Project on Concrete Mix Design for various grades of Concrete

Project on :

Concrete Mix Design for various grades of concrete

 

Aim: concrete mix design for M35 grade concrete with fly ash

Objective:  To make a concrete mix workable for the different structural members of building  by selecting proper proportion of concrete ingredients, through this selection of proper ingredients in the mix  minimize wastage of concrete material, and reduce the damages like, shrinkage, harshness, bleeding, etc. and increase the durability and the strength of concrete.

Design procedure :

Step-1. Stipulation for proportioning concrete ingredients:

• Characteristic compressive strength required in the field at 28 day : M 35
• Type of cement : OPC 43 grade confirming to IS8112
• Type of fly ash : Class F
• Maximum nominal size of aggregate : 20mm
• Shape of coarse aggregate : Angular
• Workability required at site : 100mm(slump)
• Type of exposure the structure will be subjected to (as defined in IS 456:2000) : severe
• Method of concrete placing : pumpable concrete

Step-2. Test data of Material :

The following materials are to be tested in the laboratory and result are to be ascertained for the design mix.

• Cement used : OPC 43 grade confirming to IS8112
• Fly ash : Class F
• Specific gravity of cement : 3.15
• Specific gravity of fly ash : 2.2
• Chemical admixture : Superplasticizer confirming to IS 9103
• Specific gravity :

1. Fine aggregate (sand) : 2.70
2. Coarse aggregate : 2.80

• Water absorption C.A : 0.4%
• Water absorption F.A : 1.0%
• Free surface moisture for C.A : Nil
• Free surface moisture for F.A : Nil

Aggregate assumed to be in saturated surface dry condition usually while preparing design mix

• Sieve analysis

Fine aggregate : Confirming to Zone I Table 4 IS -383

Step-3.  Determining the Target strength for Mix Proportioning :

F’ck =fck + 1.65S

Where,

fck = characteristic compressive strength at 28 days

S = Assumed standard deviation in N/mm2 = 5.0 (as per IS 10262:2009, table 1)

• F’ck = 35 + 1.65 X 5.0 = 43.25 N/mm2

Note : Under control conditions if target average compressive strength is achieved then at field, the probability of getting compressive strength of 35 Mpa is very high

Step-4. Selection of water-cement ratio :

From Table no. 5 of IS 456, Maximum water-cement ratio(w/c) = 0.45

Based on experience, water-cement ratio required for the target mean strength of 43.35Mpa is 0.33 (as per IS 10262:2009, Compressive strength vs w/c ratio graph)

• 33<0.45 Hence ok.

Note : Do not start with w/cm ratio above 0.50 even though the other desired results like strength, workability could be achieved.

Step-5. Selectin of water content :

From Table 2 of IS10262:2009, Maximum water content for 20mm size of aggregate = 186 kg (for 25mm to 50mm slump)

We have 50mm extra slump, so as per IS 10262:2009, we have to increase the water content by 6%

Hence,

• Water content = 186 + (6 X 186/100) = 197.16lit.=197lit.
• As the superplasticizer is used, Water content can be reduced up to 20%.
• Based on trial, water content reduction of 29% is achieved with plasticizers.
• Arrived water content = 197-(29 X 197)/100 = 139.87 = 140lit.

Step-6. Calculation of cement content :

• w/c = 0.33

140/C = 0.33

C = 424.24 = 425 kg/m3

425 kg/m3 > 320 kg/m3

So take, cementitious material (cement + fly ash) =425 kg/m3

Now, the proportion of mix containing fly ash the following step are suggested:

• The % of fly ash needs to be decided based on project requirement and quality of materials.
• In some cases increase in cementitious material can be warranted. Increase in cementitious material content and its percentage may be based on experience and trial (This example is with 10% increase in cementitious materials content.)
• Cementitious material content = 425 + (10 X 425)/100 = 467.5 = 468 kg/m3
• Water content = 140lit.
• w/c ratio =140/468 = 0.299 = 0.30
• flyash at 30% of total cementitious material = 468 X30/100 = 140 kg/m3
• cement OPC = 468 – 140 = 328 kg/m3
• Saving of cement while using fly ash = 425 – 328 = 97 kg/m3
• Fly ash being utilized = 140 kg/m3

Step-7. Proportion of volume of coarse aggregate and fine aggregate content :

From Table 3 of IS 10262:2009.

For Nominal Maximum size of aggregate = 20mm

Volume of coarse aggregate for water-cement ratio 0.50 is 0.60 (corresponding Zone I)

In this case  w/c= 0.4 lower by 0.10. So, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate +- 0.01 for every +-0.05 change in w/c ratio)

Therefore, corrected volume of CA for w/c ratio 0.33 = 0.654

• For pumpable concrete, this value is reduced by 10%

Volume of CA = 0.654-(10 X 0.654)/100 = 0.59

Volume of FA = 1- 0.59 = 0.41

Step-8. Mix calculation :

The mix calculation per unit volume of concrete shall be as follow :

• Volume of concrete = 1m3

 

• Volume of cement = (Mass of cement)/(specific gravity of cement) X 1/1000 = [328/3.15] X[ 1/1000] = 104m3

 

 

• Volume of fly ash = (Mass of flyash)/(specific gravity of flyash) X 1/1000 = (140/2.2) X (1/1000) = 064m3

 

• Volume of Water = (Mass of water)/(specific gravity of water) X 1/1000 = 140/1 X 1/1000 =140m3

 

 

• Volume of chemical admixture = [Mass of chemical admixture]/[specific gravity of chemical admixture]X1/1000 = [7.6/1.145] X[1/1000] = 006m3

 

• Volume of all aggregate{CA+FA}= [volume of concrete-[volume of cement + volume of fly ash + volume of water + volume of chemical admixture] = [1-{0.104 + 0.064 + 0.140 + 0.006}] = 686m3

 

 

• Mass of coarse aggregate = [volume of aggregate X volume of CA X specific gravity of CA X 1000]= [0.686 X 0.59 X 2.80 X 1000] = 1133.27 kg/m3

 

• Mass of fine aggregate = [volume of aggregate X volume of fine aggregate X specific gravity of fine aggregate X 1000] = [0.686 X 0.41 X 2.70 X 1000] = 761.25 kg/m3

Step-9. Mix Proportions :

• Cement = 328kg/m3
• Fly ash = 140kg/m3
• Water = 140kg/m3
• Fine aggregate = 761.25kg/m3
• Coarse aggregate = 1133.27kg/m3
• W/C ratio = 0.30

Mix Ratio:

Mix Ratio = 1:1.63:2.42

 

 

 

 

Q.Calculate the concrete Mix Design for M50 grade concrete without fly ash

Ans :-

 Step-1. Stipulations for proportioning :

• Grade designation : M50
• Type of cement : OPC 43 grade
• Maximum nominal size of aggregate : 20mm
• Maximum water/cement ratio : 0.45…….(From Table 5 of IS 456:2000)
• Workability : 125 mm slump
• Condition of exposure : Moderate
• Method of concrete placing : Pumping
• Type of aggregate : Crushed angular
• Chemical admixture : Superplasticizer

  Step-2. Test Data for Materials :

• Cement used : OPC 43 grade confirming to IS8112
• Specific gravity of cement : 3.15
• Specific gravity of CA = 2.75 and FA = 5.82
• Sieve analysis : Confirming to grading Zone II (Table 2&4 IS383:1970)
• Free surface moisture for CA and FA : Nil
• Water absorption for CA= 0.5% and for FA = 1.0%

Step-3. Calculation of Target strength :

F’ck =fck + 1.65S

Where,

fck = characteristic compressive strength at 28 days

S = Assumed standard deviation in N/mm2 = 5.0 (as per IS 10262:2009, table 1)

• F’ck = 50 + 1.65 X 5.0 = 58.25 N/mm2

Step-4. Selection of W/C ratio :

Based on experience, water-cement ratio required for the target mean strength of 58.25 Mpa is 0.40…….( as per IS 10262:2009, Compressive strength vs w/c ratio graph)

Maximum W/C ratio for the severe exposure condition is= 0.45 (From Table 5 of IS 456:2000)

• 40<0.45. Hence ok.

    Step-5. Selection of water content :

From Table 2 of IS10262:2009, Maximum water content for 20mm size of aggregate = 186 kg (for 25mm to 50mm slump)

We have 75 mm extra slump, so as per IS 10262:2009, we have to increase the water content by 9%

Hence,

• Water content = 186 + [9 X 186/100] = 202.74lit = 203lit.

(As per chemical admixture is not used so, we need not to be reduce the water content)

Step-6. Calculation of cement content :

• w/c = 0.40

203/C = 0.40

C= 507.5 =508 kg

From Table 5 IS 456:2000, minimum cement content for severe condition is 320 kg/m3

508kg/m3>320kg/m3…ok.

 The maximum cement content in concrete for M50 limited to 450 kg/m3 of concrete as per clause 8.2.4.2 of IS 456:2000

So, take cement content =450 kg/m3

Step-7. . Proportion of volume of coarse aggregate and fine aggregate content :

From Table 3 of IS 10262:2009.

For Nominal Maximum size of aggregate = 20mm

 

 

Volume of coarse aggregate for water-cement ratio 0.50 is 0.62 (corresponding Zone II)

In this case  w/c= 0.5 lower by 0.10. So, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate +- 0.01 for every +-0.05 change in w/c ratio)

Therefore, corrected volume of CA for w/c ratio 0.40 = 0.664

• For pumpable concrete, this value is reduced by 10%

Volume of CA = 0.664-(10 X 0.664)/100 = 0.5976 = 0.60

Volume of FA = 1- 0.60 = 0.40

 

Step-8. Mix calculation :

The mix calculation per unit volume of concrete shall be as follow :

• Volume of concrete = 1m3

 

• Volume of cement = (Mass of cement)/(specific gravity of cement) X 1/1000 = [450/3.15] X[ 1/1000] =142 m3

 

• Volume of Water = (Mass of water)/(specific gravity of water) X 1/1000 = 203/1 X 1/1000 = 203m3

 

• Volume of all aggregate{CA+FA}= [volume of concrete-[volume of cement  + volume of water ] = [1-{0.142 + 0.203}] = 655m3

 

 

• Mass of coarse aggregate = [volume of aggregate X volume of CA X specific gravity of CA X 1000]= [0.655 X 0.60X 2.75 X 1000] =1080.75 kg/m3

 

• Mass of fine aggregate = [volume of aggregate X volume of fine aggregate X specific gravity of fine aggregate X 1000] = [0.655 X 0.40 X 2.82 X 1000] =738.84 kg/m3

Step-9. Mix Proportions :

• Cement = 450 kg/m3
• Water = 203 kg/m3
• Fine aggregate = 738.84 kg/m3
• Coarse aggregate = 1080.75 kg/m3
• w/c ratio =0.40

Mix Ratio :

Mix Ratio = 1:1.64:2.40