Construction Technology_Concrete Mix Design_Project 1

1. Calculate the concrete Mix Design for M35 grade concrete with fly ash?

Ans:  Here are the following data given and assumed from the IS code books,

1. Grade Designation = M35

• Type of cement = OPC 43 grade
• Type of mineral admixtures = Fly ash conforming to IS 3812 (Part 1)
• Max nominal size of aggregates = 20MM
• Minimum cement content = 320kg/m^3 (from table 5 IS 456:2000 exposure condition)
• Max water-cement ratio = 0.45 (from table 5 IS 456:2000 exposure condition)
• workability = 100mm (Slump value for pumpable concrete)
• Exposure condition = severe (for reinforced concrete)
• Maximum cement content = 450kg/m^3
• Chemical admixture = super plasticizer Test data
• cement used = OPC 43 
• specific gravity of cement = 15
• specific gravity of fly ash = 2
• chemical admixture = super plasticizer conforming to IS 9103
• specific gravity of course aggregate = 2.74 specific gravity of fine aggregate= 74
• water absorption of course aggregates = 0.5% , water absorption of fine aggregate= 0%

Procedure:

2. Target mean strength

f'ck = fck +1.65S (from table 1 IS 10262:2009) S=5N/mm2

= 35+1.65*5 = 43.25 N/mm^2

from table 5 IS 456:2000 max w/c ratio = 0.45 Based on experience adopt w/c ratio = 0.40

0.40 < 0.45, hence OK

3. Selection of water cement ratio

From table 2 of IS 10262:2009,

Max water content for 20 mm aggregate = 186 litre , for 25 to 50 mm slump range

Estimated water content for 100 mm slump = 186+6/100 *186 = 197 litres As super plasticizer is used, the water content can be reduced by 30%

Based on water content reduction of 29% is achived with plasticizer Hence the arrived water content = 197* 0.71= 140 litres.

3. Calculation of cement content

w/c ratio = 40

cementitious material (cement+flyash)= 140/0.40= 350kg/m^3

from table 5 of IS 456:2000

Min cement content for severe exposure condition = 320kg/m^3 350kg/m^3 >320kg/m^3, Hence OK

Proportion of mix containing fly ash

•  percentage of fly ash needs to be decided based on project requirment & quality of materials
• In some cases increases in cementitious material can be warranted. increase in cementitious material content and percentage may be used on experience and trials. (this is with increase of 10% of cementitious material content)

cementitious material content = 350*1.10 =385kg/m^3 water content = 140kg/m^3

water-cement ratio= 140/385=0.364

fly ash @ 30% total cementitious material= 385*30%=115kg/m^3 cement (OPC)= 385-115=270kg/m^3

saving of cement while using flyash= 350-270=80kg/m^3 flyash being utilisez=115kg/m^3

Proportion of volume of course aggregate and fine aggregate

from table 3 of IS 10262:2009,

volume of course aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) w/c ratio = 0.6 In present case w/c ratio = 0.40

therefore volume of course aggregate needs to be increased to decreased fine aggregate content

as w/c ratio is lower by 0.10, the proportion of volume of course aggregate is increased by 0.02 at the water of -/+0.01 for every -/+0.05 change in w/c ratio corrected volume of course aggregate for w/c ratio of 0.4 =0.62

for pumpable concrete these values are reduced by 10% volume of course aggregate=0.62*0.9=0.56

volume of aggregate content=1-0.56=0.44 Mix calculation

4. Mix calculation per unit volume of concrete

1. volume of concrete= 1m^3
2.  volume of cement= mass of cement/specific gravity of cement*1/1000

                                                             = 270/3.15*1/1000=0.086m^3

    3. volume of flyash= mass of flyash/specific gravity of flyash*1/1000

                                                             = 115/2.2*1/1000=0.052m^3

    4. volume of water= mass of water/specific gravity of water*1/1000

                                                             = 140/1*1/1000=0.140m^3

    5.volume of chemical admixture= mass of chemical admixture/specific gravity of chemical admixture*1/1000

                                                             = 7.6/1.145*1/1000=0.006m^3

    6. volume of aggregate(Fine aggregate & Course aggregate)=(a-(b+c+d+e)) 

                                                             = (1-(0.086+0.052+0.140+0.006))

                                                             = 0.716m^3

    7. Mass of course aggregate= f*volume of course aggregate * specific gravity of course aggregate*1000

                                                             = 0.716*0.56*2.74*1000= 1098kg/m^3

    8. Mass of fine aggregate= f*volume of fine aggregate * specific gravity of fine aggregate*1000

                                                             = 0.716*0.44*2.74*1000= 863kg/m^3

5. Mix Proportion

cement= 270kg/m^3

 fly ash = 115kg/m^3

 water= 140kg/m^3

 fine aggregate= 863kg/m^3

 course aggregate= 1098kg/m^3

 chemical admixture= 7kg/m^3

 w/c ratio= 0.364

Trial mix ratio= 1:2.25:2.83

2. Calculate the concrete Mix Design forM50 grade concrete without flyash?

 Here are the following data given and assumed from the IS code books,

1. Grade Designation = M50

• Type of cement = OPC 43 grade
• Type of mineral admixtures = Fly ash conforming to IS 3812 (Part 1)
• Max nominal size of aggregates = 20MM
• Minimum cement content = 320kg/m^3 (from table 5 IS 456:2000 exposure condition)
• Max water-cement ratio = 0.45 (from table 5 IS 456:2000 exposure condition)
• workability = 100mm (Slump value for pumpable concrete)
• Exposure condition = severe (for reinforced concrete)
• Maximum cement content = 450kg/m^3
• Chemical admixture = super plasticizer Test data
• cement used = OPC 43 
• specific gravity of cement = 15
• specific gravity of fly ash = 2
• chemical admixture = super plasticizer conforming to IS 9103
• specific gravity of course aggregate = 2.74 specific gravity of fine aggregate= 74
• water absorption of course aggregates = 0.5% , water absorption of fine aggregate= 0%

Procedure:

1.Target mean strength

f'ck = fck +1 65S (from table 1 IS 10262:2009) S=5N/mm2

= 50+1.65*5 = 58.25 N/mm^2

from table 5 IS 456:2000,

Max w/c ratio = 0.45 Based on experience adopt w/c ratio = 0.40

0.40 < 0.45, hence OK Selection of water cement ratio

From table 2 of IS 10262:2009,

Max water content for 20 mm aggregate = 186 litre , for 25 to 50 mm slump range

Estimated water content for 100 mm slump = 186+6/100 *186 = 197 litres

As super plasticizer is used, the water content can be reduced by 30% Based on water content reduction of 29% is achived with plasticizer.

Hence the arrived water content = 197* 0.71= 140 litres

2. Calculation of cement content

w/c ratio = 40

cementitious material (cement)= 140/0.40= 350kg/m^3

 From table 5 of IS 456:2000

Min cement content for severe exposure condition = 320kg/m^3

350kg/m^3 >320kg/m^3, Hence OK

Proportion of volume of course aggregate and fine aggregate

 From table 3 of IS 10262:2009,

volume of course aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) w/c ratio = 0.6 In present case w/c ratio = 0.40

therefore volume of course aggregate needs to be increased to decreased fine aggregate content

As w/c ratio is lower by 0.10, the proportion of volume of course aggregate is increased by 0.02 at the water of -/+0.01 for every -/+0.05 change in w/c ratio corrected volume of course aggregate for w/c ratio of 0.4 =0.62

for pumpable concrete these values are reduced by 10% volume of course aggregate=0.62*0.9=0.56

volume of aggregate content=1-0.56=0.44 Mix calculation

3. Mix calculation per unit volume of concrete

 1. volume of concrete= 1m^3

 2. volume of cement= mass of cement/specific gravity of cement*1/1000

                                      = 350/3.15*1/1000=0.111m^3

 3. volume of water= mass of water/specific gravity of water*1/1000

                                      = 140/1*1/1000=0.140m^3

 4. volume of chemical admixture= mass of chemical admixture/specific gravity of chemical admixture*1/1000

                                      = 7.6/1.145*1/1000=0.006m^3

 5. volume of aggregate(Fine aggregate & Course aggregate)=(a-(b+c+d))

                                      = (1-(0.111+0.140+0.006))= 0.743m^3

 6. Mass of course aggregate= e*volume of course aggregate * specific gravity of course aggregate*1000

                                       = 0.743*0.56*2.74*1000

                                           = 1140kg/m^3

7. Mass of fine aggregate= e*volume of fine aggregate * specific gravity of fine aggregate*1000 

                                        = 0 743*0 44*2 74*1000= 896kg/m^3

Mix Proportion

 cement= 350kg/m^3

 water= 140kg/m^3

 fine aggregate= 896kg/m^3

 course aggregate= 1140kg/m^3

 chemical admixture= 7kg/m^3

 w/c ratio= 4

 Trial mix ratio= 1:2.55:3.25