Project on Concrete Mix Design for various grades of Concrete
1.Concrete Mix Design for M35 grade concrete using fly ash as replacement of cement
1.1 AIM OF THE PROJECT- Calculation of the perfect and proper mix proportion of the ingredients for the concrete for the grade of M35 to achieve the required strength i.e. 35 N/mm^2 with using the fly ash as part of replacement of the OPC with the help of design mix.
1.2 INTRODUCTION- To find the correct and perfect mix proportion of the ingredients for the concrete for the grade M35, here design mix method is used to find the mix proportion and the important of the design mix is discuss is below,
- Quality of ingredient used are rational and its neither overused nor underused.
- It is based on the actual available material to be used in the construction works.
- It is based on the experimental or laboratory trail.
- It gives the assurance of strength.
- Mix design concrete is performance based concrete.
- Admixture can be used rationally, to modify the properties of concrete.
- Slump and strength can be related for strength. Different slump can be used by changing water cement ratio with or without the use of admixture
1.2.1 Concrete mix design guidelines
- IS method: IS 10262-2009 and recommended guidelines for concrete mix design IS 10262-2982, SP- 23 handbook of civil engineering and IS 456: 2000.
- British standard0 BS EN 206-2 and complementary standard BS 8500 Part 1 and 2.
- American standard –ACI 211.
1.3 STIPULATIONS FOR PROPORTIONING
|
(a) Grade designation |
: M35 |
|
(b) Type cement |
: OPC 43 grade conforming to IS 8112 |
|
(c) Type of the mineral admixture |
: Fly ash conforming to IS 3812 (part 1) |
|
(d) Maximum nominal size of aggregate |
: 20mm |
|
(e) Minimum cement content |
: 320 kg/m^3 (From table no5, IS 456:2000- exposure condition) |
|
(f) Maximum water cement ratio |
: 0.45 (From table no5, IS 456:2000- exposure condition) |
|
(g) Workability |
: 100mm |
|
(h) exposure condition |
: Severe (Reinforced concrete) |
|
(i) Method of concrete pumping |
: Pumping |
|
(j) Degree of supervision |
: Good |
|
(k) Type of aggregate |
: Crushed angular aggregate |
|
(l) Maximum cement content |
: 450 kg/m^3 |
|
(m) Chemical admixture type |
: Super-plasticizer |
1.4 TEST DATA FOR MATERIALS
|
(a) Cement used |
: OPC 43 grade conforming to IS 8112 |
|
(b) Specific gravity of the cement |
: 3.15 |
|
(c) Fly ash |
: conforming to IS 3812 (part 1) |
|
(d) Specific gravity of Fly ash |
: 2.2 |
|
(e) Chemical admixture |
: Super-plasticizer conforming to IS 9103 |
|
(f) Specific gravity of
|
:2.74 :2.74 |
|
(g) Water absorption
|
: 0.5% : 1.0% |
|
(h) Free surface moisture
|
: Nil : Nil |
|
(i) Free surface moisture
|
: Conforming to table 7 of IS 383 : Conforming to grading zone II of table of IS 383 |
1.5 STEP 1- TARGET STRENGTH FOR MIX PROPORTIONING
Target mean strength (ft) can be calculated as below,
ft= fck+ 1.67*S
Where
fck= Characteristic compressive strength of concrete at 28 days of the curing period as per given data i.e. fck= 35 N/mm^2
S = Standard deviation from IS 456 and also from IS 10262: 2019, i.e. for M35, S= 5.
ft= 35+ 1.65*5= 43.25 N/mm^2
Hence, the target mean strength= 43.25 N/mm^2
1.6 STEP 2- APPROXIMATE AIR CONTENT
From IS 10262: 2019, Table no 3, the approximate amount of entrapped air to be expected in normal concrete for all nominal size of aggregates has been given, for 20mm maximum nominal size of aggregate is 1.0%.
1.7 STEP 3- SELECTION OF WATER-CEMENT RATIO
From Fig. 1, the free water-cement ratio required for the target strength of 43.25 N/mm2 is 0.39 for OPC 43 grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456.
0.39 < 0.45, Hence OK.
1.8 STEP 4- SELECTION OF WATER CONTENT
From table no 4, of IS 10262: 2019, maximum water content for 20mm aggregate= 186 kg (for 25mm to 50mm slump size range.
Estimated water content for 100mm slump (increase at the rate of 3% per increase in 25mm slump)
=`186+ (6/100) *186`
=197 kg
As the super-plasticizer is used, the water content can be reduced up to 30%, the water content reduction of 27% is achieved with plasticizer.
Hence the arrived water content = 197*0.73= 143.81kg Which is nearly equal to 144 kg.
1.9 STEP 5- CALCULATION OF THE CEMENT CONTENT
Water cement ratio= 0.39
Cementitious material (cement+ fly ash) = 144kg/ 0.39= 369.23 kg/m^3 which is nearly equal to 370 kg/m^3.
From table no 5 of IS 456 minimum cement content for severe exposure condition= 320 kg/m^3.
370 kg/m^3 > 320 kg/m^3, Hence OK.
To do the proportion of a mix containing fly ash the following steps are suggested:
- Decide the percentage fly ash to be used based on project requirement and quality of fly ash.
- 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 trials
Here, let’s take the increase in 10% cementitious material concrete.
|
Cementitious material content |
= 370*1.10 = 407 kg/m^3 |
|
Water content |
= 144 kg/m^3 |
|
So, water cement ratio |
=144/407 = 0.354 |
|
Fly ash @30% of total cementitious material content |
=407* 0.3 =122 kg/m^3 |
|
Cement (OPC) |
= 407-122 = 285 kg/m^3 |
|
Saving of Cement while using fly ash |
=370-285 = 85 kg/m^3 |
|
Fly ash being utilized |
= 122 kg/m^3 |
1.10 STEP 6- PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGRAGETE CONTENT
From Table 5 IS 10262: 2019, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 = 0.62.
In present case the water cement ratio is equal to 0.39,
Therefore, the volume of the coarse aggregate needs to be increased to decrease the fine aggregate content. As the water cement ratio is lower by 0.11, the proportion of the volume of coarse aggregate is increased by 0.022. (at the rate of -/+ 0.01 for every +/- 0.05 change in water cement ratio).
Therefore, the corrected value of volume of the coarse aggregate for the water cement ratio of 0.39= 0.62+0.022 = 0.642
For the pump-able concrete, these values are reduced by 10%.
Hence, the volume of the coarse aggregate = 0.642*0.9 = 0.5778, Which is nearly equal to 0.58.
Volume of the fine aggregate content = 1- 0.58= 0.42.
1.11 STEP- 7 MIX CALCULATIONS
The mix calculation for the unit volume of the concrete shall be as follows,
|
(a) Total volume of the concrete |
=`1 m^3` |
|
(b) Volume of the entrapped air in the wet concrete |
=`0.01 m^3` |
|
(c) Volume of the cement |
= (mass of the cement/ specific gravity of the cement) *(1/1000) =`(285/3.15) * (1/1000)` =`0.0904 m^3` which is nearly equal to `0.091 m^3` |
|
(d) Volume of the fly ash |
= (mass of the fly ash/ specific gravity of fly ash) *(1/1000) = `(122/2.2) * (1/1000)` =`0.0554 m^3` which is nearly equal to `0.056 m^3` |
|
(e) Volume of water |
= (mass of water/ specific gravity of water) *(1/1000) = `(144/1) * (1/1000)` =`0.144 m^3.` |
|
(f) Volume of chemical admixture (super- plasticizer ) (@1% by mass of the cementitious material i.e. 407 kg * 0.01= 4.07kg) |
= (mass of chemical admixture/ specific gravity of admixture) *(1/1000) = `(4.07/1.145) * (1/1000)` =`0.00355 m^3` which is nearly equal to `0.004 m^3.` |
|
(g) volume of all aggregate |
= [(a-b)- (c +d +e +f)] = [(1-0.01) -(0.091+ 0.056+ 0.144+ 0.004)] = 0.695 m^3 |
|
(h) mass of coarse aggregate |
= (g) * volume of coarse aggregate* specific gravity of the coarse aggregate* 1000 = 0.695* 0.58* 2.74* 1000 = 1104.494 kg which is nearly equal to 1105 kg |
|
(I ) mass of fine aggregate |
= (g) * volume of fine aggregate* specific gravity of the fine aggregate* 1000 = 0.695* 0.42* 2.74* 1000 = 799.806 kg which is nearly equal to 800 kg |
1.12 STEP – 8 MIX PROPORTIONS FOR TRIAL NUMBER 1
|
Cement |
= 285 kg/m^3 |
|
Fly ash |
= 122 kg/m^3 |
|
Water |
= 144 kg/m^3 |
|
Fine aggregate |
= 800 kg/m^3 |
|
Coarse aggregate |
= 1105 kg/m^3 |
|
Chemical admixture |
= 4.07 kg/m^3 |
|
Free water cementitious material ratio |
= 0.354 |
|
Trial mix ratio |
= 1:1.96:2.71 |
1.13 RESULT- The appropriate mix proportion of the ingredients for the concrete for the grade of M35 using the fly ash as part of replacement of the OPC with the help of design mix is calculated as 1:1.96:2.71, with the water cementitious material of 0.354.
1.14 EXPLANATION-
This is the trial mix ratio, so two more trial having variation of 10% of the water cement ratio shall be carried out and a graph between three water cement ratios and corresponding strengths shall be plotted to find out the mix proportion for given target men strength for the field trails.
2. Calculate the Concrete Mix Design for M50 grade concrete
2.1 AIM OF THE PROJECT- Calculation of the perfect and proper mix proportion of the ingredients for the concrete for the grade of M50 to achieve the required strength with the help of design mix.
2.1.1 Concrete mix design guidelines
- IS method: IS 10262-2009 and recommended guidelines for concrete mix design IS 10262-2982, SP- 23 handbook of civil engineering and IS 456: 2000.
- British standard0 BS EN 206-2 and complementary standard BS 8500 Part 1 and 2.
- American standard –ACI 211.
2.2 STIPULATIONS FOR PROPORTIONING
|
(a) Grade designation |
: M50 |
|
(b) Type cement |
: OPC 53 grade conforming to IS 8112 |
|
(c) Maximum nominal size of aggregate |
: 20mm |
|
(d) Minimum cement content |
: 320 kg/m^3 (From table no5, IS 456:2000- exposure condition) |
|
(e) Maximum water cement ratio |
: 0.45 (From table no5, IS 456:2000- exposure condition) |
|
(f) Workability |
: 100mm |
|
(g) exposure condition |
: Severe (Reinforced concrete) |
|
(h) Method of concrete pumping |
: Pumping |
|
(i) Degree of supervision |
: Good |
|
(j) Type of aggregate |
: Crushed angular aggregate |
|
(k) Maximum cement content |
: 450 kg/m^3 |
|
(a) Chemical admixture type |
: Super-plasticizer |
2.3 TEST DATA FOR MATERIALS
|
(a) Cement used |
: OPC 53 grade conforming to IS 8112 |
|
(b) Specific gravity of the cement |
: 3.15 |
|
(c) Chemical admixture |
: Super-plasticizer conforming to IS 9103 |
|
(d) Specific gravity of
|
:2.74 :2.74 |
|
(e) Water absorption
|
: 0.5% : 1.0% |
|
(f) Free surface moisture
|
: Nil : Nil |
|
(g) Free surface moisture
|
: Conforming to table 7 of IS 383 : Conforming to grading zone II of table of IS 383 |
2.4 STEP 1- TARGET STRENGTH FOR MIX PROPORTIONING
Target mean strength (ft) can be calculated as below,
ft= fck+ 1.67*S
Where
fck= Characteristic compressive strength of concrete at 28 days of the curing period as per given data i.e. fck= 50 `N/(mm^2)`
S = Standard deviation from IS 456 and also from IS 10262: 2019, i.e. for M35, S= 5.
ft= 50+ 1.65*5= 48.25 `N/(mm^2)`
Hence, the target mean strength= 58.25 `N/(mm^2)`
2.5 STEP 2- APPROXIMATE AIR CONTENT
From IS 10262: 2019, Table no 3, the approximate amount of entrapped air to be expected in normal concrete for all nominal size of aggregates has been given, for 20mm maximum nominal size of aggregate is 1.0%.
2.6 STEP 3- SELECTION OF WATER-CEMENT RATIO
From Fig. 1, the free water-cement ratio required for the target strength of 58.25 N/mm2 is 0.34 for OPC 53 grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456.
0.34 < 0.45, Hence OK.
2.7 STEP 4- SELECTION OF WATER CONTENT
From table no 4, of IS 10262: 2019, maximum water content for 20mm aggregate= 186 kg (for 25mm to 50mm slump size range.
Estimated water content for 100mm slump (increase at the rate of 3% per increase in 25mm slump)
=`186+ (6/100) *186`
=197 kg
As the super-plasticizer is used, the water content can be reduced up to 30%, the water content reduction of 27% is achieved with plasticizer.
Hence the arrived water content = 197*0.73= 143.81kg Which is nearly equal to 144 kg.
2.8 STEP 5- CALCULATION OF THE CEMENT CONTENT
Water cement ratio= 0.34
Cement content = `(144kg)/ 0.34`= 423.59 `(kg)/m^3` which is nearly equal to 424 `(kg)/m^3`.
From table no 5 of IS 456 minimum cement content for severe exposure condition= 320 `(kg)/m^3`
370 `(kg)/m^3`> 320`(kg)/m^3`, Hence OK.
2.9 STEP 6- PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGRAGETE CONTENT
From Table 5, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 = 0.62
In present case the water cement ratio is equal to 0.34,
Therefore, the volume of the coarse aggregate needs to be increased to decrease the fine aggregate content. As the water cement ratio is lower by 0.16, the proportion of the volume of coarse aggregate is increased by 0.032. (at the rate of -/+ 0.01 for every +/- 0.05 change in water cement ratio).
Therefore, the corrected value of volume of the coarse aggregate for the water cement ratio of 0.34= 0.62+0.032 = 0.652
For the pump-able concrete, these values are reduced by 10%.
Hence, the volume of the coarse aggregate = 0.652*0.9 = 0.5868, Which is nearly equal to 0.59.
Volume of the fine aggregate content = 1- 0.59= 0.41.
2.10 STEP- 7 MIX CALCULATIONS
The mix calculation for the unit volume of the concrete shall be as follows,
|
(a) Total volume of the concrete |
=1 `m^3` |
|
(b) Volume of the entrapped air in the wet concrete |
=0.01 `m^3` |
|
(c) Volume of the cement |
= (mass of the cement/ specific gravity of the cement) *(1/1000) = `(424/3.15) * (1/1000)` =0.1346 `m^3` which is nearly equal to 0.135`m^3` |
|
(d) Volume of water |
= (mass of water/ specific gravity of water) *(1/1000) = `(144/1) * (1/1000)` =0.144 `m^3`. |
|
(e) Volume of chemical admixture (super- plasticizer ) (@1% by mass of the cementitious material i.e. 407 kg * 0.01= 4.07kg) |
= (mass of chemical admixture/ specific gravity of admixture) *(1/1000) = `(4.24/1.145) * (1/1000)` =0.00370 m^3 which is nearly equal to 0.004 m^3. |
|
(f) volume of all aggregate |
= [(a-b)- (c +d +e +f)] = [(1-0.01) -(0.135+ 0.144+ 0.004)] = 0.707 m^3 |
|
(g) mass of coarse aggregate |
= (g) * volume of coarse aggregate* specific gravity of the coarse aggregate* 1000 = 0.707* 0.59* 2.74* 1000 = 1142.9362 kg which is nearly equal to 1143 kg |
|
(h) mass of fine aggregate |
= (g) * volume of fine aggregate* specific gravity of the fine aggregate* 1000 = 0.707* 0.41* 2.74* 1000 = 794.2438 kg which is nearly equal to 794 |
2.11 STEP – 8 MIX PROPORTIONS FOR TRIAL NUMBER 1
|
Cement |
= 424 `(kg)/m^3` |
|
Water |
= 144 `(kg)/m^3` |
|
Fine aggregate |
= 794 `(kg)/m^3` |
|
Coarse aggregate |
= 1143 `(kg)/m^3` |
|
Chemical admixture |
= 4.24 `(kg)/m^3` |
|
Free water cementitious material ratio |
= 0.34 |
|
Trial mix ratio |
= 1:1.87:2.69 |
2.12 RESULT- The appropriate mix proportion of the ingredients for the concrete for the grade of M50 using the fly ash as part of replacement of the OPC with the help of design mix is calculated as 1:1.87:2.69, with the water cementitious material of 0.34.
2.13 EXPLANATION-
This is the trial mix ratio, so two more trial having variation of 10% of the water cement ratio shall be carried out and a graph between three water cement ratios and corresponding strengths shall be plotted to find out the mix proportion for given target men strength for the field trails.