Challenge

Project on Concrete Mix Design for various grades of Concrete

Mix design for M35 grade concrete with fly ash

PROCEDURE :

STEP1 :Stipulations for proportioning

a) Grade designation : M35

b) type of cement : OPC 43 grade conforming to IS 8112

c) maximum nominal size of aggregate : 20mm

d) maximum cement content : 320kg/m^3 (from table 5 of IS 456:2000- exposure condition)

e) maximum water-cement ratio : 0.45 (from table 5 of IS 450:2000-exposure condition)

f) workability :100 mm (slump value for pumpable concrete)

g) exposure condition : severe (for reinforced concrete)

h) method of concrete placing : pumping

i) degree of supervision : good

j) type of aggregate : crushed angular aggregate

k) maximum cement content : 450kg/m^3

l) chemical admixture type : superplasticizer

STEP2 :TEST DATA FOR MATERIAL

a) cement used : OPC 43 grade conforming to IS8112

b) specific gravity of cement : 3.15

c) chemical admixture : superplasticizer conforming to IS 9103

d) specific gravity of:

* COARSE AGGREGATE : 2.74

* FINE AGGREGATE : 2.74

e) water absorption :

* COARSE AGGREGATE : 0.5%

* FINE AGGREGATE : 1%

f) free surface moisture :

* COARSE AGGREGATE :Nil

* FINE AGGREGATE : Nil

g) Fly ash : conforming to IS 3812 part1

h) specific gravity of fly ash : 2.2

i) seive analysis :

coarse aggregate:

IS ANALYSIS OF % OF REMARKS

SEIVE SIZE (MM) COARSE AGG. DIFFERENT FRACTION

I II I II COMBINED

60 40 100

20 100 100 60 40 100

10 0 71.20 0 28.5 28.5 CONFORMING TO TABLE

4.75 9.40 3.7 3.7 2 OF IS 383

2.36 0

FINE AGGREGATE : conforming to grading zone 1 of table 4 IS 383

STEP3 :to find target mean strength for mix proportioning :

f' ck = fck+1.65 s

=35 + 1.65 x 5

= 43.25 N/M^2

where, f'ck = target average compressive strength at 28 days

fck = characteristics compressive strength at 28 days

s = standard deviation

STEP4 : selection of water-cement ratio :

maximum water-cement ratio = 0.45 (for severe condition from table 5 of IS 456-2000)

based on experience, adopted water-cement ratio is 0.40

0.40 < 0.45, hence O.K

STEP5: calculation of water content

maximum water content for 20 mm agg. = 186 litres (table 2 of IS 1026:2009) for 25mm to 50mm slump range

for 100 mm slump =186 +(6/100)*186 =197 litres

as superplasticizer is used, the water content is reduced upto 30%

based on trials water content reduction of 29% is achieved with plasticixzer

hence arrived water content = 197*0.71 =140 litres

STEP6 : calculation of cement content :

water-cement ratio = 0.4

cement content = 140/0.4 = 350kg/m^3

from table 5 of IS 456 minimum cement content for 'severe' exposure condition = 320 kg/m^3

350 kg/m^3 >320 kg/m^3, HENCE O.K

NOW, To proportion a mix containing fly ash the following steps are suggested:

a) the percentage of fly ash needs to be decided based on project requirement and quality of materials

b) in some cases increases in cementitious material can be warranted. increase in cementitious material content and its percentage may be based on experience and trial (this example is with increse of 10% cementitious material content)

cementitious material content = 350 x 1.10 = 385 kg/m^3

water content = 140 kg/m^3

so,water cement ratio = 140/385 = 0.364

fly ash @30% total cementitious material content = 385 x 30% = 115kg/m^3

cement (OPC) = 385 - 115 = 270kg/m^3

saving of cement while using fly ash = 350 - 270 = 80kg/m^3

fly ash being utilized = 115 kg/m^3

STEP 7 :PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATES CONTENT :

From table 3 of IS 10262:2009, volume of coarse agg, corresponding to 20mm size agg.and fine agg. (zone I) for water-cement ratio of 0.5 = 0.6

in the present case the water cement ratio is 0.40

therefore, the volume of coarse agg. needs to be increased to decrease the fine agg. content.

as w/c ratio is lower by 0.10, the proportion of volume of coarse agg. is increased by 0.02 (at the rate of -/+ 0.01 for every +
0.05 change in water-cement ratio)

therefore, corrected volume of coarse agg. for the w/c ratio of 0.40 = 0.62

(note: if coarse agg. is not of angular type, then also the volume of coarse agg. may be required to be increased suitably, based on experience)

for pumpable concrete these values are reduced by 10%

therefore, volume of coares agg. = 0.62*0.9 = 0.56

volume of fine agg. content = 1-0.56 =0.44

STEP 8 : MIX CALCULATIONS:

the mix calculation per unit volume of cocrete shall be as follows:

a) volume of concrete = 1 m^3

b) volume of cement = mass of cement / specific gravity of cement *1/100

= 270/3.15 X 1/1000

= 0.086 M^3

c) volume of FLY ASH = mass of fly ash / specific gravity of fly ash *1/100

=115/2.2 X 1/1000

=0.052 M^3

d) volume of water = mass of water / specific gravity of water *1/100

=140/1 X 1/1000

=0.140 M^3

e) volume of chemical admixture = mass of chemical admixture / specific gravity of admixture *1/100

=7.6/1.145 X 1/1000

=0.006 M^3

f) volume of aggregates (FA & CA) = (a-(b +c + d + e))

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

=0.716 M^3

g) mass of coarse aggregates = f x volume of coarse aggregates x specific gravity of coarse aggregate x 1000

=0.76 x 0.56 x 2.74 x 1000

=1098 kg/M^3

g) mass of fine aggregates = f x volume of fine aggregates x specific gravity of fine aggregate x 1000

=0.716 x 0.44 x 2.74 x 1000

=863 kg/M^3

STEP 9: MIX PROPORTIONS FOR TRIAL NUMBER 1 :

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/m^3

CHEMICAL ADMIXTURE = 7 kg/m^3

WATER-CEMENT RATIO = 0.364

TRIAL MIX RATIO = 1: 2.24: 2.85

STEP 10: The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial. The mix proportion shall be reworked for the actual water content and checked for durability requirements.

STEP 11: Two more trials having variation of 10% of water cement ratio in clause x shall be carried out and a graph between three water cement ratio and corresponding strengthnshall be plotted to work out the mix proportions for the gven target strength for field trials. however, durability shall be met.

Mix design for M50 grade concrete without fly ash

PROCEDURE :

STEP1 :Stipulations for proportioning

a) Grade designation : M50

b) type of cement : OPC 43 grade conforming to IS 8112

c) maximum nominal size of aggregate : 20mm

d) maximum cement content : 320kg/m^3 (from table 5 of IS 456:2000- exposure condition)

e) maximum water-cement ratio : 0.45 (from table 5 of IS 450:2000-exposure condition)

f) workability :100 mm (slump value for pumpable concrete)

g) exposure condition : severe (for reinforced concrete)

h) method of concrete placing : pumping

i) degree of supervision : good

j) type of aggregate : crushed angular aggregate

k) maximum cement content : 450kg/m^3

l) chemical admixture type : superplasticizer

STEP2 :TEST DATA FOR MATERIAL

a) cement used : OPC 43 grade conforming to IS8112

b) specific gravity of cement : 3.15

c) chemical admixture : superplasticizer conforming to IS 9103

d) specific gravity of:

* COARSE AGGREGATE : 2.74

* FINE AGGREGATE : 2.74

e) water absorption :

* COARSE AGGREGATE : 0.5%

* FINE AGGREGATE : 1%

f) free surface moisture :

* COARSE AGGREGATE :Nil

* FINE AGGREGATE : Nil

g) Fly ash : conforming to IS 3812 part1

h) specific gravity of fly ash : 2.2

i) seive analysis :

coarse aggregate:

IS ANALYSIS OF % OF REMARKS

SEIVE SIZE (MM) COARSE AGG. DIFFERENT FRACTION

I II I II COMBINED

60 40 100

20 100 100 60 40 100

10 0 71.20 0 28.5 28.5 CONFORMING TO TABLE

4.75 9.40 3.7 3.7 2 OF IS 383

2.36 0

FINE AGGREGATE : conforming to grading zone 1 of table 4 IS 383

STEP3 :to find target mean strength for mix proportioning :

f' ck = fck+1.65 s

=50 + 1.65 x 5

= 58.25 N/M^2

where, f'ck = target average compressive strength at 28 days

fck = characteristics compressive strength at 28 days

s = standard deviation

STEP4 : selection of water-cement ratio :

maximum water-cement ratio = 0.45 (for severe condition from table 5 of IS 456-2000)

based on experience, adopted water-cement ratio is 0.40

0.40 < 0.45, hence O.K

STEP5: calculation of water content

maximum water content for 20 mm agg. = 186 litres (table 2 of IS 1026:2009) for 25mm to 50mm slump range

for 100 mm slump =186 +(6/100)*186 =197 litres

as superplasticizer is used, the water content is reduced upto 30%

based on trials water content reduction of 29% is achieved with plasticixzer

hence arrived water content = 197*0.71 =140 litres

STEP6 : calculation of cement content :

water-cement ratio = 0.4

cement content = 140/0.4 = 350kg/m^3

from table 5 of IS 456 minimum cement content for 'severe' exposure condition = 320 kg/m^3

350 kg/m^3 >320 kg/m^3, HENCE O.K

STEP 7 :PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATES CONTENT :

From table 3 of IS 10262:2009, volume of coarse agg, corresponding to 20mm size agg.and fine agg. (zone I) for water-cement ratio of 0.5 = 0.6

in the present case the water cement ratio is 0.40

therefore, the volume of coarse agg. needs to be increased to decrease the fine agg. content.

as w/c ratio is lower by 0.10, the proportion of volume of coarse agg. is increased by 0.02 (at the rate of -/+ 0.01 for every +
0.05 change in water-cement ratio)

therefore, corrected volume of coarse agg. for the w/c ratio of 0.40 = 0.62

(note: if coarse agg. is not of angular type, then also the volume of coarse agg. may be required to be increased suitably, based on experience)

for pumpable concrete these values are reduced by 10%

therefore, volume of coares agg. = 0.62*0.9 = 0.56

volume of fine agg. content = 1-0.56 =0.44

STEP 8 : MIX CALCULATIONS:

the mix calculation per unit volume of cocrete shall be as follows:

a) volume of concrete = 1 m^3

b) volume of cement = mass of cement / specific gravity of cement *1/100

= 270/3.15 X 1/1000

= 0.086 M^3

c) volume of FLY ASH = mass of fly ash / specific gravity of fly ash *1/100

=115/2.2 X 1/1000

=0.052 M^3

d) volume of water = mass of water / specific gravity of water *1/100

=140/1 X 1/1000

=0.140 M^3

e) volume of chemical admixture = mass of chemical admixture / specific gravity of admixture *1/100

=7.6/1.145 X 1/1000

=0.006 M^3

f) volume of aggregates (FA & CA) = (a-(b +c + d + e))

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

=0.716 M^3

g) mass of coarse aggregates = f x volume of coarse aggregates x specific gravity of coarse aggregate x 1000

=0.76 x 0.56 x 2.74 x 1000

=1098 kg/M^3

g) mass of fine aggregates = f x volume of fine aggregates x specific gravity of fine aggregate x 1000

=0.716 x 0.44 x 2.74 x 1000

=863 kg/M^3

STEP 9: MIX PROPORTIONS FOR TRIAL NUMBER 1 :

CEMENT = 270 kg/m^3

WATER = 140 kg/m^3

FINE AGGREGATE = 863 kg/m^3

COARSE AGGREGATE = 1098 kg/m^3

CHEMICAL ADMIXTURE = 7 kg/m^3

WATER-CEMENT RATIO = 0.364

TRIAL MIX RATIO = 1: 2.24: 2.85

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