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Materials: 105 m³ PCC 1:4:8, 725 m² plaster

Short answer

i) 105 m³ PCC (1:4:8) with dry factor 1.54: cement 358.2 bags, sand 49.75 m³, aggregate 99.51 m³. ii) 725 m² of 20 mm plaster (1:4) with dry factor 1.27: cement 106.1 bags, sand 14.73 m³.

Calculate the quantities of materials required for the following items of work: i) 105 m³ of PCC (1:4:8) in foundation ii) 725 m² of 20 mm thick cement plaster (1:4) in wall.

Assumptions: dry volume = 1.54 × wet volume for concrete and 1.27 × wet volume for mortar; 1 m³ of cement = 28.8 bags of 50 kg; no wastage is added.

(i) 105 m³ PCC (1:4:8) in foundation

  • Wet volume =105.000= 105.000 m³; dry volume =1.54×105.000=161.700= 1.54 \times 105.000 = 161.700 m³ (sum of ratio =13= 13)
  • Cement =1/13×161.700=12.438= 1/13 \times 161.700 = 12.438 m³ =12.438×28.8=358.2= 12.438 \times 28.8 = 358.2 bags (50 kg)
  • Sand =4/13×161.700=49.754= 4/13 \times 161.700 = 49.754 m³
  • Coarse aggregate =8/13×161.700=99.508= 8/13 \times 161.700 = 99.508 m³

(ii) 725 m² of 20 mm plaster (1:4)

  • Wet mortar =725.0×0.0200=14.5000= 725.0 \times 0.0200 = 14.5000 m³; dry volume =1.27×14.5000=18.4150= 1.27 \times 14.5000 = 18.4150 m³
  • Cement =1/5×18.4150=3.6830= 1/5 \times 18.4150 = 3.6830 m³ =106.07= 106.07 bags
  • Sand =4/5×18.4150=14.7320= 4/5 \times 18.4150 = 14.7320 m³

Answer: (i) cement 358.2 bags, sand 49.75 m³, aggregate 99.51 m³. (ii) cement 106.1 bags, sand 14.73 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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