Question

Ex if std no: 1abcde, 161667 (a=6, b=1, c=6, d=6, e=7)
In my case std 174702

Q1) Design a concrete mixture proportions which will be used for offshore concrete platform. Norway is a world leader on offshore concrete platforms with 1a of the world’s 30 larger offshore concrete structures located on the Norwegian continental shelf. The first of these platforms where the Ekofisk Tank installed in 1974 and the last was Troll gravity based structure installed in 1996. The Sakhalin project is located in a harsh and artic environment where concrete structures have proven excellent characteristics with a minimum of maintenance. 80 mm slump value is required for the Sakhalin project. 28-days concrete characteristic strength is 2b MPa. It is required that no more than 1 test result in 100 will fall below the specified strength. The fine aggregate has a fineness modulus of 2.60 and Zone 3. For Coarse Aggregate (Uncrushed): SSD Bulk Specific Gravity: 2.65, Absorption Capacity: 2.b %, Total Moisture: 1.c % and Dry Rodded Unit Weight: 1345 kg/m3. For Fine Aggregate (Uncrushed): SSD Bulk Specific Gravity: 2.70, Total Moisture: 3.d % and Free Moisture: 2 %. Cement: Portland Composite Cement, Specific Gravity: 2.97. For 0.e5 m3 of concrete, binder amount is calculated as 120 kg (Use ACI Method of Mix Design)

Q2) What would be the maximum aggregate size used for the given mixture properties by using ACI Method of Mix Design. 28-day specified concrete strength is 3b Mpa, standard deviation is choosen as 3.a Mpa and it is required that no more than 1 test result in 200 will fall below the specified strength.
Project Title: Detoration of wood column of a commercial building
Project summary: Wood deterioration at the lower level of a heavy timber column, which supports five floor levels, was remedied by removing a section of column and the construction of a new concrete pier and footing.
For Coarse Aggregate (Uncrushed): SSD Bulk Specific Gravity: 2.60, Absorption Capacity: 1.8 %, Total Moisture: 2.3 % and Dry Rodded Unit Weight: 1430 kg/m3.
For Fine Aggregate (Crushed): SSD Bulk Specific Gravity: 2.70, Total Moisture: 3.c % and Free Moisture: 1.d %. FM= 2.65, Zone II.
Cement: Portland Composite Cement, Specific Gravity: 3.01. For 0.2a m3 cement is calculated as 12b kg.

Q3) use the data in question 2 and determine the amount of chemical admixture required for 0.b% and 0.cd%

Answer 1

Ans 1) Let the total volume of trial mix be 1 m3 . The nominal aggregate size suitable for construction of concrete platform is 19 mm.According to ACI 211.1-91, table 6.3.3, for 80 mm slump and nominal aggregate size of 19 mm , amount of water required per cubic meter of concrete is 193 kg

=> Amount of water = 193 kg per m3 concrete

Since, concrete exposed to aggressive environment, class of exposure is severe, so air content for 19 mm aggregate size is 6 %

Now, required compressive strength as per ACI,

fcr = f'c + 8.3

=> fcr = 24 + 8.3 = 32.3 MPa

Now according to Table 6.3.4 (a) , for compressive strength of 32.3 MPa, water cement ratio is 0.43

Minimum Cement content per cubic meter concrete = 120 / 0.95 = 480 kg

Hence, amount of cement = Amount of water / w-c ratio = 193/ 0.43 = 448.84 kg < 480 kg

Hence. provide minimum cement content = 480 kg/

  Now, according to table 6.3.6 for nominal aggregate size of 19 mm and fineness modulus of 2.60 , volume of coarse aggregate is 0.64 m3

=> Amount of coarse aggregate = dry rodded density x volume = 1345 x 0.64 = 860.8 kg

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

=> Fine aggregate volume = 1 - [(193/1000) + (480 / 2.97 x 1000) + (860.8/ 2.65 x 1000) + 0.06]

=> Fine aggregate volume = 1 - 0.74 = 0.26 m3

=> Amount of fine aggregate = volume x specific gravity x water density = 0.26 x 2.7 x 1000 = 702 kg

Now, since both aggregates has moisture and has absorption capacity , amount of mixing water needs to be corrected as shown :

Net Water absorbed by coarse aggregate = (0.017 - 0.024) x 860.80 = -6.03 kg

Net Water absorbed by fine aggregate = (0 - 0.05) x 702 = -35.1 kg

=> Actual amount of water to be added = 193 - 6.03 - 35.1 = 151.87 kg

Hence, batch weight for trial mix of 1 cubic meter is as follows :

Material	Amount (kg)
Cement	480
Water	151.87
Fine aggregate	702
Coarse aggregate	860.8

Ans 2)  Let the total volume of design concrete mix be 1 m3 . The nominal aggregate size suitable for construction of concrete pier is 19 mm .According to ACI 211.1-91, table 6.3.1, design slump for the construction of concrete pier is 25 - 100 mm. From 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 193 kg

=> Amount of water = 193 kg per m3 concrete

Air content for 19 mm aggregate with moderate exposure is 5%

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

fcr = f'c + 1.34S

or

fcr = f'c + 2.33S - 3.45

Standard deviation (S) = 3.7 MPa

=> fcr = 34 + 1.34(3.7) = 39 MPa

or

=> f'cr = 34 + 2.33(3.7) - 3.45 = 39.2 MPa

Hence, f'cr = 39.2 MPa

Now according to Table 6.3.4 (a), for compressive strength of 39.2 MPa, water cement ratio is 0.35

Minimum Cement content per cubic meter concrete = 124 / 0.27 = 459.3 kg

Hence, amount of cement = Amount of water / w-c ratio = 193/ 0.35 = 551.43 kg > 459.3 kg (Hence OK)

Now, according to table 6.3.6 for nominal aggregate size of 19 mm and fineness modulus of 2.65 , volume of coarse aggregate is 0.63 m3

=> Amount of coarse aggregate = dry rodded density x volume = 1430 x 0.63 = 900.9 kg

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

=> Fine aggregate volume = 1 - [(193/1000) + (551.43 / 3.01 x 1000) + (900.9/ 2.6 x 1000) + 0.05]

=> Fine aggregate volume = 1 - 0.773 = 0.227 m3

=> Amount of fine aggregate = volume x specific gravity x water density = 0.227 x 2.7 x 1000 = 612.9 kg

Now, since both aggregates has moisture and has absorption capacity , amount of mixing water needs to be corrected as shown :

Net Water absorbed by coarse aggregate = (0.018 - 0.023) x 900.9 = -4.50 kg

Net Water absorbed by fine aggregate = (0 - 0.047) x 612.9 = - 28.80 kg

=> Actual amount of water to be added = 193 - 4.5 - 28.80 = 159.70 kg

Hence, batch weight for trial mix of 1 cubic meter is as follows :

Material	Amount (kg)
Cement	551.43
Water	159.70
Fine aggregate	612.9
Coarse aggregate	900.9

Ans 3)  Given,

Admixture dose = 0.40% and 0.70% by weight of cement

Amount ofcement in part 2 = 551.43 kg

=> Amount of chemical admixture required = 0.004 x 551.43 kg = 2.20 kg

=> Amount of chemical admixture required = 0.007 x 551.43 kg = 3.86 kg