Project on Concrete Mix Design for various grades of Concrete

AIM:
      CALCULATING THE DESIGN MIX FOR M35 CONCRETE WITH FLY ASH AND M50 WITH OUT FLY ASH AS PER IS CODES.
      GIVEN DATA:(ASSUMED)
      1.0PC 43 GRADE CEMENT(IS8112)
      2.20MM SIZE AGGREGATE,ZONE 1 FINE AGGRGATE.
      3.EXTREME ENVIRONMENT CONDION
      4.ADMIXTURE SELECTED AS PER IS3812-PART-1(FLY ASH)
      5.SLUMP VALUE 75MM

M35 DESIGN MIX WITH FLY ASH PROCEDURE:(REQUIRED DATA IS ASSUMED)

STEP:1
          f"ck=FCK+1.65(S)
          f"ck=Target mean strength
          fck =compressive strengt
              S =standed deviation (form IS10262 table-8)
          f"ck=35+1.65x5
                =43.25N/mm^2
STEP:2
          W/C RATIO
          0.45 as per IS456:2000 based on exposure conditions.
WATER CONTENT:
                         186LIT as per table-2 IS10262:2009 as per code for 75mm slump
                         =186+186x3%
                         =186+5.58=191.58lit
                         as superplastrizres are used then the water content can be reduced by 30%
                         =192-192x30%=134.4lit

CEMNT CONTENT: W/C=0.4
                           C=134.4/0.4=336KG/m^3 it is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000
                           as fly ash is to be added it is preffered to add 30% of fly based on trail and error results
                        336x30%=100.8kg
                           cement=235.2kg
                            fly ash=100.8kg
AGGREGATE VOLUME :

                                From table-3 IS10262:2009 it is 0.6 for w/c of 0.5 soo the volume os increased by 0.02 as per the codes
                                volume of coures aggregate=0.62
                                            volume of fine agg =1-0.62=0.38

                               volume of materials per unit volume of concrete:1m^3

                              (a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
                                                            =236/3.15x0.001
                                                            =0.0743m^3
                              (b) volume of fly ash=(mass of fy ash/specific graviy of fly)x1/1000
                                                            =100/2.2x0.001
                                                            =0.045m^3
                              (c) volume of water =135/1x0.001
                                                          =0.135
                              (d) total volume of aggregate =1-(a+b+c)
                                                                        =1-(0.0743+0.045+0.135)
                                                                        =0.7457m^3

                            mass of CA= (dxvolume of CA)X(specific gravity of CA)X1000
                                            =0.7457x0.62x2.74x1000
                                            =1266.7KG
                            mass of FA= (dxvolume of FA)X(specific gravity of FA)X1000
                                            =0.7457x0.38x2.74x1000
                                            =776.42m^3

MIX CONTENT:
                     CEMENT=235.2KG
                     FLY ASH=100.8KG
                            C.A=1266.7KG/M^3
                             F.A=776.42KG/M^3
                       WATER=135KG/M^3

RESULT MIX RATO:
                            1:2.3125:3.7678

M50 DESIGN MIX PROCEDURE:(REQUIRED DATA IS ASSUMED)

STEP:1
           f"ck=FCK+1.65(S)
           f"ck=Target mean strength
            fck=compressive strengt
             S =standed deviation (form IS10262 table-8)
          f"ck=50+1.65x5
                =58.25N/mm^2
STEP:2
           W/C RATIO
           0.45 as per IS456:2000 based on exposure conditions.
WATER CONTENT:
                         186LIT as per table-2 IS10262:2009 as per code for 100mm slump
                         =186+186x6%
                         =186+11.16=197.16lit
                         as superplastrizres are used then the water content can be reduced by 30%
                         =197.1-197.16x30%=137.1lit
CEMNT CONTENT:W/C=0.4
                            C=137.1/0.4=342.7KG/m^3 it is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000

AGGREGATE VOLUME :

                                  From table-3 IS10262:2009 it is 0.6 for w/c of 0.5 soo the volume os increased by 0.02 as per the codes
                                  volume of coures aggregate=0.62
                                              volume of fine agg =1-0.62=0.38

                volume of materials per unit volume of concrete:1m^3

                (a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
                                               =342.7/3.15x0.001
                                               =0.1087m^3

               (b) volume of water =137.1/1x0.001
                                              =0.137
               (c) total volume of aggregate =1-(a+b)
                                                          =1-(0.1087+0.135)
                                                          =0.7563m^3

              mass of CA= (dxvolume of CA)X(specific gravity of CA)X1000
                              =0.7563x0.62x2.74x1000
                              =1284.80KG
              mass of FA= (dxvolume of FA)X(specific gravity of FA)X1000
                              =0.7563x0.38x2.74x1000
                              =787.45m^3

MIX CONTENT:
                      CEMENT=342.7KG
                             C.A=1284.80KG/M^3
                              F.A=787.45KG/M^3
                        WATER=137.1KG/M^3

RESULT MIX RATO:
                           1:2.2977:3.7490