Question

Please show which equations you use and describe in detail.

21. concrete mix design, Design the concrete mix according to the following conditions: Design Environment Unreinforced slab, Bozeman, Montana (cold climate) Required design strength = 3500 psi Slab thickness = 12 in Statistical data indicate a standard deviation of compressive strength of 250 psi is expected (more than 30 samples). Only air entrainer is allowed Available materials Air entrainer: manufacture specification is 0.15 fl oz/1%/100 lb cement Coarse aggregate: 2 inch nominal maximum size, crushed stone Bulk over-dry specific gravity = 2.57, absorption =1.5% Oven-dry rodded density = 120 pcf Moisture content = 1% Fine aggregate: natural sand Bulk over-dry specific gravity = 2.6.4, absorption = 0.8% Moisture content = 3.76% Fineness modulus = 2.68

Answer 1

Answer:-

Given

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 slabs should be between 1 - 3 in .According to table 6.3.3 for 3 in slump and nominal aggregate size of 2 in , amount of water required per cubic yard of concrete is 285 lb/cy

=> Amount of water = 285 lb per yd3 concrete

For aggregate size 2 in, approximate air content is 0.5 % (0.135 ft3)

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

f'cr = f'c + 1.34 s or f'c + 2.33 s - 500

where, f'c = design compressive strength

s = standard deviation

Since, number of sample tested are more than 30 , standard deviation modification factor is 1

=> f'cr = 3000 + 1.34(250) or 3000 + 2.33(250) - 500

=> f'cr = 3335 psi or 3082.5 psi

fcr is larger of above two value,

=> fcr = 3335 psi

Now according to Table 6.3.4 (a) , for compressive strength of 3335 psi, water cement ratio is 0.64

Hence, amount of cement = 285 / 0.64 = 445.3 lb

Now, according to table 6.3.6 for nominal aggregate size of 2 in and fineness modulus of 2.68 , volume of coarse aggregate is 0.75 yd3 (20.25 ft3)

=> Amount of coarse aggregate = dry rodded weight x volume = 120 lb/ft3 x 20.25 ft3 = 2430 lb

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

Volume of components = Weight / (Specific gravity x Water unit weight)

=> Fine aggregate volume = 27 - [(285/62.4) (445.3 / 3.15 x 62.4) + (2430/ 2.573 x 62.4) + 0.135]

=> Fine aggregate volume = 27 - 22.08 = 4.92 ft3

=> Amount of fine aggregate = Specific gravity x Water unit weight x volume = 2.54 x 62.4 x 4.92 = 779.8 lb

Admixture dose = 0.15 fl oz / 1 % air /100 lb cement

=> Admixture needed = (0.15 fl oz / 1% /100 lb) x 445.3 lb x 0.5 % = 0.334 fl oz

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

Water provided by coarse aggregate = (0.01 - 0.001) x 2430 lb = 21.87 lb

Water provided by fine aggregate = (0.0376 - 0.002) x 779.8 = 27.76 lb

=> Actual amount of water to be added = 285 lb -21.87 lb - 27.76 lb = 235.37 lb

Hence, final batch weight of components for 1 cy concrete are :

Component Cement Welght (lb) 445.30 235.37 Water Coarse Aggregate Fine Aggregate 2430 779.80

Admixture Dose = 0.334 fl oz

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