Construction_Technology_Concrete Mix_Design_Project_1

PROJECT

 

1. Calculate the Concrete Mix Design for M35 grade concrete with fly ash

Aim : To determine and calculate mix proportions of concrete mix design for M35 grade concrete with flyash

Stipulations for Proportioning

1. Grade of concrete = M35
2. Let us assume grade of cement is 43 grade OPC conforming to IS 8112
3. Type of mineral admixture is Fly ash conforming to IS 3812 (Part I)
4. Maximum nominal size of aggregates = 20mm
5. Let us assume Exposure condition as severe for reinforced concrete
6. Minimum cement content = 320Kg/m³ (Table 5 of IS 456 :2000 – Exposure Condition)
7. Maximum water cement ratio = 0.45 (Table 5 of IS 456 :2000 – Exposure Condition)
8. Let us assume workability as 75 mm sump
9. Method of concrete placing is pumping
10. Degree of supervision is good
11. Type of aggregates is crushed angular aggregates
12. Maximum Cement content is 450 Kg/m³
13. Chemical admixture is Superplasticizer

Test Data for Materials

1. Cement Used is 43 grade OPC conforming to IS 8112
2. Fly ash conforming to IS 3812 (Part I)
3. Specific gravity of fly ash is 2.2
4. Specific gravity of CA and FA is 2.74
5. Water absorption of CA and FA are 0.5 and 1.0 % respectively
6. Chemical admixture is Superplasticizer conforming to IS 9103
7. Free Surface Moisture in both CA and FA is nil and the aggregates are in saturated surface dry (SSD) condition.
8. Coarse aggregate: Conforming to Table 2 of IS 383.
9. Fine Aggregates: Conforming to grading Zone II of Table 4 of IS 383.

Procedure

Step 1: Calculating target mean strength

f_ck is 28-day compressive strength i.e. foe M35 grade it is 35 N/mm²

Standard deviation (S) for concrete grade of M35 as per Table 1 of IS 10262:2009 is 5N/mm²

From these Target mean strength of concrete grade M50 can be calculated from clause 3.2 of IS 10262:2009

Ft = f_ck + 1.65*S

Ft = 35 + 1.65*5 = 43.25 N/mm²

So, Target Mean Strength for M35 grade of concrete is 43.25 N/mm²

                  

Step 2: Selection of water cement ratio

From table 5 IS 456 2000 we can see that for severe exposure condition and reinforced concrete the minim um cement content is 320Kg/m³, maximum free water cement ratio is 0.45

Based on experience, assume water cement ratio is 0.4

Therefore, 0.4<0.45 Hence ok.

        

Step 3: Water content

The maximum water content is calculated from table-2 of IS 10262:2000. Which will show the maximum water content for maximum nominal size of aggregates.

This table is applicable for slump range 25-50mm. If the slump value is higher than this range for every 25mm increase in slump the water content is increased by 3%.

            

For example, if 20mm maximum nominal size of aggregate is used the water content is 186kg or lts.

Given slump is 75mm i.e. 25 mm higher than standard.

Therefore, the water content is increased by 3%

               =          186 + (6/100) *186 = 191.53 Lts

As superplasticizer is used the water content is reduced by 20% and above. Based on trial and error let us assume 30% reduction using plasticizer.       

               =191.53 *0.70 =135 lts approximately

Step 4: Calculation of cement content

Water cement ratio = 0.4

Cement = 135 / 0.4 = 337.5 Kg/m³

From Table 5 of IS 456 2000 the minimum cement content is 320 Kg/m³

337.5 Kg/m³ > 320 Kg/m³ Hence Ok

Now, to proportion a mix containing fly ash the following steps are suggested:

1. Decide the percentage fly ash to be used based on project requirement and quality of materials
2. In certain situations, increase in cementitious material content may be warranted, the decision on increase in cementitious material content and its percentage may be based on experience and trial.

So, based on experience let us assume percentage increase cementitious material content is 10%.

• Cementitious Material Content = 337.5 + 0.1*337.5 = 371.25 Kg/m³
• Water content = 135
• Corrected Water cement ratio = 371.25/135 = 0.363
• Fly ash @ 30% of total cementitious material content
•                          = 371.25 Kg/m³ X 0.3 = 111.375 Kg/m³
• Cement content = 371.25 Kg/m³ X 0.7 = 259.875 Kg/m³
• Saving of cement while using fly ash = 337.5 -259.875 = 77.625 Kg/m³
• Fly ash used = 111.375 Kg/m³

Step 5: Calculation of Coarse aggregate content

From table 3 of IS 10262 2009, The volume of coarse aggregate corresponding to 20mm maximum nominal size of aggregates and Zone 2 of fine aggregates is 0.62

                      

Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content. Above table is for 0.5 water cement ratio and adapted water cement ratio in this design is 0.4.

As water cement ratio is lower by 0.1 the proportion of volume of coarse aggregates is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

Corrected coarse aggregate content for 0.4 w/c ratio is 0.64

For pumpable concrete this value is decreased by 10%

Therefore, volume of coarse aggregates is = 0.64 * 0.9 = 0.576

Volume of fine aggregate is 1-0.576 = 0.424

Step 6: Mix Calculations

The formula for volume calculations are as follows

a = Total volume of concrete is considered as 1m³

b = Volume of cement can be calculated as

         =  `(Mass of Cement )/(Specific gravity of cement) X 1/1000`

         = `(259.875 )/3.15 X 1/1000`= 0.0825 m³

c = Volume of Fly Ash can be calculated as

         = `(Mass of Fly Ash)/(Specific gravity of Fly Ash) X 1/1000`

         =`(111.375 )/2.2 X 1/1000`= 0.0506 m³

d = Volume of water can be calculated as

         =`(Mass of Water )/(Specific gravity of Water) X 1/1000`

         = `(135 )/1 X 1/1000`= 0.135 m³

e = Volume of Chemical admixture (@ 2.0 percent by mass of cementitious material) can be calculated as

         =  `(Mass of Chemical Admixture)/(Specific gravity of Chemical Admixture) X 1/1000`

         =  `(7 )/1.145 X 1/1000` = 0.006m³

f = Volume of both fine aggregates and coarse aggregates can be calculated as

         = [a – (b + c + d+e)]

         = [1-(0.0825+0.0506+0.135+0.006)]

         = 0.7259

g = Mass of coarse aggregates = e* Volume of CA * Specific gravity of CA * 1000

         = 0.7259*0.576*2.74*1000 = 1,145.64Kg/m³

h = Mass of fine aggregates = e* Volume of FA * Specific gravity of FA * 1000

         = 0.7259*0.424*2.74*1000 = 843.32 Kg/m³

Result

Step 7: Mix Proportions for Mix

      Cement = 259.875 Kg/m³

      Fly Ash = 111.375 Kg/m³

      Total Cementitious material =Cement + Fly Ash = 371.25 Kg/m³

      Water = 138 Kg/m³

      Fine Aggregates= 843.32 Kg/m³

      Coarse Aggregates= 1,145.64Kg/m³

      Chemical Admixture= 7 Kg/m³

      Water Cement ratio = 0.363

Step 8: Trail Mix ratio

The proportion of trail mix is as follows 

          = Cement : Fine Aggregate : Coarse Aggregate

          = 371.25 Kg/m^3: 843.32 Kg/m³: 1,145.64 Kg/m³

          = 1: 2.271: 3.085

Step 9: The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump based on trial, if required. The mix proportions shall be reworked for the actual water content and checked for durability requirements.

Step 10: Two more trials having variation of ±10 percent of water-cement ratio in step 9 shall be carried out and a graph between three water-cement ratios and their corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability requirement shall be met.

 

2. Calculate the Concrete Mix Design for M50 grade concrete without Fly ash

Aim : To determine and calculate mix proportions of concrete mix design for M50 grade concrete without flyash

Stipulations for Proportioning

1. Grade of concrete = M50
2. Let us assume grade of cement is 43 grade OPC conforming to IS 8112
3. Maximum nominal size of aggregates = 20mm
4. Let us assume Exposure condition as very severe for reinforced concrete
5. Minimum cement content = 340Kg/m³ (Table 5 of IS 456 :2000 – Exposure Condition)
6. Maximum water cement ratio = 0.45 (Table 5 of IS 456 :2000 – Exposure Condition)
7. Let us assume workability as 100mm sump
8. Method of concrete placing is pumping
9. Degree of supervision is good
10. Type of aggregates is crushed angular aggregates
11. Maximum Cement content is 450 Kg/m³
12. Chemical admixture is Superplasticizer

Test Data for Materials

1. Cement Used is 43 grade OPC conforming to IS 8112
2. Specific gravity of CA and FA is 2.74
3. Water absorption of CA and FA are 0.5 and 1.0 % respectively
4. Chemical admixture is Superplasticizer conforming to IS 9103
5. Free Surface Moisture in both CA and FA is nil and the aggregates are in saturated surface dry (SSD) condition.
6. Fine Aggregates: Conforming to grading Zone II of Table 4 of IS 383
7. Coarse aggregate: Conforming to Table 2 of IS 383.

Procedure

Step 1: Calculating target mean strength

f_ck is 28-day compressive strength i.e. foe M50 grade it is 50 N/mm²

Standard deviation (S) for concrete grade of M50 as per Table 8 of IS 10262:2000 is 5N/mm²

From these Target mean strength of concrete grade M50 can be calculated as

Ft = f_ck + 1.65*S

Ft = 50 + 1.65*5 = 58.25 N/mm²

So, Target Mean Strength for M50 grade of concrete is 58.25 N/mm²

                

Step 2: Selection of water cement ratio

From table 5 IS 456 2000 we can see that for very severe exposure condition and reinforced concrete the minimum cement content is 340Kg/m³, maximum free water cement ratio is 0.45.

Based on experience, assume water cement ratio is 0.4

Therefore, 0.4<0.45 Hence ok.

         

Step 3: Water content

The maximum water content is calculated from table-2 of IS 10262:2000. Which will show the maximum water content for maximum nominal size of aggregates.

This table is applicable for slump range 25-50mm. If the slump value is higher than this range then for every 25mm increase in slump the water content is increased by 3%.

          

For example, if 20mm maximum nominal size of aggregate is used the water content is 186kg or lts.

Given slump is 100mm i.e. 50 mm higher than standard.

Therefore, the water content is increased by 6%

               =          186 + (6/100) *186 = 197.16 Lts

As superplasticizer is used the water content is reduced by 20% and above. Based on trial and error let us assume 30% reduction using plasticizer.       

=197.16*0.70 =138 lts

Step 4: Calculation of cement content

Water cement ratio = 0.4

Cement = 138 / 0.4 = 345 Kg/m³

From Table 5 of IS 456 2000 the minimum cement content is 320 Kg/m³

345 Kg/m³ > 320 Kg/m³ Hence Ok

Step 5: Calculation of Coarse aggregate content

From table 3 of IS 10262 2009, The volume of coarse aggregate corresponding to    20mm maximum nominal size of aggregates and Zone 2 of fine aggregates is 0.62

         

Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content. Above table is for 0.5 water cement ratio and adapted water cement ratio in this design is 0.4

As water cement ratio is lower by 0.1 the proportion of volume of coarse aggregates is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

• Corrected coarse aggregate content for 0.4 w/c ratio is 0.64
• For pumpable concrete this value is decreased by 10%
• Therefore, volume of coarse aggregates is = 0.64 * 0.9 = 0.576
• Volume of fine aggregate is 1-0.576 = 0.424

Step 6: Mix Calculations

The formula for volume calculations are as follows

a = Total volume of concrete is considered as 1m³

b = Volume of cement can be calculated as

         =  `(Mass of Cement )/(Specific gravity of cement) X 1/1000`

         = `345/3.15 X 1/1000 = 0.1095 m3`

 

c = Volume of water can be calculated as

         =`(Mass of Water )/(Specific gravity of Water) X 1/1000`

         = `(138 )/1 X 1/1000`= 0.138 m³

d = Volume of Chemical admixture (@ 2.0 percent by mass of cementitious material) can be calculated as

         =  `(Mass of Chemical Admixture)/(Specific gravity of Chemical Admixture) X 1/1000`

         =  `(7 )/1.145 X 1/1000` = 0.006m³

e = Volume of both fine aggregates and coarse aggregates can be calculated as

         = [a – (b + c + d)]

         = [1-(0.1095+0.138+0.006)]

         = 0.746

f = Mass of coarse aggregates = e* Volume of CA * Specific gravity of CA * 1000

         = 0.746*0.576*2.74*1000 = 1,177.3 Kg/m³

g = Mass of fine aggregates = e* Volume of FA * Specific gravity of FA * 1000

        = 0.746*0.424*2.74*1000 = 866.67 Kg/m³

Step 7: Mix Quantities

      Cement = 345Kg/m³

      Water = 138 Kg/m³

      Fine Aggregates= 866.67 Kg/m³

      Coarse Aggregates= 1,177.3 Kg/m³

      Chemical Admixture= 7 Kg/m³

      Water Cement ratio = 0.4

Result:

Step 8: Trail Mix ratio

The proportion of trail mix is as follows

Cement: Fine Aggregate: Coarse Aggregate

          = 345Kg/m^3: 866.67 Kg/m³: 1,177.3 Kg/m³

          = 1: 2.512: 3.412

Step 9: The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump based on trial, if required. The mix proportions shall be reworked for the actual water content and checked for durability requirements.

Step 10: Two more trials having variation of ±10 percent of water-cement ratio in step 9 shall be carried out and a graph between three water-cement ratios and their corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability requirement shall be met.