Question

A concrete mix was designed for a concrete road pavement (i.e. heavy mass concrete). A 28-day strength of 26 MPa is required. The exposure classification for durability purposes is C. One hundred laboratory test results on the controlling mix show a standard deviation of 5.3 MPa (k=1.65). The density of crushed coarse aggregate is 2850 kg/m³ and its dry rodded density is 1600 kg/m³ and the maximum size to be used is 10 mm. The specific gravity of fine aggregate is 2750 kg/m³ and its fineness modulus is 3.0. The high early strength cement used has a density of 3150 kg/m³.

Find the weight (in kg) of cement, water, fine and coarse aggregate for this mix, required to make a 0.25 m high x 3.6 m wide x 12 m long concrete pavement using the ACI Method.

Answer 1

Ans) Let the total volume of trial mix be 1 .According to ACI 211.1-91, table 6.3.1, slump required for mass concrete platform is 25 - 50 mm . According to table 6.3.3 , for 50 mm slump and nominal aggregate size of 10 mm , amount of water required per cubic meter of concrete is 207 kg

=> Amount of water = 207 kg/

From table 6.3.3, Air content for 10 mm aggregate is 3 %

Now, required compressive strength as per ACI,shoul be larger of :

fcr = f'c + 1.34 S

f'cr = f'c + 2.33 S - 3.45

=> fcr = 26 + 1.34(5.3) = 33.1 MPa

=> f'cr = 26 + 2.33(5.3) - 3.45 = 34.9 MPa

Since, 34.9 MPa > 33.1 MPa, required compressive strength = 34.9 MPa

Now according to Table 6.3.4 (a) , for compressive strength of 34.9 MPa, water cement ratio is 0.47

Hence, amount of cement = Amount of water / w-c ratio = 207 / 0.47 = 440.42 kg /

  Now, according to table 6.3.6 for nominal aggregate size of 10 mm and fineness modulus of 3.0 , volume of coarse aggregate is 0.44 m3

=> Amount of coarse aggregate = dry rodded density x volume = 1600 x 0.44 = 704 kg/

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

=> Fine aggregate volume = 1 - [(207/1000) + (440.42 / 3150) + (704/ 2850) + 0.03]

=> Fine aggregate volume = 1 - 0.6238 = 0.3762

=> Amount of fine aggregate = volume x specific gravity x water density = 0.3762 x 2750 = 1034.55 kg/

Now,

Volume of concrete required = 0.25 m x 3.6 m x 12 m = 10.8

Since batch weight of 1 concrete is now known , batch weight for trial mix of 10.8 can be calculated as follows :

Water = 207 kg/ x 10.8 = 2235.6 kg

Cement = 440.42 kg/ x 10.8 = 4756.54 kg

Fine aggregate = 1034.55 kg/ x 10.8 = 11173.14 kg

Coarse aggregate = 704 kg/ x 10.8 = 7603.2 kg