Concrete Mix Design for M40 Grade

Concrete Mix Design for M40 Grade

Design a concrete mix for M 40 grade using fly ash with the following 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

_______________________________________

Wet density of concrete= 2481 kg/m3

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