Question

A concrete mix is required for reinforced concrete wave breaker at seashore where frequent mild freezing and thawing is dominant. A specified compressive strength of 25 MPa is required at the age of 28 days, and the size of the section and reinforcement dictate using a maximum aggregate size of 19 mm. The company had no history of testing concrete used. The coarse aggregate meets the ASTM grading requirements. It has absorption of 2%, a BSG (D) of 2.70 and a unit weight of 1550 kg/m3. The fine aggregates have absorption of 1%, a BSG (D) of 2.7 and a fineness modulus of 2.7. Both coarse and fine aggregates are used while dry. For each bag of cement (50 kg); calculate the number of containers of coarse and fine aggregate equivalent to required mass quantities. The volume of each container is 20 liters; and the loose density of the coarse and fine aggregates are 1250 and 1200 kg/m3, respectively.

Answer 1

Ans) Let the total volume of trial mix be 1 cubic meter then according to ACI 211.1.-91 , table 6.3.1 slump required for reinforced concrete wall is 25 to 100 mm. Lets design the mix for 75 mm slump

According to table 6.3.3 for 75 mm slump and nominal aggregate size of 19 mm , amount of water required per cubic meter of concrete is 184 kg

=> Amount of water = 184 kg per m^3 concrete

According to Table 6.3.3 approximate air content for mild exposure and 19 mm aggregate = 3.5% or 0.035 m^3

When history data of mixes are not available, required compressive strength (f'cr) as per ACI,

f'cr = f'c + 8.3 , where , f'c is specified strength

=> f'cr = 25 + 8.3 = 33.30 MPa

Now according to Table 6.3.4 (a) , for required compressive strength of 33.30 MPa, water cement ratio is 0.41

Hence, amount of cement = 184 / 0.41 = 448.78 kg

Now, according to table 6.3.6 for nominal aggregate size of 19 mm and fineness modulus of 2.7 volume of coarse aggregate is 0.63 m^3

=> Amount of coarse aggregate = Density x volume

Given,density of coarse aggregate as 1550 kg/m^3

=> Amount of coarse aggregate = 1550 x 0.63 = 976.50 kg

Volume of fine aggregate = Total volume of concrete - Volume of water,cement,coarse aggregate and air

=> Fine aggregate volume = 1 - [(184/1000) + (448.78 / 3.15 x 1000) + (976.50/ 2.7 x 1000) + 0.035]

=> Fine aggregate volume = 1 - 0.722 = 0.278 m3

=> Amount of fine aggregate = Volume x Specific gravity x Water density = 0.278 x 2.7 x 1000 = 750.60 kg

Now, since both aggregates has absorption capacity, so they will absorb some water from concrete so in order to maintain constant water cement ratio, amount of mixing water needs to be corrected

Water absorbed by coarse aggregate = 0.02 x 976.5 kg = 19.53 kg

Water absorbed by fine aggregate = 0.01 x 750.6 kg = 7.50

=> Corrected amount of water to be added = 184 + 19.53 + 7.50 = 211 kg

We have determined amount of each component for 1 cubic meter concrete so, now we can calculate number of containers :

We can see that for 448.78 kg cement, coarse aggregate required = 976.50 kg and fine aggregate required = 750.60 kg

=> For 50 kg cement, Coarse aggregate required = 976.50 kg x (50 /448.78) = 108.80 kg

=> Volume of coarse aggregate = Mass / Loose Density = 108.80 / 1250 = 0.087 m^3 or 87 Liters

=> For 50 kg cement, Fine aggregate required = 750.6 kg x (50 /448.78) = 83.63 kg

=> Volume of fine aggregate = Mass / Loose Density = 83.63 / 1200 = 0.0697 m^3 or 69.7 Liters

Volume of each container = 20 Liters

Hence, number of container required for coarse aggregate = 87/20 = 4.35 \approx 5

Number of container required for fine aggregate = 69.7/20 = 3.48 \approx 4