Question

In: Civil Engineering

Design a mix for a 3,000 psi concrete to be used in an indoor basement slab....

Design a mix for a 3,000 psi concrete to be used in an indoor basement slab. Your gravel is an ASTM C33C #57 stone. Your fine aggregate grades as follows: No. 4 – 3%, No. 8 – 7%, No. 16 – 20%, No. 30 – 25%, No. 50 – 20%, No. 100 – 15%, Pan – 10%.

What is your mix and list out what you can do to make a durable mix that is economically competitive? How would you adjust the workability of the mix?

Solutions

Expert Solution

Ans) Let the total volume of trial mix be 1 cubic yard (27 ft^3) then according to ACI 211.1.8 , table 6.3.1 , slump required for slab is 1 to 3 in .

ASTM C33 #57 stone has a nominal size between 1 to 0.19 in.So, according to Table 6.3.3 for 3 in slump and nominal aggregate size of 1 in , amount of water required per cubic yard of concrete is 325 lb/cy

=> Amount of water = 325 lb/ cubic yard concrete

According to Table 6.3.3 approximate air content in concret for 1 in aggregate = 1.5%

According to ACI 318, Design compressive strength is :

fcr = f'c + 1200

f'c is specified compressive strength

=> fcr = 3000 + 1200 = 4200 psi

=> fcr = 4200 psi

Now according to Table 6.3.4 (a) , for compressive strength of 4200 psi , water cement ratio is 0.55

Hence, amount of cement = 325/ 0.55 = 590.9 lb/cy

Sieve number % Retained Cumulative % retained
No.4 3 3
No.8 7 10
No.16 20 30
No.30 25 55
No.50 20 75
No.100 15 90

Fineness modulus, FM = Sum of cumulative retained /100 = (3 + 10 + 30 + 55 + 75 + 90) / 100 = 2.63

Now, according to table 6.3.6 for nominal aggregate size of 1 in and fineness modulus of 2.63, volume of coarse aggregate is 0.69 ft^3 or 18.63 cy

=> Amount of coarse aggregate = Dry rodded density x volume

Assuming dry rodded density of coarse aggregate as 95 lb/ft^3

=> Amount of coarse aggregate = 95 x 18.63 = 1769.85 lb

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

=> Fine aggregate volume = 27 - [(325/62.4) + (590.9 / 3.15 x 62.4) + (1769.85/ 2.6 x 62.4) + 0.015(27)]

=> Fine aggregate volume = 27 - 19.53 = 7.47 ft^3

=> Amount of fine aggregate = Volume x Specific gravity x Water density = 7.47 x 2.6 x 62.4 = 1212

Material Amount (lb)
Cement 325
Water 590.9
Coarse aggregate 1769.85
Fine aggregate 1212

To make durable mix that is economically competitive, follow the give steps :

1) Insure proper compaction is done

2) Use of admixtures to reduce water demand and improve workability

3) Use of lumps free durable cement

4) Choosing the best estimate of aggregate size to be used in concrete

To adjust workability of the mix :

1) Choose a w/c ratio wisely. Higher the w/c ratio higher will be the workability but lesser will be strength

2) Using of admixtures like superplastizers , retarders etc   


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