Project on Concrete Mix Design for various grades of Concrete

1. To calculate the concrete mix design for M35 grade concrete with fly ash

Grade designation =M35

Type of cement= OPC

Type of mineral =Fly ash admixture

Maximum nominal =20mm size of aggregate

Minimum cement = 320kg/m^3 (table 5 of IS 456:2000) content

Maximum water cement ratio = 0.45 (table 5 of IS 456:2000)

Workablity =100mm

Type of aggregate =crushed angular aggregate

Maximum cement content 450 kg/M^3

Chemical admixture type = super plasticizer

• STEP 2: TEST DATA FOR MATERIALS

Cement used OPC 43 Grade

Specific gravity of 3.15 cement

Specific gravity of 2.2 fly ash

Chemical admixture super plasticizer

Specific gravity of 2.74 (CA)

Specific gravity of 2.74 (FA)

Water absorption of 0.5% (CA)

Water absorption of 1.0% (FA)

Free surface moiture Nil of (CA)

Free surface moiture Nil (FA)

Sieve analysis Conforming table 2 of IS 383 of(CA)

Sieve analysis Conforming to grading zone 2 of table 4 of IS 383 of(FA)

• STEP 3: TARGET STRENGTHFOR MIX PROPORTIONING

f'ck =fck+1.65s

Where

• f'ck = target average compressive strength at 28 days
• fck = characteristics compressive strength at 28 days
• s = standdard deviation

From table 1 of IS 10262:2009,standard deviation,s =5 N/MM^2

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

Target mean strength = 43.25 N/MM^2

• STEP 4: SELECTION OF WATER -CEMENT RATIO

The maximum w/c ratio to design M35 grade concrete can be found from table 5 of IS 456-2000

• Maximum W/C ratio =0.45

Based on experience adopted water cement ratio is 0.44

0.44<0.45, hence ok

Adopted w/c ratio = 0.44

• STEP 5: SELECTION OF WATER CONTENT

From table 2 of IS 10262:2009,maximum 186 liters water content for 20 mm aggregate (25 to 50 mm slump range)

186+(6/100)*186 = 197 litre

197*0.8 = 158 litre

WATER REQUIRED 158 litres

• STEP 6: CALCULATION OF CEMENT AND FLY ASH REQUIRED

Adopted w/c Ratio = 0.44

Total cementitious (cement + fly ash)

Water content/water cement ratio

158/0.44 = 359 kg/M^3

Final cementitious content

359*1.10 395kg

From table 5 of IS 456, Minimum cement content for severe exposure conditions 320 kg /M^3

395>329 kg/M^3 Hence ok

Cementitious material content = 395

New water cement ratio = 158/395 = 0.4

Fly ash content taken as 30% of the total cementitious content = 395*0.3 = 118.5 kg/M^3

Cement required = 395-118 = 277kg/M^3

Cement quantitysaved due to fly ash addition = 350-277 = 73 kg/M^3

Cementitious material,cement and fly ash required is 395 kg/M^3, 277 kg/M^3 and 118 kg/M^3 and the new water-cement ratio is 0.4 respectively

• STEP 7: VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE C0NTENT

From table 3 of (IS 10262:2009) volume of coarse aggregate correspomding to 20 mm size aggregate and FA (zone 2) for water cement ratio of 0.50 is 0.62

In the present case water - cement ratio is 0.44 therfore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is loewer by 0.06. The proportion of volum of coarse aggregate is increased by 0.02( at the rate of -/+ 0.01 for every -/+0.05 change in water-cement ratio)

Therfore, corrected proportion of volume of coarse aggregate for the water -cement ratio of 0.44 0.632

Volume of CA 0.32*0.9 =0.568

Volume of FA 1- 0.568 = 0.431

Volume of fine aggregate and coarse aggregate is 0.431 and 0.568

• STEP 8: MIX CALCULATIONS

The mix calculations as per unit volum of concrete shall be as follows:

a) Volume of concrete 1m^3

b) Volume of cement =(mass of cement)/(specific gravity of cement )*1/1000

(277/3.15)*(1/1000) = 0.087 M^3

c) Volume of fly ash = (mass of fly ash)/(specific gravity of fly ash )*1/1000

(118/2.2)*(1/1000) = 0.053

d) Volume of water = (mass of water )/(specific gravity of water)*1/1000

(158/1)*(1/1000) = 0.158m

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

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

f) Volume ofall in aggregate = (a-(b+c+d+e))

(1-(0.087+0.053+0.158+0.006)) = 0.696 M^3

g) Mass of coarse aggregate = f*volume of CA*specific gravity of CA *1000

( 0.696*0.568*2.74*1000 )= 1077 KG/M^3

h) Mass of fine agggregate = f*volume of FA*specifi gravity of FA*1000

(0.696*0.431*2.74*1000) = 822 kg/M^3

• STEP 9: MIX PROPORTIONS

Cement 277 kg/M^3

Fly ash 118 kg/M^3

Water 158 l/M^3

Fine aggregate 822 kg/M^3

Coarse aggregate 1077 kg/M^3 20mm

Chemical admixture 7.18 kg/M^3

W/C ratio (lowest w/c 0.4 ratio is selected)

MIX PROPORTION BY WEIGHT 1:2.08:2.72

• M50 GRADE CONCRETE WITHOUT FLY ASH

• STEP 1: STIPULATIONS FOR CONCRETE DESIGN MIX

Grade designation M50

Type of cement OPC 53 Grade

Maximum nominal 20mm size of aggregate

Minimum cement 320 kg/M^3 content

Minimum water- 0.45 cement ratio

Workability 100mm

Exposure condition severe (for reinforced concrete)

Method of concrete pumping placing

Degree of supervision good

Type of aggregate crushed Angular aggregate

Maximum cement 450 kg/M^3 content

Chemical admixture super plasticizer type

• STEP 2: TEST DATA FOR MATERIALS

Cement used OPC 53 grade

Specific gravity 3.15 of cement

Chemical admixture Super plasticizer

Specific gravity of 2.74 (CA)

Specific gravity of 2.64 (FA)

Water absorption 0.5% of (CA)

Water absorption 1.0% of (FA)

Free surface moisture Nil of (CA)

Free surface moisture Nil of (FA)

Sieve analysis (CA) conforming to table 2 of IS 383

Sieve analysis (FA) conforming to Grading zone 2 of table 4 of IS 383

• STEP 3: TARGAT STRENGTH FOR MIX PROPORTIONING

f'ck =fck +1.6s

Where

• f'ck = target average compressive strength at 28 days
• fck = characteristics compressive strengthat 28 day
• s = standard deviation

From table 1 of IS 10262:2009, standard deviation, s=5 N/MM^2

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

TARGET MEAN STRENGTH = 43.25 N/MM^2

• STEP 4: SELECTION OF WATER-CEMENT RATIO

The maximum w/c ratio to design M50 grade concrete can be found from table 5 of IS 465 - 2000 for severe exposure. Maximum W/C ratio = 0.45

Based on experience adopted water cement ratio is 0.44

0.44<0.45,hence ok

Adopted W/C ratio = 0.44

• STEP 5: SELECTION OF WATER CONTENT

From table 2 of IS 10626:2009, maximum water content for 20 mm aggregate

186 litre (for 25 to 50 mm slump range )

Clause 4.2 suggests increasing of 3% of water content for every extra 25mm slump from 50mm slump.So a 6% is to be added to it

Estimated water content for 100mm slump

186+(6/100)*186 =197 litre

(Note : If super plasticizer is used, the water content can be reduced up to 20% and above.)

Based on trial with super plasticizer water content reduction of 20% has been achieved,Hence the arrived water content

Water required 158 litres

STEP 6: CALCULATION OF CEMENT REQUIRED

Adopted w/c ratio 0.44

Cement content 158/0.44 = 359 kg/M^33

From table 5 of IS 456, minimum cement content for severe exposure conditions 320kg /M^3

359 kg/M^3 > 320 kg/M^3 Hence ok

CEMENT REQUIRED IS 359 Kg/M^3

• STEP 7: VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT

From table 3 of (IS 10262:2009) volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone 2) for water cement ratio of 0.50 is 0.62

In the present case water-cement ratio is 0.44. Therefore, volume of coarse aggregate is required to be increased to decrease the FA content.As the water-cement ratio is lower by 0.06.The proportion of volum of coarse aggregate is increased by 0.02 (at the rate of -/+0.01 for every -/+0.05 change in water-cement ratio).

Therefore,corrrected proportion of volume of coarse aggregate for the w/c ratio of 0.44 is 0.632

For pumpable concrete these values should be reduced up to 10%. Therefore volume of coarse aggregate

0.632*0.9= 0.568

Volume of FA content 1-0.576 = 0.431

VOLUME OF FINE AGGREGATE AND COARSE AGGREGATE IS 0.431 AND 0.568 REASPECTIVELY

• STEP 8: MIX CALCULATION

The mix calculation per unit volume of concrete shall be s follows:

a) Volume of concrete 1M^3

b) Volume of cemen=(mass of cement)/(specific gravityof cement)*1/1000

(359/3.15)*(1/1000) = 0.114 M^3

c)Volume of water = (mass of water)/(specific gravity of water)*1/1000

(158/1)*(1/1000) = 0.158 M^3

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

(7.6/1.145)*(1/1000) = 0.006M^3

e)Volume of all in aggregate = (a-(b+c+d))

(1-(0.114+0.158+0.006))= 0.722 M^3

f) Mass of CA = e*volume of CA*specific gravity of CA

0.722*0.568*2.74*1000 = 1123kg/M^3

g) Mass of FA = e*volume of FA *specific gravity of FA *1000

0.722*0.413*2.64*1000 = 821kg/M^3

• STEP 9: MIX PROPORTIONS

Cement 359 kg/M^3

Water 158 l/M^3

Fine aggregate 821 kg/M^3

Coarse aggregate 1123 kg/M^3

Chemical admixture 7.18 kg/M^3

Water-cement ratio 0.44

MIX PPROPRTION BY WEIGHT 1:2.28:3.12