Question

Q 1: Design a concrete mix for the following performances and ingredients:

* Compressive strength at 28 days = 32 MPa, slump 80 mm

* Crushed coarse aggregate of max size 20 mm readily available

* Fine aggregate is river sand, fineness modulus = 2.5

* Cement Strength Class = 42.5

* Specific gravity of combined aggregate =2.6

* Water absorption of coarse aggregate = 0.8%

* Water absorption of fine aggregate = 0.5%

Q 2: Design a concrete mix for the following performances and ingredients:

* Characteristic compressive strength 50 N/mm2 at 7 days

* Defective rate 1 %

* Previous control data: standard deviation 5 N/mm2

* Cement: class 52.5 * Slump required, 30-60 mm

* Maximum cement content 550 kg/m3

* Coarse aggregate: Crushed single sized 10 mm

* Fine aggregate: Uncrushed with 45% passing 600 μm sieve

* Relative density of aggregate: 2.7 (assumed)

* Volume of trial mix: 0.08 m3

Answer 1

Ans)Let the total volume of trial mix be 1 cubic meter then according to ACI 211.1.8 , Table 6.3.3 for 80 mm slump and nominal aggregate size of 20 mm , amount of water required per cubic meter of concrete is 205 kg

=> Amount of water = 205 kg per m^3 concrete

According to Table 6.3.3 approximate air content in concret for 20 mm aggregate = 2%

According to ACI 318, Design compressive strength is :

fcr = f'c + 8.3

f'c is specified compressive strength

=> fcr = 32 + 8.3 = 40.3 MPa

=> fcr = 40.3 MPa

Now according to Table 6.3.4 (a) , for compressive strength of 40.3 MPa, water cement ratio is 0.42

Hence, amount of cement = 205 / 0.42 = 488.1 kg

Now, according to table 6.3.6 for nominal aggregate size of 20 mm and fineness modulus of 2.5, volume of coarse aggregate is 0.65 m^3

=> Amount of coarse aggregate = Dry rodded density x volume

Assuming dry rodded density of coarse aggregate as 1650 kg/m^3

=> Amount of coarse aggregate = 1650 x 0.65 = 1072.5 kg

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

=> Fine aggregate volume = 1 - [(205/1000) + (488.1 / 3.15 x 1000) + (1072.5/ 2.6 x 1000) + 0.02]

=> Fine aggregate volume = 1 - 0.7925 = 0.2075 m3

=> Amount of fine aggregate = Volume x Specific gravity x Water density = 0.2075 x 2.6 x 1000 = 539.5 kg

Now, since both aggregates has absorption capacity, so they will absorb some water from concrete so amount of mixing water needs to be corrected

Water absorbed by coarse aggregate = 0.008 x 1072.5 = 8.58 kg

Water absorbed by fine aggregate = 0.005 x 539.5 = 2.70 kg

=> Actual amount of water to be added = 205 + 8.58 + 2.70 = 216.28 kg

Hence, composition for trial mix for 1 m3 concrete is as follows :

Material	Amount (kg)
Cement	488.1
Water	216.28
Coarse aggregate	1072.5
Fine aggregate	539.5