Determine the batch quantities for a concrete mix design to have a 28-day compre

Question

Determine the batch quantities for a concrete mix design to have a 28-day compressive strength of 4500 psi, using angular aggregate from a No.4 sieve openingg to l/2 in. size and a medium/coarse "fine aggregate with a fineness modulus of 2.80. A : slump of 4 in. is desired. Material Descriptions are as follows 1. Coarse Aggregates: Size: Absorption: Bulk dry specific gravity: Dry-rodded unit weight: No. 4 to 1 ½ in a. b. c. d. 2.56% 2.45 94.5 lb/ft3 2. Fine Aggregates: a. Fineness modulus: b. Absorption: 2.8 2.8% 2.42 Bulk dry specific gravity (S.G.): 3. Cement: Bulk dry specific gravity (S.G.):3.14 Weight per sack: 94 lb a. b. 4. Concrete: 4 in. slump 4,500 psi 28-day compressive strength for non-air-entrained concrete. a. First, design for 1 yd5 batch of concrete, then determine the quantities required for a batch that is sufficient to produce three 6 in. diameter concrete cylinders (with 50% surplus for the losses during the mixing and casting process)

Explanation / Answer

SOLUTION:

4500 psi = 30 Mpa

Expect Standard deviation of compressive strength 3 Mpa

Target Mean Strength = 30 + 3 = 33 Mpa

ABSOLUTE VOLUME (V)

Absolute Volume = weight of material / (specific gravity of material x unit weight of water)

Take unit weight of water = 62.4 lb/ft3

Absolute Volume for Coarse Aggregate = (94.5) / (2.45 * 62.4) = 0.618 ft3 = 0.023 yd3

Fine aggreagate finess modulus 2.8 was Natural sand  unit weight = 95 lb/ft3

Absolute Volume for Fine Aggregate = (95 ) / (2.42 * 62.4) = 0.629 ft3 = 0.023 yd3

Absolute Volume for Cement = (94 ) / (3.14 * 62.4) = 0.48 ft3 = 0.018 yd3

So total Volume = 0.618 + 0.629 +0.48 = 1.727 ft3 = 0.064 yd3

Water is assumed to be 5% = 1.727 * 1.05 = 1.81 ft3

Air is only entariner is allowed so assumed as 4% = 1.81 * 1.04 = 1.88ft3

So the Volume of Air is = 1.88 - 1.81 = 0.07 ft3

To find

BULK SPECIFIC GRAVITY SATURATED SURFACE DRY (Gs(ssd)) for Course Agrregate

Gs(ssd) = (1 + Absorption) x Gs(dry)

= (1+0.0256) * 2.45 =2.51

BULK SPECIFIC GRAVITY SATURATED SURFACE DRY (Gs(ssd)) for FA

Gs(ssd) = (1 + Absorption) x Gs(dry)

= (1+0.028) * 2.42 =2.49

WATER-CEMENT RATIO (w/c) = 0.5

for ft3 Materials

Weight of Fine Aggreagate = 0.629 * 2.49 * 62.4 = 97.73 lb = (97.73 *0.037) = 3.61 lb per Yd3

Weight of Course Aggreagate = 0.618 * 2.51 * 62.4 = 96.79 lb = (96.79 *0.037) = 3.58 lb per Yd3

Weight of Cement = (0.48 * 3.14 * 62.4) / 4.2 = 22.39 lb =(22.39 *0.037) = 0.83 lb per Yd3

For 6"dia Concrete Cylinders 50% surplus for the losses with one feet depth = (3.14 * 0.25 *0.25) *1.5 =0.294ft3

Volume required for batch to produce 6"dia cylinder

Course aggreagte = 97.73 x 0.294 = 28.73 lb

Fine aggregate =  96.79 x 0.294 =28.46 lb

Cement = 22.39 x 0.294 =6.58 lb

Observations should be conduct with the above mix proportions

we should get 33Mpa. If not achived we should increase the cement and water by 2.5% and again test with obserations.

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