Project on Concrete Mix Design for various grades of Concrete

A. Concrete Mix Design for M35 grade concrete with fly ash :

Data:

1. Given grade of concrete = M35
2. Exposure condition = Severe
3. Cement grade = M43 (OPC)
4. Minimum Cement content = 320kg/cum <From table - 5 , IS 456:2000 - Exposure condition>
5. Maximum nominal size of aggrgate = 20 mm
6. Workability = 100 mm ( slump value for pumpable concrete)
7. Mathode of concrete placing = pumping
8. Type of aggregate = crushed angular aggregate
9. Chemical admixture type = superplasticizer ( conforming I9130)
10. Mineral admixture type = Fly ash ( conforming Is 3812 Part 1)

Now from test data ,

1. Cement grade = M53 (OPC) grade confirming to IS 8112
2. specific gravity of cement = 3.15
3. specific gravity of coarse aggregate = 2.74
4. specific gravity of fine aggregate = 2.74
5. Water absorption of coarse aggregate = 0.5%
6. Water absorption of fine aggregate = 1.0%
7. Free surface moisture of coarse aggregate = nil ( used in dry condition )
8. Free surface moisture of fine aggregate = nil ( used in dry condition )
9. Mineral admixture = Fly ash ( conforming to IS 3812 Part 1)
10. Specific gravity of fly ash = 2.2
11. Sieve analysis should follow for coarse aggregate as per table 2 of IS383 and for fine aggregate conforming grading zone 1 of table 4 of IS383.

Calculation :-

1. Taret mean strength of concrete, f'ck = fck + 1.65 S < fck = compressive strangth of concrete at 28 days>

= 35+(1.65*5) < S (standard deviation) = 5.0 from table IS 10262:2009, Table 1>

= 43.25 N/sqmm

2. Water Cement ratio :

We have

w/c ratio = 0.35 <From w/c ratio graph>

w/c ratio = 0.45 <From table - 5 , IS 456:2000 - Exposure condition>

But we will take w/c = 0.35 --------- (L) < Lower w/c ratio>

3. Estimated Water content :

For 25mm to 50mm slump value = 186 kg <From table 2 of IS 10262:2009>

So for 100mm slump value =186 + 186*6% <From table 2 of IS 10262:2009, 3% add for every 25mm extra slump value>

= 197.16 kg = 197.16 ltr

As superplasticizer is used, the water content can be reduced up to 20% and above. based on trials water content reduction of 29% is achived with plasticize.

Hence the arrived water content = 197.16 * 0.71= 139.98 = 140 ltr --------- (M)

4. Cement content :

as arrived water content = 140 ltr <From (M) >

and w/c ratio = 0.35 <From (L) >

So cementitious material(cement+fly ash) content =140/0.35 = 400 kg/cum

The percentage of fly ash needs to be decided based on project requirement and quality of materials. But we assume 10% increse in cementitious material content.

Therefor ,

Cementitious content = 400 x 1.10 = 440 kg/cum

Water content = 140 kg/cum

so w/c ratio = 140/440 = 0.318

Fly ash @ 30% of total cementitious material content = 440x 30% = 132 kg/cum --------(P)

Cement (OPC) = 440 - 132 = 308 kg/cum ---------- (Q)

5. Volume of coarse & fine aggregate

Proportion of volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate ( Zone 1) for water cement ratio 0.5 = 0.6

But in the present case The w/c ratio = 0.35 <From (M) >

So Water cement ratio is lower by 0.15

So the proportion of volume of coarse aggregate is incresed by 0.03 <+/- 0.01 for every +/- 0.05 change in w/c ratio>

Now the corrected volume of coarse aggregate for the w/c ratio 0.35 =0.63

For pumpable concrete these value are reduce by 10%

therefore volume of CA = 0.63*0.9 = 0.567 ----------(N)

Volume of FA = 1-0.567 = 0.433 ----------(O)

6. Mix calculation :

a) Volume of concrete = 1 cum

b) Volume of Cement = Mass of cement/(Specific gravity of cement x 1000)

= 308/3.15 x 1000

= 0.098 cum

C) Volume of Fly ash = Mass of fly ash/(Specific gravity of fly ash x 1000)

= 132/2.2x1000

= 0.06cum

d) Volume of water = Mass of water/(Specific gravity of water x 1000)

= 140/1x1000

= 0.14 cum

e) Volume of chemical admixture = Mass of chemical admixture/Specific gravity of chemical admixture x 1000

= 8.8/(1.145 x 1000) <Mass of chemical admixture= 2% of cementitous material/ 440*2%= 8.8>

= 0.007 cum

f) Volume of total aggregate ( FA & CA)= (a-(b+c+d+e))

= (1-(0.098+ 0.06+.14+.007))

= 0.695 cum

f) Mass of coarse aggregate = f x Volume of CA x Specific gravity of CA x 1000

= 0.695*0.567*2.74*1000 <From (N) >

= 1079.74 kg/cum

h) Mass of fine aggregate =f x Volume of FA x Specific gravity of FA x 1000

= 0.695*0.433*2.74*1000 <From (O) >

=824.56 kg/cum

6. Mix proportions of Trial no 1:

Cement = 308 kg/cum

Fly Ash = 132 kg/cum

water = 140 kg/cum

FA = 824 kg/cum

CA = 1079 kg/cum

Chemical admixture = 7 kg/cum

W/c ratio = 0.4

Mix Ratio = Cement : FA : CA

= 308 : 824: 1079

= 1 : 2.67 : 3.5

B. Concrete Mix Design for M50 grade concrete without fly ash :

Data:

1. Given grade of concrete = M50
2. Exposure condition = Severe
3. Cement grade = M53 (OPC)
4. Minimum Cement content = 320kg/cum <From table - 5 , IS 456:2000 - Exposure condition
5. Maximum nominal size of aggrgate = 20 mm
6. Workability = 100 mm ( slump value for pumpable concrete)
7. Mathode of concrete placing = pumping
8. Type of aggregate = crushed angular aggregate
9. Chemical admixture type = superplasticizer

Now from test data ,

1. Cement grade = M53 (OPC) grade confirming to IS 8112
2. specific gravity of cement = 3.15
3. specific gravity of coarse aggregate = 2.74
4. specific gravity of fine aggregate = 2.74
5. Water absorption of coarse aggregate = 0.5%
6. Water absorption of fine aggregate = 1.0%
7. Free surface moisture of coarse aggregate = nil ( used in dry condition )
8. Free surface moisture of fine aggregate = nil ( used in dry condition )
9. Sieve analysis should follow for coarse aggregate as per table 2 of IS383 and for fine aggregate conforming grading zone 1 of table 4 of IS383.

Calculation :-

1. Target mean strength of concrete, f'ck = fck + 1.65 S < fck = compressive strangth of concrete at 28 days>

= 50+(1.65*5) < S (standard deviation) = 5.0 from table IS 10262:2009, Table 1>

= 58.25 N/sqmm

2. Water Cement ratio :

We have

w/c ratio = 0.26 <From w/c ratio graph>

w/c ratio = 0.45 <From table - 5 , IS 456:2000 - Exposure condition>

But we will take w/c = 0.26 --------- (L) < from past experience >

3. Estimated Water content :

For 25mm to 50mm slump value = 186 kg <From table 2 of IS 10262:2009>

So for 100mm slump value =186 + 186*6% <From table 2 of IS 10262:2009, 3% add for every 25mm extra slump value>

= 197.16 kg = 197.16 ltr

As superplasticizer is used, the water content can be reduced up to 20% and above. based on trials water content reduction of 29% is achived with plasticize.

Hence the arrived water content = 197.16 * 0.71= 139.98 = 140 ltr --------- (M)

4. Cement content :

as arrived water content = 140 ltr <From (L) >

and w/c ratio = 0.26 <From (M) >

So cement content =140/0.26= 538 kg/cum

5. Volume of coarse & fine aggregate

Proportion of volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate ( Zone 1) for water cement ratio 0.5 = 0.6

But in the present case The w/c ratio = 0.26 <From (M) >

So Water cement ratio is lower by 0.24

So the proportion of volume of coarse aggregate is incresed by 0.05 <+/- 0.01 for every +/- 0.05 change in w/c ratio>

NOw the corrected volume of coarse aggregate for the w/c ratio 0.4 =0.65

For pumpable concrete these value are reduce by 10%

therefore volume of CA = 0.65*0.9 = 0.585 ----------(N)

Volume of FA = 1-0.585 = 0.415 ----------(O)

6. Mix calculation :

a) Volume of concrete = 1 cum

b) Volume of Cement = Mass of cement/(Specific gravity of cement x 1000)

= 538/3.15 x 1000

= 0.171 cum

c) Volume of water = Mass of water/(Specific gravity of water x 1000)

= 140/1x1000

= 0.14 cum

d) Volume of chemical admixture = Mass of chemical admixture/Specific gravity of chemical admixture x 1000

=10.76/(1.145 x 1000) <Mass of chemical admixture= 2% of cementitous material/ 538*2%= 10.76 >

= 0.009 cum

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

= (1-(0.171+0.140+.009))

= 0.680 cum

f) Mass of coarse aggregate = e x Volume of CA x Specific gravity of CA x 1000

= 0.680*0.585*2.74*1000 <From (N) >

= 1090 kg/cum

g) Mass of fine aggregate = e x Volume of FA x Specific gravity of FA x 1000

= 0.680*0.415*2.74*1000 <From (O) >

= 773 kg/cum

6. Mix proportions of Trial no 1:

Cement = 538 kg/cum

water = 140 kg/cum

FA = 773 kg/cum

CA = 1090 kg/cum

Chemical admixture = 10.76 kg/cum

W/c ratio = 0.26

Mix Ratio = Cement : FA : CA

= 538 : 773: 1090

= 1 : 1.44 : 2.03

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