Project on Concrete Mix Design for various grades of Concrete

* calculate the concrete mix design for M35 grade concrete with flyash and M50 without flyash?

M35 grade concrete with flyash

AIM: - the mix proportioning of M35 grade concrete with flyash.

PROCEDURE:-

I. stipulation for proportioning

1. grade of designation : M35
2. type of cement : OPC M43 conforming IS 269
3. type of mineral mixture : flyash conforming IS 3812
4. maximum nominal size of aggregate : 20mm
5. minimum cement content : very severe
6. maximum water cement ratio : 0.45
7. workability : 100mm
8. exposure : severe
9. method of concrete placing : pumping
10. supervision : good
11. type of aggregate : crushed stone
12. maximum cement content : as per IS 456
13. chemical admixture type : superplacticizer

II. test data for material

1. cement used : OPC M43
2. specific gravity cement : 3.15
3. specific gravity of flyash : 2.2
4. chemical admixture : superplacticizer

III. specific gravity of :

2. coarse aggregate : 2.74
3. fine aggregate : 2.65

IV. water absorption :

1. coarse aggregate : 0.5 %
2. fine aggregate : 0.1%

V. free surface moisture :

1. coarse aggregate : nil
2. fine aggregate : nil

VI. target mean strength of concrete(f’ck)

from table no. 2 , standard deviation , S= 5 N/mm²

f’ck = fck + 1.65 S

= 35 + 1.65 (5)

= 43.25N/mm²

VII. selection of water cement ratio

From graph of curve the W/C ratio required for target strength of 43.25 N/mm²

Is 0.38 for OPC 43 grade concrete curve

From table no. 5 from IS 456:2000 for very severe exposure water cement ratio is 0.45

0.38 < 0.45

Hence OK

VIII. selection of water content

from table no. 2 of IS 10262:2009 the maximum water content for 25mm-50mm

is 186 kg=186 lit

estimated water content slump value for 100mm = 186+6/100 *186 = 197 lit

as the superplacticizer is used, water content can be used up to 30%

Hence water content:

=197*0.7

= 137.96 lit

= 138 lit

XI. calculation of cement content

w/c = 0.45

cement content =138/0.38

= 363.15kg/m³

= 364 kg/m³

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

(a) Decide the % fly ash to be used based on project requirement and quality of flyash.

(b) In certain situation, increase in cementitious material content may be warrented. The decision on increase in cementitious material content and its % may be based on experienced and trial.

Cementitious material content (with increase in 10% of cementitious material)

= 364 x 1.10

= 400.4

= 400 kg/m³

Water content = 138 lit

So, W/C ratio :- = 138/400

= 0.34

Flyash @ 30% of total cementitious content:-

= 400 x 30%

= 120 kg/m³

Cement (OPC) = 400 – 120

= 280

From table no. 5 of IS 456, minimum cementitious content for “very severe” exposure.

= 340 kg/m³

400 > 340 kg/m³

Hence, OK

X. proportion of the volume of coarse aggregate and fine aggregate to 20mm

aggregate and fine aggregate (zone 1) for water cement ratio=0.6

in present case water cement ratio is 0.34

therefore , the volume of coarse aggregate needs to be increased to decrease

the fine aggregate content.

As W/C is lower by 0.16 , the proportion of volume of CA is increased by 0.032

( at the rate+- 0.01 for every change +- 0.05 change in W/C )

Therefore, corrected proportion of volume of CA for cementitious ratio of 0.34

= 0.6+ 0.032

= 0.632

For pumpable concrete this value are reduced by 10%

Volume of coarse aggregate= 0.632x0.9= 0.568 m³

Volume of fine aggregate = 1-0.56= 0.44 m³

XI. mix calculation

a. volume of concrete=1 m³

b. volume of cement = mass of cement/specific gravity of cement x 1/1000

= 280/3.15 x 1/1000

= 0.088 m³

c. Volume of water = mass of water/ specific gravity of water x 1/1000

= 138/1 x 1/1000

= 0.138 m³

d. volume of fly ash = mass of fly ash/ specific gravity of flyash x 1/1000

= 120 / 2.2 x 1/1000

= 0.054 m³

e. Volume of chemical admixture = mass of admixture/specific gravity of add x 1/1000

= 4.00/1.145 x 1/1000

= 0.0034 m³

f.Volume of aggregate= (a-(b+d+c+e))

= (1- ( 0.088+0.054+0.138+0.0034))

= 0.716 m³

g. mass of coarse aggregate = f x volume of coarse aggregate x specific gravity of CA x 1000

= 0.716 x 0.56 x 2.74 x 1000

= 1098.63 kg/m³

h. mass of FA = f x volume of FA x specific gravity of FA 1000

= 0.716 x0.44x2.65x1000

= 834.8 kg/m³

XII. mix proportion

cement = 280 kg/m³

water = 138 kg/m³

Fly ash = 120 kg/m³

fine aggregate = 834.8 kg/m³

coarse aggregate = 1098.63 kg/m³

chemical admixture = 4

water cement ratio = 0.34

Mix ratio = 1: 2.74 : 2.08

RESULT :- the trial mix for M35 grade of concrete with flyash is 1:2.74:2.08

AIM: - the mix proportioning of M50 grade concrete without fly ash.

PROCEDURE:-

I. stipulation for proportioning

1. grade of designation : M50
2. type of cement : OPC M53
3. maximum nominal size of aggregate : 20mm
4. minimum cement content : 320 kg/m³
5. maximum water cement ratio : 0.45
6. workability : 100mm
7. exposure : severe
8. method of concrete placing : pumping
9. supervision : good
10. type of aggregate : crushed stone
11. maximum cement content : 450 kg/m³
12. chemical admixture type : superplacticizer

II. test data for material

1. cement used :OPC M53
2. specific gravity cement : 3.15
3. chemical admixture : superplacticizer

III. specific gravity of :

2. coarse aggregate : 2.74
3. fine aggregate : 2.74

IV. water absorption :

1. coarse aggregate : 0.5 %
2. fine aggregate : 0.1%

V. free surface moisture :

1. coarse aggregate : nil
2. fine aggregate : nil

VI. target mean strength of concrete(f’ck)

f’ck = fck + 1.65 S

= 50 + 1.65 (5)

= 58.25N/mm²

VII. selection of water cement ratio

from graph curve the W/C rtaio required for target strength of 58.25 N/mm²

is 0.34 for OPC 53 grade concrete curve.

From table no. 5 from IS 456:2000 for severe exposure water cement ratio is 0.45

0.34<0.45

Hence, OK

VIII. selection of water content

from table no. 2 of IS 10262:2009 the maximum water content for 25mm-50mm

is 186 kg=186 lit

estimated water content slump value for 100mm = 186+6/100 *186 = 197 lit

as the superplacticizer is used, water content can be used up to 30%

Hence water content:

=197*0.7

= 140 lit

XI. calculation of cement content

w/c = 0.45

cement content =140/0.34

= 411.76kg/m³

= 412 kg/m³

From table 5 of IS 456 minimum cement content for ‘severe’ exposure=320 kg/m³

X. proportion of the volume of coarse aggregate and fine aggregate to 20mm

aggregate and fine aggregate (zone 1) for water cement ratio=0.6

in present case water cement ratio is 0.34

therefore , the volume of coarse aggregate needs to be increased to decrease

the fine aggregate content.

therefore , the volume of coarse aggregate needs to be increased to decrease

the fine aggregate content.

As W/C is lower by 0.16 , the proportion of volume of CA is increased by 0.032

( at the rate+- 0.01 for every change +- 0.05 change in W/C )

Therefore, corrected proportion of volume of CA for cementitious ratio of 0.34

= 0.6+ 0.032

= 0.632

For pumpable concrete this value are reduced by 10%

Volume of coarse aggregate= 0.632x0.9= 0.568 m³

Volume of fine aggregate = 1-0.56= 0.44 m³

XI. mix calculation

a. volume of concrete=1 m³

b. volume of cement = mass of cement/specific gravity of cement x 1/1000

=412 /3.15 x 1/1000

= 0.130 m³

c. Volume of water = mass of water/ specific gravity of water x 1/1000

= 140/1 x 1/1000

= 0.140 m³

d. Volume of chemical admixture = mass of admixture/specific gravity of add x 1/1000

= 8.2/1.145 x 1/1000

= 0.007 m³

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

= ( 1-( 0.130+0.140+0.007)) m³

= 0.723 m³

f. mass of coarse aggregate = e x volume of coarse aggregate x specific gravity of CA x 1000

= 0.723x 0.56 x 2.74 x 1000

= 1109.3 kg/m³

g. mass of FA = ex volume of FA x specific gravity of Fax 1000

= 0.723x0.44x2.74x1000

= 871.6 kg/m³

XII. mix proportion

cement = 412 kg/m³

water = 140 kg/m³

fine aggregate = 871.6 kg/m³

coarse aggregate = 1109.3 kg/m³

chemical admixture = 8.2

water cement ratio = 0.34

Mix ratio = 1: 2.11 : 2.69

RESULT :- the trial mix for M50 grade of concrete without flyash is 1:2.11:2.69