Project on Concrete Mix Design for various grades of Concrete

1. Ans,

Grade =M35

Cement type = 43

Aggregate size = 20 mm

Weather condition = Moderate

1. Target compressive strength

Fck =Fck' +1.65 s

Fck' = Comperssive strength attain at 28 days

S= Standerd devation

As per IS code 10262-2009 table 1

for M 35 =5

Fck= 35+(1.65*5)=43.25.

2. Water content

AS per code IS 10262-2009 table 2

From the graph between Target comperssive strength at Y-axis and Water content at X-axis.

From the above value 43.25 we can able to find the water content by drawing the line.

Water cement ratio can also be calculated by Environmental conditon

As per IS code 456-2000 Table 5

Depends on the Enviromental conditon we can calculate water content

From the Table Grades may be very but we have to check with Conditon

AS per tabel

Water cement content = .50

as per conditions =.45

Ans minimum Cement content = 300 kg

3. Cement content

It can be calculated by using aggretate size with table of IS 10262-2000 table 2

For 20 mm =186 litres

for slump value 25 mm to 50mm

our condition slump various to 125 mm.

Slump for above 50 mm , every 25 mm 3 % water content should be increased

so total 9 % has to increased = 186+(9*186/100) =202.74

As per we are using Super plasters the water content reduced to 20 %, But as per checked it can be reduced to 29 %

so =202.74-(29*202.74/100) =144.00

cement content =144/.45 = 320 kg

As per table minimum cement content 300 kg < 320 kg

hence ok.

For cementious meterial 10 % is added = 320*1.10 =352.00

the water ration =144/352 =.41

For fly ash content of 30% =352*30/100 =105.6

cement content =320-105.6 = 214.4 kg

4. Volume of Fine and Course Aggregate

As per IS code 10262-2000 table 3

for 20 mm = lies in Zone 2 =62

As per code the aggregate content =.62 where ever water cement ratio =.50

so the increase in course aggregare to decrease in Fine aggregate

in this process the volume of course aggregate to be increased for ever 5% .01 to be added to volume of course aggregate that is = .62+(.003) =.65

for pumpable concrete 10% value will be reduced

so that volume of course aggregate =.62*.9=.558

And Volume of Fine aggregate =1-.558 =.442

• Volume of cement = Mass of cement /specific gravity of cement *1000 = 214.4/3.15*1000=0.068
• volume of water =Mass of water /water specific gravity *1000=144/1*1000=.144
• volume of fly ash = mass of fly ash /specific gravity of fly ash *1000 = 105.4/2.2*1000 = .048
• volume of admixture = mass of admixture /specific gravity of admixture *1000 = 7.6/1.145*1000 =0.006
• Volume of aggregate = 1 -( volume of water+volume of fly ash + volume of admixture + volume of cement) =1-(.116+.144+.006)=.734

• Mass of Course aggregate = Total volume of aggregate *volume of course aggregate *1000*specific gravity of course aggregate =.734*.558.1000*2.74 = 1122.2 kg
• Mass of fine aggregate = total volume of aggregate * volume of course aggregate *specific gravity of fine aggregate *1000= .734*.445*2.74*1000 = 894.9 kg

Results,

Cement content =214.4 kg/m3

fly ash content = 105.6 kg/m3

water content= 144 liters/ m3

fine aggregate = 894.9 kg/m3

course aggregate =1122.2 kg / m3

admixture =7 liters per m3

water cement ratio =.41

Ratio : 1.00: 2.79 :3.50

2. Ans

M50 grade

A. Target compressive strenght

Fck= Fck' + 1.65 S

As per IS 10262-2009 table 1

S= standerd deviation = 5

Fck =50+1.65*5 = 58.52 n/mm2

B. Water cement ratio

As per code IS 456 -2000 the water content sever conditon

=.45

and cement content is 320 kg

as per conditon .45 is can be used as .40

C. Cement content

As per code IS 10262-2009 table 2

for aggregate =20 mm = 186 kg

our slump value is 100 mm

so that =186+(186*6/100)= 197.16 kg

As per condition admixter of super plasters are added which leads to reduce the water content upto 20 % so we can maximized up to 30 %

= 197.16 *7% = 138.1 kg

cement content =138.1/.4 = 345.00 kg

Well minimum cement contet 320 kg < 345.00 kg , Hence OK

D. Fine and course aggregate

As per code IS 10262-2009 table 3

Course aggregate = zone 2 =.62

from above course aggregate is more than water cement ration .4 =.62

so that the volume of aggregate is less the the water for every 10 % , .01 has to be added to aggregate that is =.64

So the volume of aggregate =.64

for pumpable concrete is reducted to 10 % =.64*.9 = .576

volume of fine aggregate =1-.576 = .424

E.

• Volume of cement = mass of cement / specific gravity of cement *1000 = 345 /3.15*1000 =.109
• Volume of water = mass of water / specific gravity of water *1000 = 138.1/1*1000 =.138
• Volume of admixture = Mass of admixture / specific gravity of admixture *1000 = 7.6 /1.145*1000 = .0066
• Volumr of aggregate = 1-( volume of cement + volume of water + volume of admixture) = 1-(.109+.138+.0066) = .746
• Mass of course aggregate = volume of aggregate *percentage of course aggregate * specific gravity of course aggregate *1000 = .746*.576*1000*2.74 = 1177.36
• Mass of fine aggregate = volume of aggregate * percentage of course aggregate * specific gravity of course aggregate *1000 = .746*.424*2.74*1000 = 866.67

Results,

Cement content = 345 Kg/m3

water cement ratio = 0.40

water content = 138.1 liters /m3

Fine aggregate = 866.67 kg/m3

Course aggregate = 1177.36 kg/m3

admixture = 7 litres/m3

Ratio = 1: 2.51 : 3.41