Challenge

Project on Concrete Mix Design for various grades of Concrete

VVishnu Kumar23 Jun 2023

DESIGN

1. CALCULATE THE CONCRETE MIX DESIGN FOR M35 GRADE CONCRETE WITH FLY ASH:

AIM:

Concrete mix design for M35 grade concrete with fly ash.

PROCEDURE:

Step1: Stipulations for proportioning

Grade of designation : M35
Types of cement : OPC 43 grade conforming to IS 8112
Types of mineral admixture : Fly ash conforming to IS 3812 part-1
Maximum nominal size of aggregate : 20mm
Minimum cement content : Minimum cement content 320 kg/m^3 (Table 5 of IS 456-2000, exposure condition)
Maximum w/c ratio : 0.45 (Table 5 of IS 456-2000, exposure condition
Workability : 100mm slump value of pumpable concrete
Eposure condition : Severe (for reinforced concrete)
Methods of concrete placing : Pumping
Degree of supervision : Good
Types of aggregate : Crushed angular aggregate
Maaximum cement content : 450 kg/m^3
Chemical admixture type : Superplasticizer

Step2: Test data for materials

Cement used : OPC 43 grade conforming to IS 8112
Specific gravity of cement : 3.15
Fly ash : Conforming to IS 3812 part-1
Specific gravity of flyash : 2.2
Chemical admixture : Superplasticizer
Specific gravity of : Coarse aggregate - 2.74, Fine aggregate - 2.74
Water absorption : Coarse aggregate - 0.5%, Fine aggregate - 1.0%
Free surface moisture : Coarse aggregate - Nil, Fine aggregate - Nil
Sieve analysis :

1.Coarse aggregate : Conforming to IS 383

2.Fine aggregate : Conforming to grading Zone 1 of Table-4 IS 383

Step3: To find Target mean strength for proportioning

Target mean strength f'ck = fck + 1.65s

where,

fck - Characteristic compressive strength
s - standard deviation

From the table-1 of IS 10262:2009, where s = 5 N/mm^2

f'ck = 35+(1.65*5) = 43.25 N/mm^2

Target mean strength, f'ck = 43.25 N/mm^2

Step4: Selection of water-cement ratio

From table IS 456:2000 for severe exposure condition,

severe condition, Maximum w/c ratio = 0.45

Based on experience w/c ratio is 0.4

0.40 < 0.45, Hence OK.

Step5: Calculation of Water content

Maximum water content for 20mm aggregate = 186 litres (for 25mm to 50m slump range)

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

A superplasticizer is used the water content reduced 30%

Based on trails water content reduction 29% in with plasticizer

Hence the water content = 197*0.71 = 140 litres

Step6: Calculation of cement content

w/c ratio = 0.4

Cementitious material content = 140/0.4 = 350 kg/m^3

In IS 456:2000, the cement content for severe exposure = 320 kg/m^3

350 kg/m^3 > 320 kg/m^3, Hence OK.

Now proportion of mix containing flyash the following steps:

The percentage of flyash need to be based on project requirement and quality materials
In some cases, Increase cementitious materials and its percentage based on experience and its trial

Increase 10% cementitious material = 350*1.1 = 385 kg/m^3

water content = 140 kg/m^3

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

Flyash at 30% cemetitious material = 385*30% = 115 kg/m^3

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

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

Flyash utilized = 115 kg/m^3

Step7: Proportion of volume of coarse aggregate and fine aggregate content

From IS 10262, Volume of coarse aggregate 20mm size and fine aggregate(zone1) for w/c ratio = 0.6

In Present w/c ratio is 0.4

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

As w/c ratio is lower by 0.1, the volume of coarse aggregate is increased by 0.02

Hence corrected value of coarse aggregate w/c ratio = 0.64

(If coarse aggregate is not angular type, then also volume of coarse aggregate to be increased suitably)

For pumpable concrete these values reduced 10%

Volume of coarse aggregate = 0.64*0.9 = 0.576

Volume of fine aggregate = 1-0.576 = 0.424

Step8: Mix calculation

Mix calculation per unit volume of concrete:

Volume of concrete = 1 m^3
Volume of cement = (Mass of cement/Specific gravity 0f cement)*(1/1000) = (270/3.15)*(1/1000) = 0.086 m^3
Volume of flyash = (Mass of flyash/Specific gravity of flyash)*(1/1000) = (115/2.2)*(1/1000) = 0.052 m^3
Volume of water = (Mass of water/Specific gravity of water)*(1/1000) = (140/1)*(1/1000) = 0.140 m^3
Volume of chemical admixture = (Mass of chemical admixture/Specific gravity of chemical admixture)*(1/1000) = (7.6/1.14)*(1/1000) = 0.006 m^3
Volume of aggregate(Coarse and Fine) = a-(b+c+d+e) = [1-(0.086+0.052+0.140+0.006)] = 0.716 m^3
Mass of coarse aggregate(CA) = f*Volume of CA*Specific gravity of CA*1000 = 0.716*0.576*2.74*1000 = 1130 kg/m^3
Mass of fine aggregate(FA) = f*Volume of FA*Specific gravity of FA*1000 = 0.716*0.424*2.74*1000 = 831 kg/m^3

Step9: Mix proportion

Cement = 270 kg/m^3
Fly ash = 115 kg/m^3
Water = 140 kg/m^3
Coarse aggregate = 1130 kg/m^3
Fine aggregate = 831 kg/m^3
Chemical admixture = 7 kg/m^3;
Water-cement ratio = 0.364

Mix ratio 1:2.15:2.94

Step10:

The slump shall be measured and water content and admixture can be adjusted the slump on trial. The mix proportion shall be reworked for actual water content and check for durability.

Step11:

Two or more trails having variation of 10% w/c ratio in step10 shall be carried out and graph between three w/c ratios and corresponding strength plotted to mix proportion for given target strength for field trials.

RESULTS: The mix design for M35 grade of concrete with flyash is 1:2.15:2.94

2. CALCULATE THE CONCRETE MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLY ASH: