Concrete Mix Design

(1)

Data:-28 MPa

30mm 20mm 2.55 2.65 180 mm

Step

1:-Target Mean Strenth = Specific Charateristics Strenth + Himsworth Constant × Standard deviation

fm= fmin+ks

= 28+1.64×4 34.56MPa Here,

s=

= 4

K=

Step

2:-Determination of Water Content

Ratio:-Water / Cement Ratio = 0.65

Step

3:-180 mm

So Final Water Content : 225 - 15 Determination of Water Content :

more than 20% Flyash. Assuming Fly ash content 20% . Water to be reduced according to the table 11.13 considering slump = 125mm

Mix of Fly Ash in Cement: 20%

By Table 11.2 water content is 225 Kg/m^3

** Reduction of Water:

Since portland composite cement will be used and most of the cement brands uses Considering Slump:

Maximum Size of Aggregate : 20mm (Crushed) Maximum Size of Aggregate:

Exposure Condition:

at the age of 28 days

By Using Table: 11.11 & Figure: 11.3 from concrete technology by M.S. Shetty

1.64 ( where 5% result are allowed to fall below specific design strength)

Clear Cover to be used :

Standard Deviation

225-15 = 210 kg/m^3 Reduction per CBM: 15kg/m^3

Concrete Mix Design

Specific Characteristics Strerngth:

very severe( In Pile)

Specific Graivity of Coarse Aggregate Targeted Slump

Specific Gravity of Fine Aggregate:

(2)

(3)

StrengthStep 4

:-Here, = 301 Kg pC 100 - p 20*337 100 - 20 = 75.25 kg

Step

5:-Step

= 3.76 p = percent of flyash W = water content =

Determination of Cement Content:

Cement Content = ( 100 - p)*w ( 100 -0.7P)* W/C

Refferring to table 9.20, for very severe exposure condition concrete cover 25mm the maximum ( 100 - 20)*210

(100 -0.7*20)* 0.65

the fly ash content =

free water/cemetitious material ratio is 0.50 and minimum cement content is 350 kg/m^3 Hence the total cementitious material content is 301 + 75.25 = 376.25 kg/m^3

Say, Cement Content = 376 kg/m^3 =

Cement = 376 kg

Amount of Coarse Aggregate = 1837 - 735 = 1114 kg/ m^3

So Final Estimated Quantities:

CA = 1114 Kg FA = 742 kg Water = 169 kg

Thus water content = 376*0.50 = 188 kg/m^3

Admixure @ 1% of Count weight 2404.76kg

Quantity of Fine Aggregate = 742 kg/ m^3 Determination of Total Aggregate:

Bulk Specific gravity of 20mm Crushed Aggregate = 2.65

Determination of Fine Aggregate:-Water/ Cement Ratio = 0.50

*** Water also can be reduced by use wate reducing Admixture or high range water reducer Rheobuild 623 is Prescibe to be used and water reducer considered 10%

Maximum size of Aggregate = 20mm Ratio of Fine Aggregate = 40%

Free Water Content = 188 kg/m^3 By using figure 11.4 , Density of wet Concrete is 2401 kg/m^3

So weight of total Aggrgate = 2401 - ( 376+ 169) = 1856kg/m^3

So Water = 188 - 18.8

= 169.2 kg/m^3≈ 169 kg/m^3

(4)

Step 4

:-0.55 189 kg/m^3 Water /Cement Ratio=

Cement Content = 189/0.55 = 358kg/m^3

So, 344 Kg/m^3

According to the durabelity and exposure condition table 9.20, Minimum cement content is = 325 kg/m^3

Maximum Content is Accepted

Determination of Cement Content: Total Water =

(5)

(6)

Data:-30 MPa

30mm 20mm 2.55 2.65 150 mm

Step

fm= fmin+ks

=

30+1.64×4

36.56MPa

Here,

s=

= 4

K=

Step

2:-Determination of Water Content

Ratio:-Water / Cement Ratio = 0.62

Step

3:-125mm

So Final Water Content : 225 - 15

Concrete Mix Design

Specific Characteristics Strerngth:

at the age of 28 days

Exposure Condition:

Severe( In Pier)

Clear Cover to be used : Maximum Size of Aggregate: Specific Gravity of Fine Aggregate: Specific Graivity of Coarse Aggregate Targeted Slump

Determination of Targated Mean

Strength:-Standard Deviation

1.64 ( where 5% result are allowed to fall below specific design strength)

By Using Table: 11.11 & Figure: 11.3

Determination of Water Content :

Considering Slump:

Maximum Size of Aggregate : 20mm (Crushed)

By Table 11.2 water content is 225 Kg/m^3

** Reduction of Water:

Since portland composite cement will be used and most of the cement brands uses more than 20% Flyash. Assuming Fly ash content 20% . Water to be reduced according to the table 11.13 considering slump = 125mm

Mix of Fly Ash in Cement: 20% Reduction per CBM: 15kg/m^3

(7)

(8)

Step 4

:-Here, = 315 Kg pC 100 - p 20*315 100 - 20 = 78.75 kg

Step

5:-Step

6:-Determination of Cement Content:

Cement Content =

**( 100 - p)*w**

**( 100 -0.7P)* W/C**

p = percent of flyash =

**( 100 - 20)*210**

W = water content

(100 -0.7*20)* 0.62 the fly ash content =

=

Hence the total cementitious material content is 315 + 78.75 = 395.75 kg/m^3 Say, Cement Content = 394 kg/m^3

Refferring to table 9.20, for Severe exposure condition concrete cover 30mm the maximum free water/cemetitious material ratio is 0.50 and minimum cement content is 350 kg/m^3

So Water = 197 - 19.7

= 177.3 kg/m^3≈ 177 kg/m^3

Determination of Total Aggregate:

Bulk Specific gravity of 20mm Crushed Aggregate = 2.65 Free Water Content = 177 kg/m^3 By using figure 11.4 , Density of wet Concrete is 2398 kg/m^3

So weight of total Aggrgate = 2398- ( 394+ 177) = 1827 kg/m^3

Determination of Fine

Aggregate:-Water/ Cement Ratio = 0.50

Quantity of Fine Aggregate = 731 kg/ m^3

So Final Estimated Quantities:

Cement = 394 kg

Admixure @ 1% of Count weight FA = 731 kg

CA = 1096 Kg Water = 177 kg

(9)

= 3.94 2401.94 kg

(10)

Data:-32 MPa

30mm 20mm 2.55 2.65 125 mm

Step

fm= fmin+ks

=

32+1.64×4

38.56MPa

Here,

s=

= 4

K=

Step

2:-Determination of Water Content

Ratio:-Water / Cement Ratio = 0.61

Step

3:-125mm

Concrete Mix Design

Specific Characteristics Strerngth:

at the age of 28 days

Exposure Condition:

Severe( In Pier)

Determination of Targated Mean

Strength:-Standard Deviation

1.64 ( where 5% result are allowed to fall below specific design strength)

Determination of Water Content :

Considering Slump:

By Table 11.2 water content is 225 Kg/m^3

** Reduction of Water:

(11)

(12)

Step 4

:-Here, = 320 Kg pC 100 - p 20*320 100 - 20 = 80 kg

Step

5:-Step

6:-Determination of Cement Content:

Cement Content =

**( 100 - p)*w**

**( 100 -0.7P)* W/C**

**( 100 - 20)*210**

W = water content

(100 -0.7*20)* 0.61 the fly ash ontent =

=

Hence the total cementitious material content is 320 + 80 = 400 kg/m^3 So, Cement Content = 400 kg/m^3

Refferring to table 9.20, for Severe exposure condition concrete cover 30mm the maximum free water/cemetitious material ratio is 0.50 and minimum cement content is 350 kg/m^3

So Water = 200 - 30

= 170 kg/m^3

Determination of Total Aggregate:

So weight of total Aggrgate = 2390- ( 400+ 170) = 1820 kg/m^3

Determination of Fine

Quantity of Fine Aggregate = 728 kg/ m^3

So Final Estimated Quantities:

Cement = 400 kg

CA = 1092 Kg Water = 170 kg

(13)

= 4.00 2394 kg

(14)

Data:-35 MPa

30mm 20mm 2.55 2.65 125 mm

Step

fm= fmin+ks

=

35+1.64×4

41.56MPa

Here,

s=

= 4

K=

Step

2:-Determination of Water Content

Ratio:-Water / Cement Ratio = 0.58

Step

3:-125mm

Concrete Mix Design

Specific Characteristics Strerngth:

at the age of 28 days

Exposure Condition:

Moderate( In Deck)

Determination of Targated Mean

Strength:-Standard Deviation

1.64 ( where 5% result are allowed to fall below specific design strength)

By Using Table: 11.11 & Figure: 11.3 from concrete technology by M.S. Shetty

Determination of Water Content :

Considering Slump:

By Table 11.2 water content is 225 Kg/m^3

** Reduction of Water:

(15)

(16)

Step 4

:-Here, = 337 Kg pC 100 - p 20*337 100 - 20 = 84.25 kg

Step

5:-Step

6:-Determination of Cement Content:

Cement Content =

**( 100 - p)*w**

**( 100 -0.7P)* W/C**

**( 100 - 20)*210**

W = water content

=

Refferring to table 9.20, for moderate exposure condition concrete cover 25mm the maximum free water/cemetitious material ratio is 0.50 and minimum cement content is 350 kg/m^3

Thus water content = 421*0.50 = 210.5≈ 211 kg/m^3

So Water = 211 - 31.5

= 179.35 kg/m^3≈ 179 kg/m^3

Determination of Total Aggregate:

So weight of total Aggrgate = 2400 - ( 421+ 179) = 1800 kg/m^3

Determination of Fine

Quantity of Fine Aggregate = 720 kg/ m^3

So Final Estimated Quantities:

Cement = 421 kg FA = 720 kg CA = 1080 Kg Water = 179 kg

(17)

= 4.21

Step 4

:-0.55 2404.21 kg

Determination of Cement Content:

Total Water = 189 kg/m^3 Water /Cement Ratio=

Cement Content = 189/0.55 = 358kg/m^3

So, 344 Kg/m^3

According to the durabelity and exposure condition table 9.20, Minimum cement content is

= 325 kg/m^3

(18)

Data:-35 MPa

30mm 20mm 2.55 2.65 150 mm

Step

fm= fmin+ks

=

35+1.64×4

41.56MPa

Here,

s=

= 4

K=

Step

2:-Determination of Water Content

Ratio:-Water / Cement Ratio = 0.58

Step

3:-150mm

Concrete Mix Design

Specific Characteristics Strerngth:

at the age of 28 days

Exposure Condition:

Moderate( In Girder)

Determination of Targated Mean

Strength:-Standard Deviation

1.64 ( where 5% result are allowed to fall below specific design strength)

Determination of Water Content :

Considering Slump:

By Table 11.2 water content is 225 Kg/m^3

** Reduction of Water:

(19)

(20)

Step 4

:-Here, = 337 Kg pC 100 - p 20*337 100 - 20 = 84.25 kg

Step

5:-Step

6:-Determination of Cement Content:

Cement Content =

**( 100 - p)*w**

**( 100 -0.7P)* W/C**

**( 100 - 20)*210**

W = water content

=

Refferring to table 9.20, for moderate exposure condition concrete cover 25mm the maximum free water/cemetitious material ratio is 0.50 and minimum cement content is 350 kg/m^3

Thus water content = 421*0.50 = 210.5≈ 211 kg/m^3

So Water = 211 - 21.1

= 189.9 kg/m^3≈ 190 kg/m^3

Determination of Total Aggregate:

So weight of total Aggrgate = 2400 - ( 421+ 190) = 1789 kg/m^3

Determination of Fine

Quantity of Fine Aggregate = 716 kg/ m^3

So Final Estimated Quantities:

Cement = 421 kg

CA = 1073 Kg Water = 190 kg

(21)

= 4.21 2404.21 kg