Project on Concrete Mix Design for various grades of Concrete

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

Ans-

Aim: - To Calculate the Concrete Mix Design for M35 grade concrete with fly ash.

Introduction: - Calculate mix design with fly ash, using IS Code- 10262:2019(update version) & IS Code- 456:2000.

Procedure: -

I- STIPULATIONS FOR PROPORTIONING

a)Grade designation: M35
b) Type of cement: OPC 43 Grade conforming IS 12269
c) Maximum nominal size of aggregate: 20mm
d) Minimum cement content : 300kg/m³ (IS 456:2000)
e) Maximum water-cement ratio : 0.45 (Table 5 of IS 456:2000)
f) Workability: 100mm slump
g) Exposure condition: Moderate (For Reinforced Concrete)
h) Method of concrete placing: Pumping
j) Degree of supervision: Good
k) Type of aggregate: Crushed Angular Aggregates
m) Maximum cement content : 360 kg/m³
n) Chemical admixture type: Superplasticizer

2- Test data

1. cement used = OPC 43 conforming to IS 8112
2. the specific gravity of cement = 3.15
3. Fly ash = conforming to IS 3812 (part 1)
4. the specific gravity of fly ash = 2.2
5. chemical admixture = super plasticizer conforming to IS 9103
6. specific gravity of course aggregate = 2.67
7. specific gravity of fine aggregate= 2.65
8. specific gravity of chemical admixture = 1.145
9. Sieve analysis- coarse aggregate conforming to table 2 of IS 383

fine aggregate conforming to grading (zone 1) of table 4 of IS 383

Sol.-

3-TARGET MEAN STRENGTH

Fck=Fck+1.65*S

=53+1.65*5

=43.35 N/mm2

Were,
Fck = target average compressive strength at 28 days,
Fck = characteristics compressive strength at 28 days, and
S= standard deviation. ( Using IS 10262:2019)

4- SELECTION OF WATER-CEMENT RATIO

water-cement ratio = 0.39

(From Table 5 of IS 456 for severe maximum Water Cement Ratio is 0.45

0.39< 0.45 Hence ok.)

5-SELECTION OF WATER CONTENT.
From Table 2 of IS 10262:2019, maximum water content for 20 mm aggregate= 186 lit. (for 25 to 50 mm slump range)

Estimated water content for 100mm slump = 186+ (6/186)* 186

= 197 lit.

(Note: If Superplasticizer is used, the water content can be reduced up to 20% to 30%)

Superplasticizer water content reduction of 29% has been achieved, Hence the arrived water content = 197-(29/100)*197

=140 lit.

6-CALCULATION OF CEMENT CONTENT

w/c ratio = 0.39

cementitious material (cement +fly ash)= 140/0.39= 359 kg/m^3

(From Table 5 of IS 456, Minimum cement content for ‘severe’ exposure conditions 320kg/m3, 359kg/m3> 320kg/m³ hence ok.)

The proportion of mix containing fly ash

1. percentage of fly ash needs to be decided based on project requirements & quality of materials
2. In some cases, increases in cementitious material can be warranted. increase in cementitious material content and percentage may be used on experience and trials. (this is with an increase of 10% of cementitious material content)

cementitious material content = 359*1.1

=394 kg/m3

water content = 140kg/m^3

water-cement ratio= 140/394

=0.36

fly ash @ 30% total cementitious material = 394*30%

=118kg/m3

cement (OPC)= 394-118

=276 kg/m3

saving of cement while using fly ash = 359-276

=83 kg/m3

fly ash being utilizes=118 kg/m3

7.-VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT.

From Table 5 of (IS 10262:2019) Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for the water-cement ratio of 0.36 =0.60

In the present case, the water-cement ratio is 0.36. Therefore, the volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.06. 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).

Therefore, corrected proportion of volume of coarse aggregate for the water-cement ratio is =((0.5-w/c ratio/0.05)*0.01)+unit vol.

=((0.5-0.36/0.05)*0.01)+0.60

=0.628 kg/m3

For pumpable concrete, these values should be reduced up to 10%. Therefore, volume of coarse aggregate =0.628 x 0.9

=0.5652 m3

Volume of fine aggregate content = 1 – 0.5652

= 0.4348 m3

8.-MIX CALCULATIONS

The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3

b) Volume of cement = [Mass of cement] / {[Specific Gravity of Cement] * 1/1000}
= [276/3.15] * [1/1000]

=0.088 m3

c) Volume of fly ash = [Mass of fly ash] / {[Specific Gravity of fly ash] * 1/1000}
= [118/2.2] * [1/1000]

=0.054 m3

d) Volume of water = [Mass of water] / {[Specific Gravity of water] * 1/1000}
= [140/1] * [1/1000]

=0.14 m3

e) Volume of chemical admixture = [Mass of chemical admixture] / {[Specific Gravity of chemical admixture ] * 1/1000}

= [7.8/1.145] * [1/1000]

=6.64 m-3

=0.007 m3

f) Volume of all in aggregate = [a-(b + c + d + e)]
= [1-(0.088+0.054+0.14+0.007)]
= 0.858 kg/m3

g) Mass of coarse aggregate= f x Volume of Coarse Aggregate x Specific Gravity of Fine Aggregate x 1000
= 0.858 x 0.5652 x 2.67 x 1000

=1295 kg/m3

h) Mass of fine aggregate= f x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000
= 0.858 x 0.4348 x 2.65 x 1000

=989 kg/m3

9.-MIX PROPORTIONS
Cement =276 kg/m3
Water = 140kg/m3

Fly ash= 118 kg/m3
Fine aggregate = 989 kg/m3

Coarse aggregate =1295 kg/m3
Chemical admixture = 7.8 kg/m3
Water-cement ratio = 0.36

10.- Mix Proportion By weight-

= C : S: A
= 276: 989 : 1259

= (276/276) : (989/276) : (1259/276)

Result = 1 : 3.59 : 4.57

Problem 2.- To Calculate the Concrete Mix Design for M50 grade concrete.

Aim: - To Calculate the Concrete Mix Design for M50 grade concrete.

Introduction: -Calculate mix design without fly ash, using IS Code- 10262:2019(update version) & IS Code- 456:2000.

Procedure: -

I- STIPULATIONS FOR PROPORTIONING

a)Grade designation: M50
b) Type of cement: OPC 53Grade conforming IS 12269
c) Maximum nominal size of aggregate: 20mm
d) Minimum cement content : 320 kg/m³ (IS 456:2000)
e) Maximum water-cement ratio : 0.45 (Table 5 of IS 456:2000)
f) Workability: 75mm slump
g) Exposure condition: severe(For Reinforced Concrete)
h) Method of concrete placing: Pumping
j) Degree of supervision: Good
k) Type of aggregate: Crushed Angular Aggregates
l) Maximum cement content : 350 kg/m³
m) Chemical admixture type: Superplasticizer

2-TEST DATA FOR MATERIALS:-
a) Cement used: OPC 53Grade conforming IS 12269

3. b) Specific gravity of cement: 3.15
4. c) Chemical admixture: Super Plasticizer
5. d) Specific gravity of-

1) Coarse aggregate 20mm : 2.74
2) Fine aggregate: 2.74

3)Chemical admixture: 1.145

1. e) Sieve analysis:

1) Coarse aggregate: Conforming to all in aggregates of Table 2 of IS 383
2) Fine aggregate: Conforming to Grading Zone I of Table 4 of IS 383

3-TARGET MEAN STRENGTH

Fck=Fck+1.65*S

=50+1.65*5

=58.25 N/mm2

where
Fck = target average compressive strength at 28 days,
Fck = characteristics compressive strength at 28 days, and
S= standard deviation.( Using IS 10262:2019)

4- SELECTION OF WATER-CEMENT RATIO

water-cement ratio = 0.39

(From Table 5 of IS 456 for severe maximum Water Cement Ratio is 0.45

0.39< 0.45 Hence ok.)

5-SELECTION OF WATER CONTENT.
From Table 2 of IS 10262:2019, maximum water content for 20 mm aggregate= 186 lit. (for 25 to 50 mm slump range)

Estimated water content for 75mm slump = 186+ (3/186)

= 192lit.

(Note: If Superplasticizer is used, the water content can be reduced upto 20% to 30%)

Superplasticizer water content reduction of 23% has been achieved, Hence the arrived water content = 192-(23/100)*192

=148 lit.

6-CALCULATION OF CEMENT CONTENT
Adopted w/c Ratio = 0.39
Cement Content = 148/0.39 = 380 kg/m3

(From Table 5 of IS 456, Minimum cement content for ‘severe’ exposure conditions 320kg/m3, 380kg/m3> 320kg/m³ hence ok.)

7.-VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT.

From Table 5 of (IS 10262:2019) Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for the water-cement ratio of 0.39 =0.60

In the present case, the water-cement ratio is 0.39. Therefore, the volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.06. The proportion of the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

Therefore, corrected proportion of volume of coarse aggregate for the water-cement ratio is =((0.5-w/c ratio/0.05)*0.01)+unit vol.

=((0.5-0.39/0.05)*0.01)+0.60

=0.622kg/m3

For pumpable concrete, these values should be reduced up to 10%. Therefore, volume of coarse aggregate =0.622 x 0.9

=0.5598 m3

Volume of fine aggregate content = 1 – 0.5598

= 0.4402 m3

8.-MIX CALCULATIONS

The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3

b) Volume of cement = [Mass of cement] / {[Specific Gravity of Cement] * 1/1000}
= [380/3.15] * [1/1000]

=0.120 m3

c) Volume of water = [Mass of water] / {[Specific Gravity of water] * 1/1000}
= [148/1] * [1/1000]

=0.148 m3

d) Volume of chemical admixture = [Mass of chemical admixtur] / {[Specific Gravity of chemical admixture ] * 1/1000}

= [7.6/1.145] * [1/1000]

=6.64 m-3

=0.006 m3

e) Volume of all in aggregate = [a-(b+c+d)]
= [1-(0.120+0.148+0.006)]
= 0.726 kg/m3

f) Mass of coarse aggregate= e x Volume of Coarse Aggregate x Specific Gravity of Fine Aggregate x 1000
= 0.726 x 0.5598x 2.74 x 1000

=1114 kg/m3

g) Mass of fine aggregate= e x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000
= 0.726 x 0.4402 x 2.74 x 1000

=875 kg/m3

9.-MIX PROPORTIONS
Cement =380 kg/m3
Water = 148 kg/m3
Fine aggregate = 876 kg/m3

Coarse aggregate =1114 kg/m3
Chemical admixture = 7.6 kg/m3
Water-cement ratio = 0.39

10.- Mix Proportion By weight-
= 380 : 876 : 1114

= (380/380) : (876/380) : (1114/380)

Result = 1 : 2.4 : 2.94