Question

Use the following information to complete the concrete mixture designs for questions 4.

The following materials are available for a concrete mixture design:

ASTM C 150 Type I/II Cement

            Relative density of 3.15

Coarse Aggregate: well-graded ¾ in. maximum-size aggregate (MSA), crushed limestone, angular

            Oven-dry specific gravity: 2.45

            Absorption: 0.80%

            Oven-dry rodded bulk density: 97 lb/ft³

            Moisture content of coarse stockpile: 0.63%

Fine Aggregate: well-graded natural sand

            Oven-dry specific gravity: 2.64

            Absorption: 1.4%

            Moisture content of fine stockpile: 2.8%

            Fineness Modulus: 3.00

An air entraining admixture will provide enough air for the design mixture at a dosage rate of 3 oz/ft³, if the system needs to be air entrained.

Question 4.

Concrete is required for an 8 in thick exterior concrete slab in central Texas. A specified compressive strength, f’_c , of 5700 psi is required at 28 days using an ASTM C 150 Type II portland cement. The design calls for a minimum of 2 in. of concrete cover over the reinforcing steel. The minimum distance between reinforcing bars is 4 in. A slump of 6 inches should be the target. The sidewalk is located in an environment that does not require air entraining. The concrete is required to have low permeability when exposed to water and moderate sulfates. The concrete will be exposed water soluble sulfates in soil at 1.14%. No statistical data on previous mixes are available. Determine:

1. The theoretical mixture proportions of all concrete constituents (fine aggregate, coarse aggregate, water, cement, chemical admixtures) based on 1 yd³ of concrete on an oven dry basis for aggregates.  

2. The as-batched mixture proportions based on the moisture contents provided in the materials available section.

3. How much of each material would be required to make a 20 ft long by 15 ft slab from the concrete mixture described above?

Answer 1

Solution:-

Ans a)

Let the total volume of trial mix be 1 yd3 (27 ft3) then according to ACI 211.1.8 Table 6.3.3 for 6 in slump and nominal aggregate size of 3/4in , amount of water required per cubic yard of concrete is 325 lb/cy

=> Amount of water = 325 lb per yd3 concrete

Since , class of exposure is moderate, air content for 3/4 in aggregate size is 5 %

Now, required compressive strength as per ACI 318 and specified strength more than 5000 psi os :

=> fcr = 1.1 f'c + 700

=> fcr = 1.1(5700) + 700

=> fcr = 6970 psi

Now according to Table 6.3.4 (a) , for compressive strength of 6970 psi , water cement ratio will be 0.35

Hence, amount of cement = 325 / 0.30 = 928.5 lb

Now, according to table 6.3.6 for nominal aggregate size of 3/4 in and fineness modulus of 3 , volume of coarse aggregate is 0.60 yd3 or 16.2 ft3

=> Amount of coarse aggregate = dry density x volume = 97 x 16.2 = 1571.4 lb/ft3

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

=> Fine aggregate volume = 27 - [(325/62.4) + (928.5 / 3.15 x 62.4) + (1571.4/ 2.45 x 62.4) + 0.05(27)]

=> Fine aggregate volume = 27 - 20.84 = 6.16 ft3

=> Amount of fine aggregate = Specific gravity x Unit weight of water x Volume = 2.64 x 62.4 x 6.16 = 1014.7 lb

Now, amount of admixture = 3 oz/ft3 x 27 ft3 = 81 oz

Hence, theoratical mix proportions of constituents are :

Constitutient	Amount
Cement	928.5 lb
Water	325 lb
Coarse aggregate	1571.4 lb
Fine aggregate	1014.7
Admixture	81 oz

Ans b)

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

Water provided by coarse aggregate = (0.0063 - 0.008) x 1571.4 = -2.67 lb

Water provided by fine aggregate = (0.028 - 0.014) x 1014.7 = 14.20 lb

=> Actual amount of water to be added = 325 + 2.67 - 14.20 = 313.47 lb

Now,

Since both the coarse and fine aggregates have moisture content, actual amount of aggregates also needed to be corrected

Correct amount of coarse aggregate = 1.0063 x 1571.4 lb = 1581.30 lb

Correct amount of fine aggregate = 1.028 x 1014.7 lb = 1043.10 lb

Hence, as-batched mixture proportions per cubic yard concrete based on moisture are :

Constitutient	Amount
Cement	928.5 lb
Water	313.47 lb
Coarse aggregate	1581.30 lb
Fine aggregate	1043.10 lb
Admixture	81 oz

Ans c)

Volume = L x B x t = 20 ft x 15 ft x 8 in/ 12 in/ft = 201 ft^3 or 7.45 yd^3

Hence, weight of each material required are as follows :   

Constituents	Amount
Cement	928.5 lb/cy x 7.45 cy = 6917.32 lb
Water	325 lb/cy x 7.45 cy = 2421.25 lb
Coarse aggregate	1581.3 lb/cy x 7.45 cy = 11780.68 lb
Fine aggregate	1043.1 lb/cy x 7.45 cy = 7772 lb
Admixture	81 oz/cy x 7.45 cy = 603.45 lb