Construction Technology_Concrete Mix Design_Project 1

 

1) M35 Grade Concrete with Fly ash 

Aim: To design the concrete mix for M35 using Fly ash

Introduction :

The Proportions used for the mix design are to be calculated as per the assumption of the values that are suited to get the required compressive strength.

Therefore the following details are to be assumed:

Type of cement	OPC 43 Grade
Type of Fly ash	F type conforming to IS 3812 (Part 1)
Max size of aggregate (MSA)	20mm (From table 2 IS 456-2000)
Minimum Cement Contact	320kg/m3 (From table 5 IS 456-2000)
Maximum w/c ratio	0.45
Workability	125 mm slump (assume)
Exposure Condition	Severe (RCC)
Method of placing concrete	Pumping
Chemical Admixture	Superplasticizer
The specific gravity of cement	3.15
The specific gravity of Fly ash	2.2
The specific gravity of C.A	2.78
The specific gravity of F.A	2.70
Water absorption
Coarse aggregate	0.5%
Fine aggregate	Nil
Free surface moisture
Coarse aggregate	Nil
Fine aggregate	1.5%
Grading of coarse aggregate	Conforming to table 2 of IS 383
Grading of fine aggregate	Zone I

 

Target mean strength :

f'ck = fck + 1.65 * standard deviation           (For standard deviation From Table 1 of IS 10262-2009)

       =35 + 1.65*5

       =43.25 N/mm2

Selection of W/C ratio :(From table 5 of IS 456-2000)

Experience says 43.25 N/mm2 can be achieved in 28 days by using a w/c ratio of 0.46;

But as per the table 5 of IS 456, a maximum w/c ratio permitted is 0.45

Adopt w/c ratio of 0.45

Selection of water Content :

Maximum water content for MSA 20mm aggregate is 186 litre (for slump up to 50mm and w/c ratio of 0.5)

Estimated water content for 125 mm slump = 186+186*(9/100)=203 liter

As Superplasticizer is used, it Is assumed that water content can be reduced to the extent of 25%

So,The net amount of water required to be used = 203* 0.75 = 153 liters

Calculation of cement and fly ash content :

w/c ratio = 0.45

Cementitious material (cement + fly ash) content = 153/0.45 = 340 kg/m3

Since 340 > 320 it is ok

IS 456 table number 5 permits minimum cement content 320 kg/m3 for severe exposure condition.

Since fly ash is not as active as that of cement, it is usual to increase the cementitious material by some percentage, based on experience and trails.

In this example an increase of 10% in considered.

Cementitious material content = 340 * 1.1 = 374kg/m3

Assume percentage of fly ash as 25%

Fly ash content = 25/100 * 374 = 93.5kg/m3

Cement = 374-93.5 = 280.5 kg/m3

w/cementious ratio = 153/374 = 0.41

The proportion of the volume of C.A and F.A content :(Table 3 IS 10262-2009)

The volume of coarse aggregate for 20mm MSA and fine aggregate falling on the zone I and W/C of 0.5 = 0.60.

Here the w/c is 0.45. Therefore the volume of Coarse aggregate is required to be increased to decrease the F.A. content.

As the w/c is lower by 0.15, the proportion of the volume of coarse aggregate is increased by (0.01/0.05)*0.05 = 0.01.

Therefore corrected proportion of volume of coarse aggregate for the water cement ratio of 0.45 = 0.60 + 0.01 = 0.61.

For pumpable concrete, C.A. may be reduced by 10 %.

Volume of C.A. = 0.61*0.9 = 0.59

Volume of F.A = 1-0.59 = 0.41

Mix Calculation :

Volume of concrete = 1m3

Volume of cement = 280.5/(3.15*1000) = 0.089 m3

The volume of fly ash = 93.5/(2.2*1000) = 0.0425 m3

The volume of water = 0.153 m3

The volume of chemical admixture, assuming dosage of 1.2% by weight of cementitious material and sp. gr of 1.1

                                                             = (1.2*374)/100 = 4.49

                                                             = 4.49/(1.1*1000) = 0.0041

The total volume of all in aggregate = 1- (0.089+0.0425+0.153+0.0041)

                                                    = 0.71 m3

Weight of coarse aggregate = Total volume of aggregates*Total volume of coarse aggregate*Specific gravity of coarse aggregate*1000

                                        = 0.71*0.59*2.78*1000

                                        = 1165Kg/m3

Weight of fine aggregate = Total volume of aggregates*Total volume of fine aggregate*Specific gravity of fine aggregate*1000

                                    = 0.71*0.41*2.7*1000

                                    = 786Kg/m3

Mix proportions :

Cement = 280.5 Kg/m3

Fly ash = 93.5 Kg/m3

Water =153 Kg/m3

Fine aggregate = 786 Kg/m3

coarse aggregate = 1165 Kg/m3 

Water/cementitious ratio = 0.41 Kg/m3

Coarse aggregate can be further divided into 10mm size and 20mm size, depending upon the grading.

we may divide the total aggregate into 40 per cent of 10mm size and 60 per cent of 20 mm size.

In that case quantity of 10mm size = 1165*0.4 = 466 Kg/m3

Quantity of 20mm in size = 1125*0.6 = 700 Kg/m3

 

    

 

 

2) M50 Grade Concrete without Fly ash

 

Aim: To design the concrete mix for M50 without using Fly ash

Introduction :

The Proportions used for the mix design are to be calculated as per the assumption of the values that are suited to get the required compressive strength.

Therefore the following details are to be assumed:

Type of cement	OPC 43 Grade
Max size of aggregate (MSA)	20mm (From table 2 IS 456-2000)
Minimum Cement Contact	320kg/m3 (From table 5 IS 456-2000)
Maximum w/c ratio	0.45
Workability	100 mm slump (assume)
Exposure Condition	Severe (RCC)
Method of placing concrete	Pumping
Chemical Admixture	Superplasticizer
The specific gravity of cement	3.15
The specific gravity of C.A	2.78
The specific gravity of F.A	2.70
Water absorption
Coarse aggregate	0.5%
Fine aggregate	1%
Free surface moisture
Coarse aggregate	Nil
Fine aggregate	Nil
Grading of coarse aggregate	Conforming to table 2 of IS 383
Grading of fine aggregate	Zone II

 

Target mean strength :

f'ck = fck + 1.65 * standard deviation           (For standard deviation From Table 1 of IS 10262-2009)

       =50 + 1.65*5

       =58.25 N/mm2

Selection of W/C ratio :(From table 5 of IS 456-2000)

Experience says 43.25 N/mm2 can be achieved in 28 days by using a w/c ratio of 0.40;

The water-cement ratio is to be selected from Strength consideration and the maximum W/C ratio as per table 5 of IS 456and lesser of two is to be considered for durability requirement.

a maximum w/c ratio permitted as per code is 0.45

W/C ratio proposed is 0.40

So, Adopt w/c ratio of 0.40

Selection of water Content :

Maximum water content for MSA 20mm aggregate is 186 litre (for slump up to 50mm and w/c ratio of 0.5)

Estimated water content for 100 mm slump = 186+186*(6/100)=197 liter

As Superplasticizer is used, it Is assumed that water content can be reduced to the extent of 20%

So, The net amount of water required to be used = 197* 0.80 = 158 litres

Calculation of cement content :

w/c ratio = 0.40

cement content = 158/0.40 = 395 kg/m3

Since 395 > 320 it is ok

IS 456 table number 5 permits minimum cement content 320 kg/m3 for severe exposure condition.

Cement = 395 kg/m3

water/cement ratio = 158/395 = 0.40

The proportion of the volume of C.A and F.A content :(Table 3 IS 10262-2009)

The volume of coarse aggregate for 20mm MSA and fine aggregate falling on the zone II and W/C of 0.5 = 0.62.

Here the w/c is 0.40. Therefore the volume of Coarse aggregate is required to be increased to decrease the F.A. content.

As the w/c is lower by 0.1, the proportion of the volume of coarse aggregate is increased by (0.01/0.05)*0.10 = 0.02.

Therefore corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.45 = 0.62 + 0.02 = 0.64.

For pumpable concrete, C.A. may be reduced by 10 %.

Volume of C.A. = 0.64*0.9 = 0.576

Volume of F.A = 1-0.576 = 0.424

Mix Calculation :

Volume of concrete = 1m3

Volume of cement = 395/(3.15*1000) = 0.125 m3

The volume of water = 0.158 m3

The volume of chemical admixture, assuming dosage of 1.2% by weight of cement material and sp. gr of 1.1

                                                             = (1.2*395)/100 = 4.74

                                                             = 4.74/(1.1*1000) = 0.0043

The total volume of all in aggregate = 1- (0.125+0.158+0.0043)

                                                    = 0.713 m3

Weight of coarse aggregate = Total volume of aggregates*Total volume of coarse aggregate*Specific gravity of coarse aggregate*1000

                                        = 0.713*0.576*2.78*1000

                                        = 1142Kg/m3

Weight of fine aggregate = Total volume of aggregates*Total volume of fine aggregate*Specific gravity of fine aggregate*1000

                                    = 0.713*0.424*2.7*1000

                                    = 817Kg/m3

Mix proportions :

Cement = 395 Kg/m3

Water =158 Kg/m3

Fine aggregate = 817 Kg/m3

coarse aggregate = 1142 Kg/m3 

Water/cementitious ratio = 0.40 Kg/m3