Question

Q1// Use the ACI method to design a concrete mix which is required to obtain f'c28= 34.5MPa and a slump of 75mm. the fineness modulus of the aggregate is 3.0, the bulk density of coarse aggregate is 1600 kg/m' and the maximum size is 25. Both the specific gravity of fine and coarse aggregate are 2.68. neglect the absorption and total moisture content.

Answer 1

Complete mix design as per ACI is shown below.

Design of non air entrained concrete FA - Fine aggregate CA - Coarse aggregate Given: Target strength= 34.5 Mpa Dry rodded unit wt CA= 1600 kg/m3 Sp. Gr. of cement= 3.15 Assume Sp. Gr. of CA= 2.68 Sp. Gr. of FA= 2.68 Fineness modulus of FA= 3.0 Slump value required is = 75 mm Maximum size of aggregate = 25 mm For 75 mm slump and 25 mm aggregates, from table 1 Approximate mixing water = For non air entrained concrete 195 kg 2 Air content= 1.5 % Entrapped air 3 Water cement ratio Mpa w/c 35 0.47 34.5 0.48 30 0.54 w/c= 0.48 For 34.5 Mpa non air entrained concrete by interpolation w/c= 0.48 4 Weight of cement= weight of water/ (w/c) Weight of cement= 406.25 kg

5 Coarse aggregate calculation For 25 mm aggregate with fineness modulus of fine aggregate = 3.0 Volume of CA= 0.65 (From table) Weight of CA = Volume * density Weight of CA= 1040 kg 6 Weight of fine aggregate Sp. Gr. Of FA* {1000-(wt. of water/sp. gr. of water)- Weight of FA = (wt of cement/sp. gr. of cement)-(wt of CA/sp. gr. of CA)-(1000*Air content)} Where absolute volume of, Water = (wt. of water/sp. gr. of water) = 195 Cement = (wt of cement/sp. gr. of cement) = 129 CA= (wt of CA/sp. gr. of CA) = 388 Air content= (1000* Air content) = 15 Now, substituting all values in the above formula Weight of FA= 818.916 kg For 1m3 of concrete Cement = 406.25 kg CA= 1040.00 kg FA= 818.92 kg Water 195.00 kg