Question

1) A concrete trial batch was calculated according to ACI method and to comply with the project requirements. The weights calculated per batch were: W required/net) = 2.1 kg, C = 5.9 kg, CA moist = 22.2 kg, FA moist = 16.9 kg While mixing the batch, we noticed that the mixture is a little harsh so the expected slump could not be satisfactory. Therefore, we preferred to add 0.2 kg of water to the batch. After mixing, we measured the volume and the fresh unit weight of concrete: V = 0.018m3, Unit Weight of concrete=2425 kg/m3. The slump obtained was perfect. It is required to calculate the new adjusted proportion in kg/m3 and in site/moist and dry conditions, so as to obtain the suitable workability with the required strength. Given: coarse fine cement Spec. grav. (g/cm3) 2.71 2.48 3.1 Absorption % 2.4 1.6 Moisture % 1.4 3.6 Unit weight (g/cm3) 1.48

Answer 1

Complete solution is given below.

Density of CA= 1480 kg/m3 Sp. Gr. of cement= 3.1 Sp. Gr. of CA= 2.71 Sp. Gr. of FA= 2.48 AC of CA (ACCA)= 0.024 MC of CA (MCCA)= 0.014 AC of FA (ACFA)= 0.016 MC of FA (MCFA)= 0.036 Weight of cement = 5.9 kg Moist CA weight = 22.2 kg Moist FA weight = 16.9 kg Weight of water = 2.1 kg Additional water = 0.2 kg Dry weight of CA = Moist weight/(1+AC+MC) Dry weight of CA = 21.39 kg Dry weight of FA = Moist weight/(1+AC+MC) Dry weight of FA = 16.06 Additional water from CA= Additional water from FA= Dry weight * MC = Dry weight * MC = 0.30 0.58 Therefore, Total water content required= 2.1+0.2+0.3+0.58 Total water content required= 3.18 kg Materials required for 0.018 m3 in dry condition, Cement = 5.9 kg Water = 3.18 kg CA = 21.39 kg FA = 16.06 kg For 1m3 Cement = 327.778 kg Water = 176.542 kg CA= 1188.18 kg FA= 892.48 kg Materials required for 0.018 m3 in moist condition, Cement = 5.9 kg Water = 2.30 kg CA = 22.2 kg FA = 16.9 kg For 1m3 Cement = 327.778 kg Water = 127.778 kg CA = 1233.33 kg FA= 938.889 kg