Concrete Mix Design for M40 Grade
Concrete Mix Design for M40 Grade
Concrete Mix Design for M 40 grade using fly ash given below data:
(a) Type of cement: OPC 43 grade
(b) Type of mineral admixture: Fly ash conforming to IS: 3812 (Part-l)
(c) Maximum nominal size of Aggregate: 20 mm
(d) Minimum cement content: 320 kg/ m3
(e) Maximum water-cement ratio: 0.45
(f) Workability: 100 mm slump
(g) Exposure condition: Severe (For RCC)
(h) Method of placing concrete: Pumping
(I) Degree of supervision: Good
(j) Type of aggregate: Crushed angular aggregate
(k)Maximum cement content: 450 kg/m3
(1) Chemical admixture type: Super plasticizer
(m) Specific gravity of cement: 3.15
Specific gravity of C.A.: 2.74
Specific gravity of F.A.: 2-74
Specific gravity of Fly ash: 2.20
(n) Water absorption
Coarse aggregate: 0.5%
Find aggregate: Nil
(0) Free surface moisture
Coarse aggregate: Nil
Find aggregate 1.5%
Grading of C.A. conforming Table-2 of IS: 383
Grading of EA. Conforming to grading Z one-Io f Table-4o f IS: 383
SOLUTION:
Step-l: Target mean strength
fck_= 40 N/mm2
fck’=fck + 1.65 From Table 6.4 for M40 concrete,
= 40+ 1.65x 5 standard deviatio S = 5 N/mm2
= 48.25 N/mm2
Step-2: Selection of w/c ratio
From Table-5 of IS: 456-2000 Table 6.5
Maximum free w/c ratio = 0.45
Based on experience of the mix design w/c, ratio is taken as 0.40.
Adopt smaller of the two values,
w/c = 0.40
Step-3 : Selection of water content
Maximum water content. From Table-6.6 for nominal maximum size of aggregate 20 mm
= 186 litre.
This is for 50 mm slump.
Increase 3% water for every 25 mm slump over and above 50 mm slump.
We have slump value of 100 mm, hence increase water content by 6%.
Estimated water content for 100 mm slump
=186+6/100×186
= 197 litre
As super plasticizer is used the water content can be reduced up to 20% and above.
Assume 25% reduction in water content due to super plasticizer,
Actual water to be used = 197 x 0.75
= 148 litre.
Step-4 : Calculation of cement + fly ash content
w/c ratio = 0.40
Water used = 148 litre
Cement + fly ash content, w/c = 0.40
C = 148/0.40
= 370 kg/m3
As per IS: 456_2000. Table_5. Minimum cement content for severe exposure condition
370 kg/m3 > 320 kg/m3 hence O.K.
Since fly ash is not as active as that of cement. It is usual to increase the cementites
materially some percentage.
In this example an increase of 10% is considered.
Cementitious material (Cement + fly ash) content
= 370 x 1.10
= 407 kg/m3
Water content = 148 litre
w/c ratio = 148/407 = 0.364
Let us the percentage of fly ash as 30%
Fly ash content = 407 x 0.30 = 122.0 kg/m3
Cement content = 407 – 122 = 285 kg/m3
Therefore. saving of cement while using fly ash: 370 – 285 = 85 kg/m3
Step-5 : Coarse aggregate and find aggregate content
From Table-6.8. volume of coarse aggregate corresponding to 20 mm maximum size of
Aggregate and fine aggregate (Zone-I) for water cement ratio 0.50 = 0.60.
In the present case w/c = 0.40. i.e. less by 0.10. As the w/c ratio is reduced it is desirable
to increase the CA. content to decrease F.A. content.
For every decrease of w/c ratio by 0.05, the CA. volume may be increased by 1.0%
(Le. 0.01).
As the w/c ratio is less by 0.10, the CA. volume is increased by 0.02.
Corrected proportion of volume of CA. = 0.60 + 0.02 = 0.62
For pumpable concrete, C.A. volume can be reduced by 10%.
Final volume of CA. = 0.62 X 0.90 = 056
Volume of EA. = l – 0.56 = 0.44
Step-6: Calculation of Mix Proportions
(1) Volume of concrete = 1 m3
(2) Volume of cement mass of cement = mass of cement/specific gravity of cement x 1/100=
= 285/3.15 x 1/100 = 0.090 m3
3) Mass of fly ash = mass of fly / specific gravity of fly ash x 1/100 = 122x 1/100
= 0.055 m3
4) Volume of water = mass of water / specific gravity of water x 1/100 = 148/1 x 1/100
= 0.148 m3
(5) Volume of chemical admixture (supper plasticizer) @ 2.0% by mass of cementitious
= mass of chemical admixture / specific gravity of chemical admixture x 1/100
= 8.14/1.145 x 1/100
= 0.0071 m3
Absolute volume of all the materials except total aggregates
= 0.09 + 0055 + 0.148 + 0.0071
= 0.3 m3
Absolute volume of total aggregate:- 1 – 0.3
- = 0.7 m3
Mass of CA. = V. x Volume of CA. x Sp.gravity of CA. x 1000
= 0.7 x 0.56 x 2.74 x 1000
= 1074 kg
Mass of EA. = V. x Volume of FA. x Sp. Gravity of EA. x 1000
= 0.7 x 0.44 x 2.74 x 1000
= 844 kg
Step-7 : Mix Proportions for Trial No. l :
Cement : 285 kg/m3
Fly ash : 122kg
Water : 148 litre
Fine aggregate : 844 kg/m3
Coarse aggregate : 1074 kg/m3
Chemical admixture : 8.14 kglm3
Wet densityof concrete= 2481 ltglm3
w/c ratio= 148/407
= 0.364
We may use 40% 0f 10 mm sizeand60% of 20 mm size of aggregate.
Quantity of mm size aggregate =ga%tex1074= 429.k6g /m3
Quantiotyf20mmsizeaggreg=at£ex1074 = 644.4k g/m3
Step_8: Site corrections for water absorption and surface moisture
Quantity of F.A. = 844 kg
Absorption = Nil
Surface moisture = 1.5%
Quantity of surface moisture = 1.5/100×844 = 12.66 kg
Mass of FA. In field condition = 844 + 12.66 = 856.66 kg/m3
Absorption of C.A= 0.5/100×844=5.37g
Mass of CA. in field condition = 1074-5.37 = 1068.63 kg/m3
As regards to water, 12.66 kg of water is contributed by F.A. and 5.37 kg of water is
Absorbed by CA.
Therefore, 12.66-5.37 = 7.29 kg of extra water is contributed .This quantity of water
Is to be deducted from total water.
Net quantity of water required= 148-7.29 = 140.71 kg/m3
Step-9: Mix Proportions (by mass):
Water Cement+ Fly ash F.A. CA.
140.71 lit. 285kg+ 122= 407kg 856.66 kg 1068.6k3g
0.345 1 2.105 2.625
Final quantities of materials:
Cement: 285 kg/m3.
Fly ash: 122 kg/m3
Water; 140.71 kg/m3
Fine aggregate 856.66 kg/m3
Coarse aggregate 1068.63 kg/m3
Chemical admixture =8.14 kg/m3
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Wet density of concrete= 2481 kg/m3
