Project on Concrete Mix Design for various grades of Concrete
1. Aim :- To find the proportion for the mix design having the strength of M35 grade concrete and fly ash used as the part of cement.
Procedure :-
- Stipulation for proportioning :-
a) Grade designation = M35
b) Type of cement = OPC 43 grade conforming to IS 8112
c) Type of mineral admixture = fly ash conforming to IS 3812 (Part 1)
d) Maximum nominal size of aggregate = 20 mm
e) Minimum cement content = 340 kg/`m^3` (From table 5 of IS 456:2000 - exposure condition)
f) Maximum Water-Cement ratio = 0.45 (From table 5 of IS 456:2000 - exposure condition)
g) Workability = 100 mm slump value for pumpable concrete
h) Exposure condition = Very sevre (For reinforced concrete)
i) Method of concrete placing = pumping
j) Degree of supervision = Good
k) Type of aggregate = Crushed angular aggregate
l) Maximum content of cement = 450 kg/`m^3`
m) Chemical admixture type = Superplastisizer
- Test data for materials :-
a) Cement used = OPC 43 grade conforming to IS 8112
b) Specific gravity of cement = 3.15
c) Fly ash = Conforming to IS 3812 (Part 1)
d) Specific gravity of fly ash = 2.2
e) Chemical admixture = Superplastisizer conforming to IS 9103
f) Specific gravity of aggregate,
> Coarse aggregate = 2.74
> Fine aggregate = 2.74
g) Water absorption,
> Coarse aggregate = 0.5%
> Fine aggregate = 1.0%
h) Free surface moisture,
> Coarse aggregate = Nil
> Fine aggregate = Nil
i) Sieve analysis,
> Coarse aggregate = Conforming to Table 2 of IS 383
> Fine aggregate = Conforming to grading zone 1 of Table 4 of IS 383
- Target strength of mix proportioning :-
F'ck = Fck + 1.65S
From Table 1 of IS 10262:2009, Standard Deviation (S) = 5.0 N/`mm^2`
Therefore, Target Strength = 35 + (1.65*5.0) = 43.25 N/`mm^2`.
- Selection of Water-Cement ratio :-
a) From Table 5 of IS 456:2000 for very severe exposure max W/C ratio = 0.45
b) From graph-1 we assume W/C ratio = 0.42
c) 28-days compressive strength is 43.25 N/`mm^2`,
So, for graph-2 we use curveline E,
So, W/C ratio = 0.41
So, Final W/C ratio = 0.41 as O.41<0.42<0.45, Hence o.k.
- Selection of water content :-
From Table 2 of IS 10262:2009, Maximum water content for 20 mm aggregate = 186 litre (For 25mm to 50 mm slump range)
Estimate water content for 100 mm slump = 186 + ((6/100)*186) = 197 litre
As Superplastisizer is used, the water content can be reduced to 30%, Based on trials water content reduction of
29% is achieved with the help of plastisizers,
Hence, the arrived water content = 197*0.71 = 140 litre
- Calculation of cement content :-
Water-Cement ratio = 0.41
Cementitious material (Cement + Fly ash) content = 140/0.41 = 341 kg/`m^3`.
From Table-5 of IS 456:2000 minimum cement content for 'very severe' exposure condition = 340 kg/`m^3`.
341 kg/`m^3` > 340 kg/`m^3`, Hence Ok.
Now, to proportion a mix containing fly ash the following steps are suggested.
a) The percentage of fly ash used to be decided based on project requirement and quality of material.
b) 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 calculation is with increase of
10% cementitious material content).
-Cementitious material content = 341*1.10 = 375 kg/`m^3`.
-Water content = 140 kg/`m^3`
-So, W/C ratio = 140/375 = 0.373,
-Consider, fly ash at 30% total cementetious material content = 375*0.3 = 113 kg/`m^3`.
-Cement (OPC) = 375 - 113 = 262 kg/`m^3`.
-Saving of cement while using fly ash = 341 - 262 = 79kg/`m^3`.
-Fly ash being utilized = 113 kg/`m^3`.
- Proportion of volume of Coarse aggregate and Fine aggregate content :-
From Table-3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size aggregate and fine
aggregate (zone 1) for W/C ratio of 0.5 = 0.6,
In the present case the W/C ratio is 0.41,
- Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
- As W/C rati is lower by 0.09, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of
-/+0.01 for every +0.05 change in W/C ratio.)
Therefore, corrected volume of coarse aggregate for the W/C ratio of 0.41 = [(0.09*0.01)/0.05]+0.6 = 0.618.
For, pumpable concrete these values are reduced by 10%,
Therefore, the volume of coarse aggregate = 0.618*0.9 = 0.556,
The volume of fine aggregate = 1 - 0.556 = 0.444.
- Mix calculation :-
The mix calculation per unit volume of concrete shall be as follows,
a) Volume of concrete = 1 `m^3`.
b) Volume of cement = [(mass of cement / Specific gravity of cement)*(1/1000)] = [(262 / 3.15)*(1 / 1000)] = 0.083`m^3`.
c) Volume of fly ash = [(mass of fly ash / Specific gravity of fly ash)*(1/1000)] = [(113 / 2.2)*(1 / 1000)] = 0.051`m^3`.
d) Volume of water = [(mass of water / Specific gravity of water)*(1/1000)] = [(140 / 1)*(1 / 1000)] = 0.14`m^3`.
e) Volume of chemical admixture = [(mass of chemical admixture / Specific gravity of admixture)*(1/1000)]
= [(7 / 1.145)*(1 / 1000)] = 0.006`m^3`.
f) Volume of aggregate (Fine aggregate & coarse aggregate) = [a-(b+c+d+e)] = [1-(0.083+0.051+0.14+0.006)] = 0.72 `m^3`.
g) Mass of coarse aggregate = f*(volume of coarse aggregate)*(specific gravity of coarse aggregate)*1000
= 0.72*0.556*2.74*1000 = 1097 kg/`m^3`.
h) Mass of fine aggregate = f*(volume of fine aggregate)*(specific gravity of fine aggregate)*1000
= 0.72*0.444*2.74*1000 = 876 kg/`m^3`.
- Mix proportion based on the calcutions :-
- Cement = 262 kg/`m^3`,
- Fly ash = 113 kg/`m^3`,
- water = 140 kg/`m^3`,
- Fine aggregate = 876 kg/`m^3`,
- Coarse aggregate = 1097 kg/`m^3`,
- Chemical admixture = 7 kg/`m^3`,
- W/C ratio = 0.41
Trial mix ratio = 1 : 2.34 : 2.93
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2. Aim :- To find the proportion for the mix design having the strength of M50 grade concrete.
Procedure :-
- Stipulation for proportioning :-
a) Grade designation = M50
b) Type of cement = OPC 43 grade conforming to IS 8112
c) Maximum nominal size of aggregate = 20 mm
d) Minimum cement content = 360 kg/`m^3` (From table 5 of IS 456:2000 - exposure condition)
e) Maximum Water-Cement ratio = 0.40 (From table 5 of IS 456:2000 - exposure condition)
f) Workability = 100 mm slump value for pumpable concrete
g) Exposure condition = Extreme (For reinforced concrete)
h) Method of concrete placing = pumping
i) Degree of supervision = Good
j) Type of aggregate = Crushed angular aggregate
k) Maximum content of cement = 450 kg/`m^3`
l) Chemical admixture type = Superplastisizer
- Test data for materials :-
a) Cement used = OPC 43 grade conforming to IS 8112
b) Specific gravity of cement = 3.15,
c) Chemical admixture = Superplastisizer conforming to IS 9103
d) Specific gravity of aggregate,
> Coarse aggregate = 2.74
> Fine aggregate = 2.74
e) Water absorption,
> Coarse aggregate = 0.5%
> Fine aggregate = 1.0%
f) Free surface moisture,
> Coarse aggregate = Nil
> Fine aggregate = Nil
h) Sieve analysis,
> Coarse aggregate = Conforming to Table 2 of IS 383
> Fine aggregate = Conforming to grading zone 1 of Table 4 of IS 383
- Target strength of mix proportioning :-
F'ck = Fck + 1.65S
From Table 1 of IS 10262:2009, Standard Deviation (S) = 5.0 N/`mm^2`
Therefore, Target Strength = 50 + (1.65*5.0) = 58.25 N/`mm^2`.
- Selection of Water-Cement ratio :-
a) From Table 5 of IS 456:2000 for very severe exposure max W/C ratio = 0.40
b) From graph-1 we assume W/C ratio = 0.30
c) 28-days compressive strength is 58.25 N/`mm^2`,
So, for graph-2 we use curveline F,
So, W/C ratio = 0.35
So, Final W/C ratio = 0.30 as O.30<0.35<0.40, Hence o.k.
- Selection of water content :-
From Table 2 of IS 10262:2009, Maximum water content for 20 mm aggregate = 186 litre (For 25mm to 50 mm slump range)
Estimate water content for 100 mm slump = 186 + ((6/100)*186) = 197 litre
As Superplastisizer is used, the water content can be reduced to 30%, Based on trials water content reduction of
29% is achieved with the help of plastisizers,
Hence, the arrived water content = 197*0.71 = 140 litre
- Calculation of cement content :-
Water-Cement ratio = 0.30
Cement content = 140/0.30 = 467 kg/`m^3`.
From Table-5 of IS 456:2000 minimum cement content for 'Extreme' exposure condition = 360 kg/`m^3`.
467 kg/`m^3` > 360 kg/`m^3`, Hence Ok.
- Proportion of volume of Coarse aggregate and Fine aggregate content :-
From Table-3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size aggregate and fine
aggregate (zone 1) for W/C ratio of 0.5 = 0.6,
In the present case the W/C ratio is 0.30,
- Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
- As W/C rati is lower by 0.2, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of
-/+0.01 for every +0.05 change in W/C ratio.)
Therefore, corrected volume of coarse aggregate for the W/C ratio of 0.30 = [(0.2*0.01)/0.05]+0.6 = 0.64.
For, pumpable concrete these values are reduced by 10%,
Therefore, the volume of coarse aggregate = 0.64*0.9 = 0.576,
The volume of fine aggregate = 1 - 0.576 = 0.424.
- Mix calculation :-
The mix calculation per unit volume of concrete shall be as follows,
a) Volume of concrete = 1 `m^3`.
b) Volume of cement = [(mass of cement / Specific gravity of cement)*(1/1000)] = [(467 / 3.15)*(1 / 1000)] = 0.148`m^3`.
c) Volume of water = [(mass of water / Specific gravity of water)*(1/1000)] = [(140 / 1)*(1 / 1000)] = 0.14`m^3`.
d) Volume of chemical admixture = [(mass of chemical admixture / Specific gravity of admixture)*(1/1000)]
= [(7 / 1.145)*(1 / 1000)] = 0.006`m^3`.
e) Volume of aggregate (Fine aggregate & coarse aggregate) = [a-(b+c+d)] = [1-(0.148+0.14+0.006)] = 0.706 `m^3`.
f) Mass of coarse aggregate = f*(volume of coarse aggregate)*(specific gravity of coarse aggregate)*1000
= 0.706*0.576*2.74*1000 = 1115 kg/`m^3`.
i) Mass of fine aggregate = f*(volume of fine aggregate)*(specific gravity of fine aggregate)*1000
= 0.706*0.424*2.74*1000 = 820 kg/`m^3`.
- Mix proportion based on the calcutions :-
- Cement = 467 kg/`m^3`,
- water = 140 kg/`m^3`,
- Fine aggregate = 820 kg/`m^3`,
- Coarse aggregate = 1115 kg/`m^3`,
- Chemical admixture = 7 kg/`m^3`,
- W/C ratio = 0.30
Trial mix ratio = 1 : 1.76 : 2.39