Question

The columns for a building in central Long Island are to be made of concrete with the following specifications: • Design compressive strength: 4500 psi, based on 20 tests of samples having a standard deviation of 600 psi • Non-air entrained, and clear space between rebar: 1.5 inch • Fine aggregate: G = 2.350, FM = 2.5 • Coarse aggregate: G = 2.65, bulk density = 3000 lb/cy Determine the following: i. w/c ratio, largest nominal max. coarse aggregate size, % coarse aggregate, % air (2 pts.) ii. weight of water, weight of cement, and weight of coarse aggregate to be used per cubic yard of concrete (1 pt.) iii. weights and volumes of water, cement, air, coarse aggregate, and fine aggregate needed to mix a 15-yard batch of this mix (3 pts.) Also: • Specify two admixtures that might be used for this mix based on the provided information, and why to use it. (2 pts.) • Why we might create a composite of this concrete poured into a PVC (polyvinyl chloride) sleeve for these columns? (Give specific reasons based on the characteristics of the individual materials.) (2 pts.)

Answer 1

Ans i) According to ACI 318 , when number of samples tested is less than 30 , a modification factor is needed to multiplied by standard deviation to obtain adjusted standard deviation.

According to Table 11, ACI 318 , for number of test 20 modification factor is 1.08

=> Adjusted standard deviation , s = 1.08 x 600 = 648 psi

According to table 12 , ACI 318 , for strength < 5000 psi , required compressive strength (f'cr) is larger of :

f'cr = f'c + 1.34 s

or

f'c + 2.33 s - 500

Hence, f'cr = 4500 + 1.34(648) = 5368 psi

or

f'cr = 4500 + 2.33(648) - 500 = 5510 psi

=> f'cr = 5510 psi

According to ACI 211.1-91, Table 6.3.4(a), for F'c = 5510 psi and non air entrained concrete, w/c ratio = 0.45

Now, nominal maximum size of aggregate is three fourth of spacing between bars

=> Nominal maximum size of aggregate = (3/4) x 1.5 = 1.12 in 1 in

Now, according to table 6.3.6 for nominal aggregate size of 1 in and fineness modulus of 2.5 , volume of coarse aggregate is 0.70 yd3 or 70%

According to table 6.3.3, for 1 in aggregate, amount of air = 1.5%

Ans ii) According to table 6.3.3, for a slump of 3 in - 4 in, nominal aggregate size of 1 in and non-air entrained concrete, amount of water required per cubic yard of concrete is 325 lb

=> Amount of water = 325 lb per yd3 concrete

Hence, amount of cement = Water / w/c ratio = 325 / 0.45 = 722.22 lb/cy

Now,

Amount of coarse aggregate = dry rodded density x volume = 3000 lb /cy x 0.70 ft3 = 2100 lb/cy

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

Volume of water = Weight / 62.4 = 325/62.4 = 5.208 ft3

Volume of cement = Weight / (62.4 x specific gravity) = 722.22 / (3.15 x 62.4) = 3.674 ft3

Volume of coarse aggregate = Weight / (62.4 x spcific gravity) = 2100 / (2.65 x 62.4) = 12.70 ft3

Volume of air = 1.5% = 0.015 x 27= 0.405 ft3

=> Fine aggregate volume = 27 - (5.208 + 3.674 + 12.70 + 0.405)

=> Fine aggregate volume = 27 - 21.987 = 5.013 ft3/cy

=> Amount of fine aggregate =volume x specific gravity x unit weight of water= 5.013 ft3 x 2.35 x 62.4 pcf = 735.1 lb/cy

Now, we have calculated the amount of each component of concrete for batch of 1 cy so we can determine batch weight for 15 cy batch as shown below :

Material	Weight per 15 cy (lb)	Volume per 15 cy (ft3)
Cement	15 x 722.22 = 10833.33	15 x 3.674 = 55.11
Water	15 x 325 = 4875	15 x 5.208 = 78.12
Fine aggregate	15 x 735.1 = 11026.5	15 x 5.013 = 75.2
Coarse aggregate	15 x 2100 = 31500	15 x 12.7 = 190.5
Air	-	15 x 0.405 = 6.075

Ans iv)  Following are the two admixture which can be used :

1) To reduce the amount of mixing water to make more economical mix , water reducing admixture (plasticizers and super plasticizers) can be used to improve workability with less amount of water

2) Since, the required strength is high, it is recommended to use air entraining admixtures to achieve durable mix