Challenge

Project on Concrete Mix Design for various grades of Concrete

RRahul Sharma20 Apr 2023

Aim : Mix design for M 35 grade concrete with fly ash

Procedure:

Step :1 Data required

Grade designation : M35
Type of cement : OPC 43 grade
Type of mineral admixture: Fly ash
Maximum nominal size of aggregates : 20mm
Minimum cement content : 320 kg/m^3
Maximum water-cement ratio : 0.45
Workability :100mm slump value of pumpable concrete
Exposure condition : Severe
Method of concrete placing : Pumping
Type of aggregate: Crushed angular aggregate
Maximum cement content : 450 kg/m^3.
Chemical admixture type : Superplaticizer.

Step :2

Cement used : OPC 43 grade.
Specific gravity of cement : 3.15
Specific gravity of flyash : 2.2
Chemical admixture : Superplaticizer
Specific gravity of :
- Coarse aggregate : 2.74
- Fine aggregate : 2.74
Water absorption :
- Coarse aggregate : 0.5%
- Fine aggregate : 1%
Free surface moisture
- Coarse aggregate : Nil
- Fine aggregate : Nil

Step :3

Target mean strength:

Target mean strength f'ck = fck +1.65s

= 35 +1.65*5 =43.25 N/mm^2.

Step:4

Water-cement ratio:

Maximum water-cement ratio =0.45.

On experience water- cement ratio =0.4

0.4 <0.45, Hence OK.

Step :5

Water content

Maximum water content for 20mm aggregate =186 litres

For 100 mm slump ,= 186+(6/100)*186 =197 litres

As superplaticizer is used, the water content is reduced upto 30%

Based on trials water content reduction of 29% is achieved with plasticizer

Hence arrived water content = 197*0.71 = 140 litres

Step :6

Cement content

Water - cement ratio = 0.4

Cement +fly ash= 140/0.4 =350 Kg/m^3

Increase of 10% cementatious material content is considered.

Cementatious material content = 350*1.1 =385 Kg/m^3

Water content = 140 Kg/mm^3

So, water- cement ratio = 140/385 =0.364

Fly ash at 30% total cementatious material content= 385*30/100=115.5 Kg^m^3

Cement (OPC) =385-115 = 270 kg/m^3

Saving of cement while using flyash = 350-270=80 kg/m^3

Fly ash beign utilized = 115 kg/m^3

Step: 7

Proportion of volume of coarse aggregate and fine aggregate content:

Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone 1 ) for water- cement ratio =0.6

In present case the water - cement ratio is 0.4

Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content .

As the water- cement ratio is lowered by 0.1 ,the proportion of volume of coarse aggregate is increased by 0.02( at the rate of +/- 0.01 for every +/- 0.05 change in water- cement ratio)

Hence the corrected value of coarse aggregate for the water-cemnt ratio = 0.64

For pumpable concrete, these are reduced by 10 %

Volume of coarse aggregate = 0.64*0.9 = 0.576

Volume of fine aggregate = 1-0.576 =0.424

Step :8

Mix calculations:

Mix calculation per unit volume of concrete:

Volume of concrete = 1 m^3.
Volume of cement = (270/3.15 )*1/1000 =0.086 m^3
Volume of flyash= (115/2.2) *1/1000 = 0.05 m^3
Volume of water =(140/1)*1/1000 =0.140 m^3
Volume of chemical admixture =( 7.6/1.14)*1/1000 =0.006 m^3
Volume of aggregate ( Fine and coarse) =1-(0.086+0.052+0.140+0.006) =0.716 m^3
Mass of coarse aggregate =0.716*0.576*2.74*1000 =1130 kg/m^3
Mass of fine aggregate =0.716*0.424*2.74*1000= 831kg/m^3

Step :9

Mix proportion

Cement = 270kg/m^3

Fly ash = 115kg/m^3

Water = 140kg/m^3

Fine aggregate = 831kg/m^3

Coarse aggregate = 1130kg/m^3

Chemical admixture = 7kg/m^3

Water- cement ratio = 0.364

Result : The mix design for M 35 grade of concrete with flyash is 1: 2.15:2.94

Ans 2:

Aim : Mix design for M 50 grade concrete