with mortar $= 0.240 \times 0.120 \times 0.065 = 1872.0\times 10^{-6}$ m³; bricks per m³ $= 1/(1872.0\times10^{-6}) = 534.19$
Bricks for 10.00 m³ $= 534.19 \times 10.00 = 5,342$ nos
Volume of bricks only $= 0.230\times0.110\times0.055 = 1391.5\times10^{-6}$ m³ each; wet mortar per m³ $= 1 - 534.19\times1391.5\times10^{-6} = 0.2567$ m³
Wet mortar for 10.00 m³ $= 2.567$ m³; dry $= 1.27\times2.567 = 3.260$ m³
Cement $= 1/7\times3.260 = 0.466$ m³ $= 13.4$ bags
Sand $= 6/7\times3.260 = 2.794$ m³

(ii) 10 m³ PCC (1:3:6)

Wet volume $= 10.000$ m³; dry volume $= 1.54 \times 10.000 = 15.400$ m³ (sum of ratio $= 10$ )
Cement $= 1/10 \times 15.400 = 1.540$ m³ $= 1.540 \times 28.8 = 44.4$ bags (50 kg)
Sand $= 3/10 \times 15.400 = 4.620$ m³
Coarse aggregate $= 6/10 \times 15.400 = 9.240$ m³

Answer: (i) bricks 5,342, cement 13.4 bags, sand 2.79 m³. (ii) cement 44.4 bags, sand 4.62 m³, aggregate 9.24 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

Materials: 10 m³ brick 1:4, 100 m³ PCC 1:3:6 · Estimating and Costing, chapter 4 · asked 2×
Materials for 10 m³ brick masonry 1:3 · Estimating and Costing, chapter 4
Well foundation: excavation, sinking, brick, PCC curb · Estimating and Costing, chapter 6 · asked 2×
Well foundation: sinking, brick, sand filling, PCC · Estimating and Costing, chapter 6 · asked 2×
Materials: 75 m³ brick 1:3, 115 m² PCC floor · Estimating and Costing, chapter 4
Materials for 115 m³ brick masonry 1:3 · Estimating and Costing, chapter 4
Brick (1:6) materials; PCC (1:2:4) rate · Estimating and Costing, chapter 4
Mortar properties; cement and sand for 10 cum · Building Technology, chapter 3
Single-storey external wall: stress, brick, mortar · Concrete Technology and Masonry Structure, chapter 8
Stepwise preparation of cement mortar · Civil Engineering Materials, chapter 6 · asked 4×
Single-room wall: brick strength and mortar · Concrete Technology and Masonry Structure, chapter 8

All Estimating and Costing chapter 4 questions →

Materials: 10 m³ brick 1:6, 10 m³ PCC 1:3:6

(i) 10 m³ brickwork in 1:6 mortar

(ii) 10 m³ PCC (1:3:6)

Related questions