Project on Concrete Mix Design for various grades of Concrete

LCalculation of mix design for M35 grade concrete with fly ash.

Aim:- calculation of mix design for M35 grade concrete with fly ash.

Procedure

1) Stipulation of Data

• Grade of concrete: M35
• Type of cement - OPC 43 grade conforming to 15 8112.
• Type of mintal admixture fly ash conforming to IS 3512 (part 1).
• Maximum size of aggregate = 20
• Minimum cement content-320 kg /m^3. (From table 5 of I$ 456:2000 exposure condition).
• Maximum water cement ratio=0.45 (From table 5 of 15 456 2000 condition is exposure).
• Workability=100 mm (slump for pumpable concrete).
• Exposure condition=severw for reinforcement concrete.
• Method of concrete placing= pumping
• Degree of supervision - Good
• Maximum cement content = 450 kg /m^3,
• Admixture=superplastisizer

2) Test data for materials

• cement = OPC 43 grade conformning to IS 8112.
• Specific gravity of cement=3.15
• Fly ash conforming to 3812 part 1. Specific gravity of fly ash 2.2 . .
• Admixture = superplastisizer contorming to IS 5103.
• Specific gravity of course aggregate and fine aggregate = 2.74,
• Water absorption
• Course aggregate = 0,5 %
• Fine aggregate = 10%
• Slive analysis course aggregate conforming to table 2 of I$ 393 and fine aggregate conforming to grading zone 1 of the table IS 383

3) To calculate the target mean strongth

• As per IS 10262:2009 the standard deviations = 5 N/MM2
• Fok Fok +1.65 *S
• Fck 35+165x5-4325 N/MMA2
• Target mean strength = 43.25 N/mm^2

4) Selection of water cement ratio

• from table 5 15 456:2000 for severe exosure, the maximum w/c ratio is 0.45.
• Based on experience adopted water cement ratio is 0.40.
• 0.40 < 0.45 hence value is ok.

5) Selection of water content

• from table 2 of 15 10282-2009 the maximum water content of 20 mm aggregate in 180 ...

(For 25 mm to 50 mm slump).

• Additional 3 % slump is increased for per 25 mm slump.
• Estimated water content for 100 mm slump=186-6/100 x 186 = 197 lit
• Due to superplastisizer 30 % water content can be reduced
• Based on trial 29 % achieved with plasticizer
• Hence, arrived water content 1970.71 = 140 liter

6) Calculation of cement content

• w/c ratio = 0.40
• Cementious material (cement fly ash) Cement content (140/04) -350 kg/m^3
• From table 5 of 15 456 2000 minimum cement content for severe exposure condition is 320 kg/m^3

Now,

Preparing the mix containing fly ash the following steps are fallow

• Persentage of fly ash needed to be deside.
• in this design mix we increase the 10 % of cementious material contant
• Cementious material content=350-1.10 =385 kg/m^3.
• Water content = 140 kg/m^3
• So w/c ratio= (140/385) = 0.364.
• Fly ash @ 30 % of cementious material content = 385 30 % = 115 kg/m^3
• Cement OPC-365-115 kg /m^3.
• Saving of cement with using fly ash= 350-270- 80 kg/m^3
• Fly ash being utilised=115 kg/m^3.

7) Proportion of fine and course aggregate

• fine aggregate content from table 3 of IS 10262 2009 volume course aggregate corresponding to 20 mm size of aggregate and fine aggregate (zone 1) from w/c ratio is 0.6
• In present case w/c ratio = 0.4
• Therefore increase the volume of course aggregate and decrease the fine aggregate.
• W/cratio is lower by G.10 the proportion of volume of course aggregate is increasing by 0.02
• Therefore corrected volume of coarse aggregate for the w/c ratio 0.4 = 0.62
• For pumpable concrete this volume is reduced by 10% Therefore
• Volume of coorse aggregate = 0.62 x 0.9 = 0.56
• Volume of fine aggregate 1- 0.56 = 0.44

8) Mix calculation

a) Volume of concrete: 1 m^3)

b) Volume of cement: (mass of cement/ specific gravity of cement) x (1/1000)

(270/315) x (1/1000) = 0.0066 m43

c) Volume of cement: 0.0086 m^3

d) volume of fly ash: (mass of fly ash/ specific gravity of fly ash)*(1/1000)

(115/2.2) (1/1000)-0,052 mA3

Volume of fly ash: 0.052 m^3

e) volume of water: (mass of water/ specific gravity of water) x (1/1000)

(140/1) x (1/1000) 0.140 m^3,

Volume of water = 0.140 m3.

f) volume of chemical admixture:

(mass of chemical admixture/ specific gravity of chemical admixture) x (1/1000)

(7.6/1145) x (1/1000) = 0.006 m^3

Volume of chemical admbeture: 0.006 m 3.

g) volume of aggregate (FA &CA)=

(volume of concrete - (voume of cement + volume of fly ash + volume of water + volume of chemical admixture))

(1-(0.086+0.052+0.140+0.006))= 0.716 mA3

Volume of aggregate=0.716 m^3

h) mass of course aggregate:-

(volume of aggregate xvourne of coarse aggregate specific gravity of coure aggregatex 1000)

I.e 0.716 x 0.56 x 2.74 x 1000 = 1088 kg/m^3

Mass of course aggregate = 1098 kg/m^3

i) mass of fine aggregate:-

(volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x 1000)

0.716x0.44 274-1000-863 kg/m3

Mass of tine aggregate= 863 kg/m^3

9) Mix proportion for trial

• Cement: 270 kg/m^3
• Fly ash 115 kg/m^3
• Water: 140 kg/m^3
• Fine aggregate: 863 kg/m^3
• Coarse aggregate: 1098 kg/m3
• Chemical admixture: 7 kg/m^3
• Water cement ratio: 0.364

10) Results

Trial mix ratio :- 1 : 2.24 : 1.85

2. Calculation of concrete mix design for M50 grade of concrete without fly ash.

1) Stipulation of data

• Type of cement: OPC 43 grade contorming to 15 8112.
• Maximum size of aggregate: 20 mmm
• Minimum cement content: 320 kg/m3 (From table 5 of IS456 2000 exposure condition).
• Maximum w/c ratio 0.45 (From table 5 of iS 456:2000 exposure condition).
• Workobility: 100 (Slump value of pumpable concrete)
• Exposure condition severe for reinforced concrete
• Method of concrete placing pumping Degree of supervision: Good
• Typeof aggregate: crushed angular aggregate
• Maximum cement content: 450 kg/m^3
• Chemical admixture: superplastisizer

2) Test data for materials:

• Cement used OPC 43 grade conforming to 15 8112.
• Specific gravity of cement: 3.15
• Chemical admixture superplastisizer conforming to IS 9103
• Specific gravity of course aggregate and fine aggregate:274 Water absorption
• Course aggregate:0.5 %
• Fine aggregate 10%
• Free super moisture
• Course aggregate Nill
• Fine aggregate: Nill
• Slive analysis: The course aggregate conforming to table 2 of 18 383 and fine aggregate conforming to grading zone 1 of table of IS 383

3) calculation of target mean strength for mix:

• Standard deviations:5 N/mm^2
• Fck-Fox +1.65.5
• Fck = 50+ 1.65×5= 58.25 N/mm/2
• Target mean strength: 58.25 N/mm^2

4) Selection of water cement ratio:

• from table 5 of 5 455 2000 for severe exposure maximum w/c ratio is 0.45
• Based on experience adopted w/c ratio is 0.40
• 0.40 < 0.45 hence ok

5) Slection of Water Content

• From table 2 of IS 10262: 2009 maximum water content for 20
• mm aggregate is 180 lit... (For 25 mm to 50 mm slump range).
• Estimated water content for 100 mm slump = 186+ (6/100)x 186-197 mm
• As per superplastisizer is used to water content can be reduced up to 20 % and above based on trials water content reduced of 29 % is achieved with plasticizer

Hence, the arrived water content: 197% 0.7) = 140 lit

6) Calculation of cement content

• Water cement ratio: 0.4
• Cement content: 140/04-350 kg/m^3
• From table 5 IS 456 minimum cement content for servere exposure condition is 320 kg /m^3

350 kg/m^3 > 320 kg/m^3 Hence ok

7) Proportion of volume of coarse aggregate and fine aggregate content

• from table 3 of IS 10282: 2009 volume of course aggregate corresponding to 20 mm size aggregate and fine aggregate zone I toe w/c ratio is 0.6.
• In present case w/c ratio is 0.40 Therefore the volume of course aggregate needs to increase to decrease the fine aggregate
• As per w/c ratio is lower by the 0.10 and proportion of volume of coarse aggregate is increased by 0.02.
• Therefore corrected volume of coarse aggregate for the water cement ratio of 0.40 -0.62
• For pumpable concrete value is reduced by 10 persent
• Volume of coarse aggregate = 0.62*0.9 = 0.56
• Volume of fine aggregate-1-0.56 = 0.44

8) Mix calculation:

a) volume of concrete:-1 m^3

b) Volume of cement:- (mass of cement /specific gravity of cement Xx (1/1000)

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

Volume of cement is 0.111 m^3

c) volume of water = (mass of water / specific gravity of water ) x (1/1000)

(140/1) (1/1000) = 0.140 m^3

Valume of water is 0.140 m/3

d) volume of chemical admixture:

(mass of chemical admixture/ specific gravity of chemical admixture )* (1/1000)

(7.6/1145) (1/1000) = 0.006m^3

Volume of chemical admixture is 0.006m^3

e) volume of aggregate (Fa& CA):-

(volume of concrete - ( volume of cement+ volume of water + volume of chemical admixture ))

( 1 - ( 0.11 + 0.140 + 0.000 ) )

Volume of aggregate is = 0.743 m/3

f) mass of course aggregate:

= (volume of aggregate x volume of course aggregate x specific gravity of course aggregate x 1000)

0.743 x 0.56 x 2.74 x 1000 = 1140 kg/m^3

Mass of coarse aggregate is 1140 kg/m^3

g) Mass of fine aggregate:

= (volume of aggregate x volume of tine aggregate x specific gravity of fine aggregate x 1000

0.743 x 0.44 274 x 1000 = 896 kg/m^3

Mass of fine aggregate is 896 kg/m^3

h) mix proportion:

• Cement: 350 kg/m^3
• Water: 140 kg/m^3 Fine aggregate: 896 kg/m^3
• Coarse aggregate 1140kg/m^3
• Chemical admixture: 7 kg /m^3
• Water cement ratio: 0.4

10) Result:

Trial mix = 1 : 2.56 : 3.26