Title Concrete Mix Design 346.3 KB 21
##### Document Text Contents
Page 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 p = percent of flyash

=
( 100 - 20)*210 W = water content

(100 -0.7*20)* 0.58

the fly ash content =

=

Hence the total cementitious material content is 337 + 84.25 = 421.35 kg/m^3

Say, Cement Content = 421 kg/m^3

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

*** 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%

So Water = 211 - 21.1 = 189.9 kg/m^3≈ 190 kg/m^3

Determination of Total Aggregate:
Bulk Specific gravity of 20mm Crushed Aggregate = 2.65

Free Water Content = 211 kg/m^3

By using figure 11.4 , Density of wet Concrete is 2400 kg/m^3

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

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

Maximum size of Aggregate = 20mm

Ratio of Fine Aggregate = 40%

Quantity of Fine Aggregate = 716 kg/ m^3
Amount of Coarse Aggregate = 1789 - 716 = 1073 kg/ m^3

So Final Estimated Quantities:
Cement = 421 kg

Admixure @ 1% of Count weight

FA = 716 kg

CA = 1073 Kg

Water = 190 kg