Project on Concrete Mix Design for various grades of Concrete

Calculate the Concrete Mix Design for M35 grade concrete with fly ash.

Aim of the project:- To design a concrete mix of M35 grade with the use of fly ash as a replacement for cement with the help of indian standard codes.

Introduction of the project:-

Data collection:-

Grade designation=M35

Type of cement=43 grade conforming to IS8112

Maximum nominal size of aggregate=20mm

minimum cement content=320kg/m³

Maximum water cement ratio=0.45(from table 5 of IS 456:2000 based on exposure condition, here the adopted exposure condition is severe for RCC)

workability=100mm (for pumpable concrete)

Method of concrete placing=pumping

Type of aggregate=crushed angular aggregate

maximum cement content=320kg/m³

Chemical admixture type= superplasticizer

Test data for materials:-

1. Specific gravity of cement=3.15
2. Fly ash - Conforming to IS 3812(part 1)
3. Specific gravity of fly ash=2.2
4. Chemical admixture:superplasticizer conforming to IS 9103
5. Specific gravity of

chemical admixture=1.145

coarse aggregate=2.74

fine aggregate=2.74

f. water absorption

coarse aggregate=0.5%

fine aggregate=1.0%

g. free surface moisture

coarse aggregate & fine aggregate=nil

Concrete Mix Design Procedure

• Target mean strength of concrete

f’_ck=f_ck+1.65S

From table 1 of IS 10262:2009, for M35 grade of concrete

S= standard deviation=5.0N/mm²

F’_ck=35+1.65*5

=43.25 N/mm²

• Selection of water cement ratio:-

From table 5 of IS 456:2000 for severe exposure adopted for RCC maximum w/c ratio =0.45

but, based on experience adopted w/c ratio is =0.40

and, 0.40<0.45,hence,ok

• Selection of water content :-

From table 2 of IS 10262:2009,maximum water content for 20mm aggregate size=186 lit.(for 20 to 50mm slump range)

but, our target slump is 100mm(code specifies that for each 25mm increment in slump the water content should be increased by 3%)

so, the estimated water content for 100mm slump

=186+(6/100*186)=197lit.

As superplasticizer is used the water content can be reduced upto 30%

Based on trials water content reduction achieved is 29%

so, =197*0.71=140lit.

• Calculation of cement content:-

water cement ratio=0.40

cementitious material=(cement +fly ash)=140/0.4= 350kg/m³

from table 5 of IS 456:2000 minimum cement content for severe exposure condition =320kg/m³

here, 350>320,hence,ok.

now,proportioning a mix containing fly ash

based on the project requirement, experience and trials, fly ash replacement @30% of cementitious material is recommended

fly ash content=0.3*350=105kg.

cement content=350-105=245kg.

As the chemical admixture is also used=mass of chemical admixture=2% of cementitious material

so,350*(2/100)=7kg.

• Proportion of volume of fine aggregate and volume of coarse aggregate content.

from table 3 of IS10262:2009,

volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for w/c ratio of 0.5=0.6

but, in the present case the w/c ratio =0.40

therefore in this case, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content

As the w/c ratio is lower by 0.10 the proportion of volume of coarse aggregate needs to be increased by 0.02(i.e., @-/+0.01 for every +/-0.05 change in w/c ratio)

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

but, for pumpable concrete these values are reduced by 10%, therefore

volume of coarse aggregate=0.62*0.9=0.56

volume of fine aggregate=1-0.56=0.44.

• The mix calculation per unit volume of concrete shall be as follows

1. volume of concrete=1m³
2. volume of cement=(mass of cement/sp.gravity of cement)*(1/1000)=(245/3.15)*(1/1000)=0.077m³
3. volume of fly ash=(mass of fly ash/sp.gravity of fly ash)*(1/1000)=(105/2.2)*(1/1000)=0.042m³
4. volume of water=(mass of water/sp.gravity of water)*(1/1000)=(140/1)*(1/1000)=0.140m³
5. volume of chemical admixture=(mass of chemical admixture/sp.gravity of admixture)*(1/1000)

=(7/1.145)*(1/1000)=0.00611m³

f. Volume of aggregate(F.A&C.A)=(a-(b+c+d+e)) =(1-(0.077+0.042+0.140+0.00611))=0.73489m³

g. mass of coarse aggregate=f.*volume of coarse aggregate*sp.gravity of CA*1000=1128kg/m³

h.mass of fine aggregate=f.*volume of FA*sp.gravity of FA*1000=0.73489*0.44*2.74*1000=886kg./m³

MIX PROPORTION OF THE ABOVE TRIAL

cement=245 kg./m³

fly ash=105 kg./m³

water=140kg/m³

Fine aggregate=886kg./m³

Coarse aggregate=1128kg./m³

Chemical admixture=7kg./m³

TRIAL MIX RATIO=1:2.53:3.22

RESULT:-

The mix ratio suggests that for 1 bag of cement that is 50 kg. 126.5kg. of fine aggregate will be used and 161kg. of coarse aggregate shall be used. The water content for the mix is 140lits.

^{Calculate the Concrete Mix Design for M50 grade concrete without Fly ash.}

Aim of the project:- To design a concrete mix of M35 grade with the use of fly ash as a replacement for cement with the help of indian standard codes.

Introduction of the project:-

DATA COLLECTION:-

Grade designation= M50

Type of cement =OPC 43 grade conforming to IS 8112

Maximum nominal size of aggregate=20mm

Minimum cement content=320kg/m³ (from table 5 of IS 456:2000 based on the exposure condition)

Maximum water-cement ratio=0.45 (from table 5 of IS 456:2000 based on the exposure condition)

workability=100mm (slum value for pumpable concrete)

method of concrete placing=pumping

Exposure condition = severe (for RCC structures)

Degree of supervision=good

Type of aggregate=crushed angular aggregate

maximum cement content=450kg./m³

chemical admixture - superplasticizers

TEST DATA FOR MATERIALS:-

Cement used-OPC 43 grade conforming to IS 8112

Specific gravity of cement=3.15

Chemical admixture-superplasticizer conforming to IS 9103

Specific gravity

• coarse aggregate=2.74
• fine aggregate=2.74

water absorption

• coarse aggregate=0.5%
• fine aggregate=1.0%

free moisture content

• coarse aggregate=nil
• fine aggregate=nil

sieve analysis

• coarse aggregate-conforming to table 2 of IS 383
• fine aggregate-conforming to grading zone 1 of table 4 of IS 383

CONCRETE MIX DESIGN PROCEDURE

• TARGET MEAN STRENGTH

f’_ck=f_ck+1.65S

From table 1 of IS 10262:2009, for M50 grade of concrete

S= standard deviation=5.0N/mm²

F’_ck=50+1.65*5

=58.25 N/mm²

• SELECTION OF WATER CONTENT:-

From table 5 of IS 456:2000 for severe exposure codition

maximum water-cement ratio=0.45

but, based on experience, adopted w/c ratio=0.40

0.40<0.45,hence ,ok.

• SELECTION OF WATER CONTENT

From table 2 of IS 10262:2009,maximum water content for 20mm aggregate =186lit.(for 25 to 50mm sdlump range)

but, here workability required is 100mm so, the code has suggested to increase the water content by 3% for every 25mm increment of slump value

Estimated water content for 100mm slump

186+((6/100)*186)=197 lit.

Code hass also suggested that if the plasticizer is used water content can be reduced by 20%

Based on trials water content reduction of 29% is achieved

so,the arrived water content=197*0.71=140 lit.

CALCULATION OF CEMENT CONTENT:-

W/C ratio =0.40

Cement content=140/0.4=350kg/m³

From table 2 of IS 456:2000, minimum cement content suggested for severe condition is =320kg./m³

since, 350>320kg./m³ hence,ok

As the chemical admixture is also used=mass of chemical admixture=2% of cementitious material

so,350*(2/100)=7kg.

PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT

From table 3 of IS 10262:2009,

volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for w/c ratio of 0.5=0.60

but, in the present situation we have,the w/c ratio =0.40

therefore, the code has suggested to increase the volume of coarse aggregate to decrease the volume of fine aggregate content.

As the w/c ratio is lower by 0.10 the proportion of volume of coarse aggregate needs to be increased by 0.02(i.e., @-/+0.01 for every +/-0.05 change in w/c ratio)

so,the corrected volume of coarse aggregate for the w/c ratio of 0.40=0.62

for pumpable concrete, these values are reduced by 10%

therefore,volume of coarse aggregate =0.62*0.9=0.56

volume of fine aggregate=1-0.56=0.44

MIX CALCULATIONS:-

The mix calculation per unit volume of concrete

1. volume of concrete=1m³
2. volume of cement=(mass of cement/sp.gravity of cement)*(1/1000)=(350/3.15)*(1/1000)=0.111m³
3. volume of water=(mass of water/sp.gravity of water)*(1/1000)=(140/1)*(1/1000)=0.140m³
4. volume of chemical admixture=(mass of chemical admixture/sp.gravity of admixture)*(1/1000)

=(7/1.145)*(1/1000)=0.006m³

e. volume of aggregate(FA&CA)=(a-(b+c+d))

=(1-(0.111+0.140+0.006))=0.743m³

f. mass of coarse aggregate=e.*volume of coarse aggregate*sp.gravity of CA*1000=0.743*0.56*2.74*1000

=1140kg./m³

g.mass of fine aggregate=e.*volume of FA*sp.gravity of FA*1000=0.743*0.44*2.74*1000=896kg./m³

MIX PROPORTIONING

cement=350 kg./m³

water=140kg/m³

Fine aggregate=896kg./m³

Coarse aggregate=1140kg./m³

Chemical admixture=7kg./m³

hence, THE ARRIVED MIX RATIO

1:2.56:3.26

RESULT:-

The mix ratio suggests that for 1 bag of cement that is 50 kg. 128kg. of fine aggregate will be used and 163kg. of coarse aggregate shall be used. The water content for the mix is 140lits.