Question

Design a concrete mixture that has a 28 day compressive strength of 4000 psi and a maximum size aggregate of 0.75 in. The concrete would be placed in a column exposed to freezing and thawing and will be in contact with a soil that has a sulfate content of 0.3%.

Step 1: Assemble information from specifications and local materials

Cement:

Use any type and assume a specific gravity of 3.15

Coarse Aggregate:

Bulk Specific Gravity (BSG) – 2.65

Absorption Capacity – 1.5%

Surface Moisture – 1.0%

Dry-Rodded Unit Weight = 105 lb/ft^3

Fine Aggregate:

Bulk Specific Gravity (BSG) – 2.75

Absorption Capacity – 1.0%

Surface Moisture – 3.0%

Fineness Modulus – 2.7

Step 2: Select Slump

Step 3: Determine the maximum size of the aggregate

Step 4: Estimate the mix water and the air content

Step 5: Choose the water-to-cement ratio

Step 6: Determine the Cement Content

Step 7: Estimate the Coarse Aggregate Content

Step 8: Estimate the Fine Aggregate Content for a cubic yard

Step 9: Adjust for Moisture

Step 10: Summarize batch proportions in a table

Answer 1

Ans) Following is required mix design procedure :

Step 1) Selection of slump :

Let the total volume of trial mix be 1 yd3 (27 ft3) then according to ACI 211.1.8 Table 6.3.1 ,slump required for column is between 1 to 4 in .Let us choose 3 in slump .

Step 2) Maximum size of aggregate is given as 0.75 in

Step 3) Estimation of mixing water and air content

According to table 6.3.3 for a slump of 3 in and nominal aggregate size of 0.75 in , amount of water required per cubic yard of concrete is 305 lb

=> Amount of water = 360 lb per yd3 concrete

Since,concrete is exposed to freezing/thawing and sulphates, class of exposure is moderate so according to table 6.3.3, air content for 0.75 in aggregate size is 5 %

Step 4) Estimation of water-cement ratio

Now according to Table 6.3.4 (a) , for compressive strength of 4000 psi , water cement ratio is 0.48 .But since concrete is exposed to suphates, according to Table 6.3.4 (b) , maximum permissible water cement ratio is 0.40

Step 5) Estimation of cement content :

Hence, amount of cement = Amount of water / water cement ratio

=> Cement requied = 305 lb / 0.40 = 762.50 lb

Step 6) Estimation of Coarse aggregate

Now, according to table 6.3.6 for nominal aggregate size of 0.75 in and fineness modulus of 2.7 , volume of coarse aggregate is 0.63 yd3 or 17.01 ft3

=> Amount of coarse aggregate = dry rodded density x volume = 105 x 17.01 = 1786 lb

Step 7) Estimation of fine aggregates

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

Volume of material = Amount / (Specific gravity x Unit weight of water)

=> Fine aggregate volume = 27 - [(305/62.4) + (762.5 / 3.15 x 62.4) + (1786/ 2.65 x 62.4) + 0.05(27)]

=> Fine aggregate volume = 27 - 20.91 = 6.09 ft3

=> Amount of fine aggregate = Specific gravity x unit weight of water x volume = 2.75 x 62.4 x 6.09 = 1045 lb

Step 8) Adjustment of moisture

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

=> Water provided by coarse aggregate = (0.01 - 0.015) x 1786 = -8.93 lbs

=> Water provided by fine aggregate = (0.03 - 0.01) x 1045 = 20.90 lbs  

Corrected amount of mixing Water = Uncorrected weight - Water provided by aggregates

=> Corrected mixing Water needed = 305 - 20.90 + 8.93 = 293 lbs

Step 9) Final batch weigts for 1 cubic yard concrete are as follows :

Material	Amount (lb)
Cement	762.50
Water	293
Coarse aggregate	1786
Fine aggregate	1045