CALCULATION OF QUANTITIES OF CEMENT, FINE AGGREGATE, & COARSE AGGREGATE IN CONCRETE OF GRADE M20 FOR 1 CUBIC METRE VOLUME

CALCULATION OF CEMENT,FINE AGGREGATE,COARSE AGGREGATE QUANTITIES :-

For calculation of  quantities of cement , fine aggregate and coarse aggregate we need to know some data related to ratio of different grades of concrete and density of cement ,fine aggregate, and coarse aggregate.
                GRADES                                                                 RATIO
                 M5                                                                            1:5:10
                M7.5                                                                          1:4:8
                M10                                                                           1:3:6
                M15                                                                           1:2:4
                M20                                                                           1:1.5:3
               M25                                                                           1:1:2


DENSITY OF CEMENT :-1440 Kg/m^3
DENSITY OF FINE AGGREGATE(SAND):-1450 to 1600 Kg/m^3
DENSITY OF COARSE AGGREGATE :-1450 to 1500 kg/m^3

STEP 1:-first of all find the total volume of concrete for which we have to calculate quantities of its ingredients.we are calculating for 1 cubic metre of volume of concrete.So this 1 cubic metre volume of concrete will be the WET VOLUME OF CONCRETE i.e after mixing all ingredients with water we need 1 cubic metre volume of concrete .But we have the ingredients cement ,fine aggregate , coarse aggregate in dry form so we need to calculate the dry volume of  required concrete.
THE DRY VOLUME = WET VOLUME ✖ 1.54
DRY VOLUME OF CONCRETE =1 ✖ 1.54 =1.54 M^3
Now we have to calculate the quantity of different ingredients in total of 1.54 m^3 volume


Now we need the ratio of grade of concrete ,as we have grade of concrete as M20 for this grade the ratio is 1:1.5:3
            ↙   ↓   ↘
cement   fine    coarse
         aggregate  aggregate
     
FORMULA FOR CALCULATING QUANTITIES:-
QUANTITY= ( RATIO OF INGREDIENT WHOSE QUANTITY IS TO BE FOUND) ➗(  SUM OF RATIO OF ALL INGREDIENTS)✖ 1.54

                        
                               ( WHERE 1.54 IS DRY VOLUME)

STEP 2:-CALCULATION OF QUANTITY OF CEMENT:-

CEMENT

          Quantity of cement=(1/5.5) ✖ 1.54  = 0.28 m^3

where 1= ratio of cement 

          5.5 =sum of all ratios (1+1.5+3=5.5)

          1.54=dry volume 

so quantity of cement we found out as 0.28 m^3 if we want this volume in cubic feet we can get by multiplying meter cube with 35.315 then we can get result in cubic feet i.e 0.28 ✕ 35.315 =9.88 cft.
now we have to calculate mass of cement ,the formula for calculating mass is
 MASS = VOLUME ✕ DENSITY
             =0.28 ✕ 1440 kg/m^3
            =403.2 kg
Now we have to calculate number of bags of cement and 1 bag of cement contain mass of 50 kg so we will now divide total mass by 50 to get number of bags of cement.
No. of bags of cement =403.2 /50
                                   = 8.06 bags
STEP 3:-CALCULATION OF QUANTITY OF FINE AGGREGATE:-

FINE AGGREGATE

    quantity of fine aggregate=(1.5 /5.5)✕1.54  = 0.42 m^3

where 1.5= ratio of fine aggregate

          5.5 =sum of all ratios (1+1.5+3=5.5)

          1.54=dry volume 

so quantity of fine aggregate we found out as 0.42 m^3 if we want this volume in cubic feet we can get by multiplying meter cube with 35.315 then we can get result in cubic feet i.e 0.42✕ 35.315 =14.83 cft.

now we have to calculate mass of fine aggregate,the formula for calculating mass is
 MASS = VOLUME ✕ DENSITY
             =0.42 ✕ 1500 kg/m^3
            =630 kg
STEP 4:-CALCULATION OF QUANTITY OF COARSE AGGREGATE:-

         quantity of coarse aggregate=(3 /5.5)✕1.54  = 0.84 m^3
where 3= ratio of coarse aggregate

          5.5 =sum of all ratios (1+1.5+3=5.5)

          1.54=dry volume 

so quantity of coarse aggregate we found out as 0.84 m^3 if we want this volume in cubic feet we can get by multiplying meter cube with 35.315 then we can get result in cubic feet i.e 0.84✕ 35.315 =29.66 cft.

now we have to calculate mass of fine aggregate,the formula for calculating mass is
 MASS = VOLUME ✕ DENSITY
             =0.84 ✕ 1500 kg/m^3
            =1260 kg
In this way we can calculate mass/volume of different ingredients of concrete.
  we found the quantities as :-
               INGREDIENTS                         VOLUME                    MASS
                 CEMENT                                     0.28 M^3                      403.2 kg
            FINE AGGREGATE                        0.42 M^3                      630 kg
           COARSE AGGREGATE                 0.84 M^3                      1260 kg
 AND TO CHECK THE CALCULATION WE HAVE TO ADD ALL THE INGREDIENTS VOLUME AND THE SUM MUST BE EQUAL TO THE DRY VOLUME OF CONCRETE.
     0.28 +0.42 + 0.84  =1.54 WHICH IS EQUAL TO DRY VOLUME OF CONCRETE SO CALCULATIONS ARE CORRECT.