Project on Concrete Mix Design for various grades of Concrete
The Concrete Mix Design for M35 grade concrete
Grade Designation: M35
Exposure: Severe considered
Type of cement: OPC 43 considered
Type of mineral admixture: Fly ash considered (30% of total cementitious material content)
Maximum size of aggregate: 20 mm considered
1. Target mean strength of concrete: ft=fck + 1.65S
Ft = 35+(1.65*5)
ft= 43.25 N/mm2
2. Water cement Ratio : 0.45 Lowest adopted
a. graph = w/c = 0.45
b. graph = w/c = 0.45
c. Exposure= w/c = 0.60
3. Selection of water content: (Nominal Max size of aggregate considered as 20 mm)
From table2 of IS 10262:2009,
The maximum water content for 25 mm to 50 mm slump
= 186 Kg = 186 lit
4. Selection of Cement content:
W = 186 Kg
W/C = 0.45
C = W/0.45
C = 186/0.45
C = 414 kg
Minimum Cement content based on exposure moderate is 320 kg/m3
Considered as Cement content = 415 Kg
Increase in Cemetntious material content as 10%
Cemetnious material content = 415*1.10 = 456 kg
Water content = 186/456=0.400
Fly Ash @ 30 % total cemetious content= 456 *0.30 = 136 kg/m3
cement = 456-136 = 320 Kg/m3
Saving of cement while using fly ash = 415 -320 = 95 kg/m3
Fly ash being utilized = 136 kg/m3
5. Volume of Coarse & Fine Aggregate:
The proportion of Volume of CA & FA can be found in Table-3 of IS 10262:2009
Considered as Zone IV: and the proportion is 0.66 for w/c of 5
But as we considered W/C =0.45
As W/C ratio is decreased by -0.05 so the volume is increased by +0.01
Therefore,
Volume of coarse Aggregate is considered as = 0.66+0.01=0.67
Volume of fine aggregate = 1- 0.67 = 0.33
6. Mix Calculation:
Volume of concrete = 1 m3
volume of cement= (320 /3.15) * (1/1000) = 0.101 m3
volume of fly ash = (136/2.2)*(1/1000)=0.062 m3
Volume of water = 186/1000 = 0.186 m3
Volume of all in Aggregate = (FA & CA)
= 1-(0.101+0.062+0.186)
= 0.651 m3
Mass of coarse aggregate = 0.651*0.67*2.74*1000
=1195 kg/m3
Mass of Fine Aggregate = 0.651 * 0.33*2.74*1000
= 588 kg/m3
Therefore Trail Mix ratio is :
1 : 1.83 : 3.73
M50 GRADE WITHOUT FLY ASH:
Grade Designation: M50
Exposure: Severe considered
Type of cement: OPC 43 considered
Maximum size of aggregate: 20 mm considered
1. Target mean strength of concrete: ft=fck + 1.65S
Ft = 50+(1.65*5)
ft= 58.25 N/mm2
2. Water cement Ratio : 0.34 Lowest adopted
a. graph = w/c = 0.34
b. Exposure= w/c = 0.45
3. Selection of water content: (Nominal Max size of aggregate considered as 20 mm)
From table2 of IS 10262:2009,
The maximum water content for 25 mm to 50 mm slump
= 186 Kg = 186 lit
4. Selection of Cement content:
W = 186 Kg
W/C = 0.34
C = W/0.34
C = 186/0.34
C = 547 kg
Minimum Cement content based on exposure moderate is 360 kg/m3
5. Volume of Coarse & Fine Aggregate:
The proportion of Volume of CA & FA can be found in Table-3 of IS 10262:2009
Considered as Zone IV: and the proportion is 0.66 for w/c of 0.50
But as we considered W/C =0.34
As W/C ratio is decreased by -0.16 so the volume is increased by +0.032
Therefore,
Volume of coarse Aggregate is considered as = 0.66+0.032=0.692
Volume of fine aggregate = 1- 0.692 = 0.308
6. Mix Calculation:
Volume of concrete = 1 m3
volume of cement= (547 /3.15) * (1/1000) = 0.173 m3
Volume of water = 186/1000 = 0.186 m3
Volume of all in Aggregate = (FA & CA)
= 1-(0.173+0.186)
= 0.641 m3
Mass of coarse aggregate = 0.641*0.692*2.74*1000
=1215 kg/m3
Mass of Fine Aggregate = 0.641 * 0.308*2.74*1000
= 540 kg/m3
Therefore Trail Mix ratio is :
1 : 1 : 2.2