Project on Concrete Mix Design for various grades of Concrete
Aim: (i) To calculate Concrete Mix Design for M35 grade concrete with fly ash.
(ii) To calculate Concrete Mix Design for M50 grade concrete without fly ash.
Steps:
(a) Data Collection
(b) Obtain the w/c ratio
(c) Selection of water content
Calculation of cement content
(e) Proportion of volume
(f) Mix Calculations
Procedure: (i) To calculate Concrete Mix Design for M35 grade concrete with fly ash.
(a) Data Collection:
Grade of concrete = M35
Type of Mineral admixture = Flyash
Type of Chemical admixture = Super Plasticizer
Exposure condition = Sever
consider Slump = 75mm
Aggregates = 20mm
- Target mean strength:
fck = 35 N/mm^2
S = 5 N/mm^2
f’ck = fck + 1.65 S
= 35 + (1.65*5)
= 43.25 N/mm^2
(b) Obtain the w/c ratio:
- From the graph of IS 10262:2019,
w/c ratio = 0.38
- Based on the given exposure condition, from table 5 of IS 456,
w/c = 0.45
- Therefore, the lower value is taken, w/c = 0.38
(c) Selection of water content:
- From table 2 of IS 10262,
Max water content = 186 litres
- Considering slump as 75 mm = 186 + (3/100)186
= 191.58 litres
- By adding super plasticizer, there will be 20%-30% reduction in water content.
water content= 191.58 * {(100-30)/100}
= 134.10 litres
- Consider water content = 134.10 litres
(d) Calculation of cement content:
- cement content = Water content/ water-cement ratio
= 134.10/0.38
= 352.89 Kgs
- Cementetious material = cement + flyash
= cement * 10% increase
= 352.89 * 1.1
= 388.18 Kgs
- Flyash = 30% of cementetious material
= 0.3 * 388.18
= 116.45 Kgs
- Therefore, cement = cementetious material - flyash
= 388.18 - 116.45
= 271.73 Kgs
(e) Proportion of volume:
- Volume of CA corresponding to 20mm size aggregate,
Vol of CA = 0.62 [from 10262, clause 5.5]
Vol of CA = {[(0.5 - w/c)/0.05] * 0.01} + 0.62
= {[(0.5 - 0.35)/ 0.05] * 0.01} + 0.62
= 0.65
- If concrete is pumpable concrete, then reduces by 10%
vol of CA = 0.65 * {(100-10)/100}
= 0.585 = 0.59
- vol of FA = 1 - vol of CA
= 1 - 0.59
= 0.41
(f) Mix Calculations:
- Volume of concrete = 1m^3
- Vol of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)
= (271.73/3.15) * 0.001
= 0.086m^3
- Vol of Flyash = (Mass of flyash/ Sp. Gravity of flyash) * (1/1000)
= (116.45/2.2) * 0.001
= 0.053m^3
- Vol of water = (Mass of water/ Sp. Gravity of water) * (1/1000)
= (134.1/1) * (1/1000)
= 0.134m^3
- Vol of Super plasticizer = (Mass of Super Plasticizer/ Sp. Gravity of Super Plasticizer) * (1/1000)
Mass of Super plasticizer= 2% of 388.18
= 7.76
Vol of Super plasticizer = (7.76/1.145) * 0.001
= 0.0067m^3
- Vol of Aggregates = 1- (Vol of cement + Vol of Flyash + Vol of water + Vol of Super plasticizer)
= 1 - (0.086+0.053+0.134+0.0067)
= 0.72m^3
- Mass of CA = Vol of aggregates * Vol of CA * Sp. Gravity of CA *1000
= 0.72*0.59*2.75*1000
= 1168.2 Kg/m^3
- Mass of FA = Vol of aggregates * Vol of FA * Sp. Gravity of CA *1000
= 0.72*0.41*2.75*1000
= 811.8 Kg/m^3
(g) Mix proportions:
Cement = 271.73 Kg = 272 Kg
Flyash = 116.45 Kg = 116.5 Kg
CA = 1168.2 Kg
FA = 811.8 Kg
Water = 134.1 Kg
Mix ratio = 1:4.29:2.98 or 1:4.3:3 (Cement: Coarse aggregate: Fine aggregate)
(ii) To calculate Concrete Mix Design for M50 grade concrete without fly ash.
(a) Data Collection:
Grade of concrete = M50
Exposure condition = Sever
consider Slump = 75mm
Aggregates = 20mm
Zone = Zone 2
- Target mean strength:
fck = 50 N/mm^2
S = 5 N/mm^2
f’ck = fck + 1.65 S
= 50 + (1.65*5)
= 58.25 N/mm^2
(b) Obtain the w/c ratio:
- From the graph of IS 10262:2019,
w/c ratio = 0.34
- Based on the given exposure condition, from table 5 of IS 456,
w/c = 0.45
- Therefore, the lower value is taken, w/c = 0.34
(c) Selection of water content:
- From table 2 of IS 10262,
Max water content = 186 litres
- Considering slump as 75 mm = 186 + (3/100)186
= 191.58 litres
(d) Calculation of cement content:
cement content = Water content/ water-cement ratio
= 191.58/0.34
= 563.47 kg/m^3
(e) Proportion of volume:
- Volume of CA corresponding to 20mm size aggregate,
Vol of CA = 0.62 [from 10262, clause 5.5]
Vol of CA = {[(0.5 - w/c)/0.05] * 0.01} + 0.62
= {[(0.5-0.34)/0.05] * 0.01} + 0.62
= 0.652
- Vol of FA = 1 - Vol of CA
= 1 - 0.652
= 0.348
(f) Mix Calculations:
- Volume of concrete = 1m^3
- Vol of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)
= (563.47/3.15) * 0.001
= 0.178 m^3
- Vol of water = (Mass of water/ Sp. Gravity of water) * (1/1000)
= (191.58/1) * 0.001
= 0.191 m^3
- Vol of Aggregates = 1- (Vol of cement + Vol of water)
= 1 - (0.178+0.191)
= 0.63 m^3
- Mass of CA = Vol of aggregates * Vol of CA * Sp. Gravity of CA *1000
= 0.63*0.65*2.74*1000
= 1122.03 Kg/m^3
- Mass of FA = Vol of aggregates * Vol of FA * Sp. Gravity of CA *1000
= 0.63*0.348*2.74*1000
= 600.71 Kg/m^3
(g) Mix proportions:
Cement = 563.47 Kg/m^3
CA = 1122.03 Kg/m^3
FA = 600.71 Kg/m^3
Water = 191.58 Kg
Mix ratio = 1:2:1.06 (Cement: Coarse aggregate: Fine aggregate)