Skip to main content

Materials for PCC roof 1:1.5:3 with rebar

Short answer

Net concrete volume 49.6 m³ (after deducting 0.8% steel), dry volume 76.384 m³. For 1:1.5:3 mix: cement 400 bags (50 kg), sand 20.83 m³, coarse aggregate 41.66 m³, steel 3,140 kg.

Calculate the quantities of materials required for PCC (1:1.5:3) for RCC roof 0.10 m thick, 20 m wide and 25 m long [assuming rebar 0.8% of volume of PCC].

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

Volumes

  • Gross volume of roof slab =0.10×20×25=50.00= 0.10 \times 20 \times 25 = 50.00 m³
  • Steel (0.8%) =0.008×50.00=0.400= 0.008 \times 50.00 = 0.400 m³, which is deducted because the steel occupies this space.
  • Net volume of concrete =50.00−0.400=49.600= 50.00 - 0.400 = 49.600 m³

Materials for concrete 1:1.5:3

  • Wet volume =49.600= 49.600 m³; dry volume =1.54×49.600=76.384= 1.54 \times 49.600 = 76.384 m³ (sum of ratio =5.5= 5.5)
  • Cement =1/5.5×76.384=13.888= 1/5.5 \times 76.384 = 13.888 m³ =13.888×28.8=400.0= 13.888 \times 28.8 = 400.0 bags (50 kg)
  • Sand =1.5/5.5×76.384=20.832= 1.5/5.5 \times 76.384 = 20.832 m³
  • Coarse aggregate =3/5.5×76.384=41.664= 3/5.5 \times 76.384 = 41.664 m³

Steel

Mass of steel =0.400×7850=3,140= 0.400 \times 7850 = 3,140 kg =3.140= 3.140 tonne.

Answer: cement = 400.0 bags (13.89 m³); sand = 20.83 m³; coarse aggregate = 41.66 m³; steel = 3,140 kg.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Estimating and Costing chapter 4 questions →