Project on Concrete Mix Design for various grades of Concrete
1) M35 GADE CONCRETE WITH FLY ASH.
Size of Aggregate = 20 mm
Specific gravity of cement = 3.15
Specific gravity of Coarse and Fine Aggregate = 2.74
Specific gravity of Fly Ash = 2.2
- Target mean strength f'ck = fck + 1.65 * S
fck = characteristic compressive strength at 28 days
S = standard deviation (IS 10262:2019, Table 1)
S = for 35 is 5.0 `N/mm^2`
Target mean strength = 35 + 1.65 * 5
= 43.25 `N/mm^2`
- Selection of water cement ratio (w/c) : From target mean strength by using the graph.
W/C ratio taken from the graph = 0.42, 0.48
- Selection of water cement ratio (w/c) : Table -5 of IS 456:2000
According exposure condition Moderate = 0.50
- Selection of water content : Table - 2 of IS-10262:2009
The maximum water content = 186 kg
- Selection of Cement Content
W = 186 kg
W/C = 0.4
C = W/0.4
= 186/0.4
= 465 kg
- Volume of coarse aggregate and fine aggregate : Table 3 of IS 10262:2009 = 0.60
10% of cementatios material need to be increased
Cementitious material content = 465 x 1.10 = 511.6 `(kg)/m^3`
Water Content = 186 Litre
Water Cement ratio = 186 / 511.6 = 0.36
Fly ash @ 30% total cementitious material = 511.6 x 0.30 = 153.5 `(kg)/m^3`
Cement = 511.6 - 153.5 = 358 `(kg)/m^3`
Cement saved = 465 - 358 = 107 `(kg)/m^3`
Volume of coarse aggregate to be increase to due to change of water content = 0.62
For pumpable concrete the volume of coarse aggregate is reduced by 10%
Volume of coarse aggregate = 0.62 * 0.9 = 0.56
Volume of Fine Aggregate = 1 - 0.56 = 0.44
Mix Calculations:
a) Volume of Concrete = 1 `m^3`
b) Volume of Cement = `((Mass of Cement)/(Specific gravity of cement))*(1/1000)`
= `(358/3.15)*(1/1000)`
= 0.114 `m^3`
c) Volume of Fly ash = `((Mass of fly ash)/(Specific gravity of fly ash))*(1/1000)`
= `(153.5/2.2)*(1/1000)`
= 0.07 `m^3`
d) Volume of water = `((Mass of water)/(Specific gravity of water))*(1/1000)`
= `(186/1)*(1/1000)`
= 0.186 `m^3`
e) Volume of chemical admixture = `((Mass of Chemical Admixture)/(Specific gravity of admixture))*(1/1000)`
= `((2*358/100)/1.45) * (1/1000)`
= 0.005 `m^3`
f) Volume of aggregate (FA & CA) = (a-(b+c+d+e))
= (1-(0.114+0.07+0.186+0.005))
= 0.625 `m^3`
g) Mass of Coarse Aggregate = f * Volume of Coarse aggregate * Specific gravity of coarse aggregate * 1000
= 0.625 * 0.56 * 2.74 * 1000
= 959 `(kg)/m^3`
h) Mass of Fine Aggregate = f * Volume of Fine aggregate * Specific gravity of Fine aggregate * 1000
= 0.625 * 0.44 * 2.74 * 1000
= 753 `(kg)/m^3`
Mix proportions:
Cement = 358 `(kg)/m^3`
fly ash = 153.5 `(kg)/m^3`
Water = 186 `(kg)/m^3`
Fine Aggregate = 753 `(kg)/m^3`
Coarse Aggregate = 959 `(kg)/m^3`
Chemical Aggregate = 7 `(kg)/m^3`
Water-Cement Ratio = 0.36
Trial Mix Ratio = 1 : 2.103: 2.679
2) M50 GADE CONCRETE WITHOUT FLY ASH.
Size of Aggregate = 20 mm
Specific gravity of cement = 3.15
Specific gravity of Coarse and Fine Aggregate = 2.74
- Target mean strength f'ck = fck + 1.65 * S
fck = characteristic compressive strength at 28 days
S = standard deviation (IS 10262:2009, Table 1)
S = for M50 is 5.0 `N/mm^2`
Target mean strength = 50 + 1.65 * 5
= 58.25 `N/mm^2`
- Selection of water cement ratio (w/c) : From target mean strength by using the graph.
W/C ratio taken from the graph = 0.40
- Selection of water cement ratio (w/c) : Table -5 of IS 456:2000
According exposure condition Moderate = 0.50
- Selection of water content : Table - 2 of IS-10262:2009
The maximum water content = 186 kg
- Selection of Cement Content
W = 186 kg
W/C = 0.4
C = W/0.4
= 186/0.4
= 465 kg
- Volume of coarse aggregate and fine aggregate : Table 3 of IS 10262:2009 = 0.60
Mix Calculations:
a) Volume of Concrete = 1 `m^3`
b) Volume of Cement = `((Mass of Cement)/(Specific gravity of cement))*(1/1000)`
= `(465/3.15)*(1/1000)`
= .148`m^3`
c) Volume of water = `((Mass of water)/(Specific gravity of water))*(1/1000)`
= `(186/1)*(1/1000)`
= 0.186 `m^3`
d) Volume of chemical admixture = `((Mass of Chemical Admixture)/(Specific gravity of admixture))*(1/1000)`
= `((2*465/100)/1.45) * (1/1000)`
= 0.006 `m^3`
e) Volume of aggregate (FA & CA) = (a-(b+c+d))
= (1-(0.148+0.186+0.006))
= 0.660 `m^3`
f) Mass of Coarse Aggregate = e * Volume of Coarse aggregate * Specific gravity of coarse aggregate * 1000
= 0.660 * 0.60 * 2.74 * 1000
= 1085 `(kg)/m^3`
g) Mass of Fine Aggregate = e * Volume of Fine aggregate * Specific gravity of Fine aggregate * 1000
= 0.660 * (1-0.60) * 2.74 * 1000
= 723 `(kg)/m^3`
Mix proportions:
Cement = 465 `(kg)/m^3`
Water = 186 `(kg)/m^3`
Fine Aggregate = 723`(kg)/m^3`
Coarse Aggregate = 1085`(kg)/m^3`
Chemical Aggregate = 7 `(kg)/m^3`
Water-Cement Ratio = 0.4
Trial Mix Ratio = 1 : 1.554: 2.333