Provide me with the mix design for M30 concrete with the exact quantity of materials. What is the concoction or mix ratio for the M30 concrete?
Step 1: For the degree of quality control specified, namely, good the value of standard deviation σ read from Table 11.7 = 6 N/mm2. Hence, the target mean strength for the desired characteristic compressive strength
= 30 + 1.65 × 6 = 39.9 N/mm2
Step 2: Corresponding to this target mean strength the water-cement ratio is read from the appropriate curve corresponding to the 28day strength of cement (Fig. 11.2). For a cement strength of 51 N/mm2 curve D is selected and the water-cement ratio 0.4 obtained. This value has now to be checked against the maximum limit of the water-cement ratio for the given exposure condition. Table 11.1 for moderate exposure and reinforced concrete the maximum water-cement ratio recommended is 0.50. Hence the value of 0.40 obtained is acceptable.
Step 3: For a maximum size of aggregate of 20mm, the air content is taken as 2.0 per cent from Table 11.11. Since the required grade of concrete is M30 which is lower than the M35 grade, water content per m3 of concrete = 186 litres and sand as a percentage of total aggregate by absolute
volume = 35 (Table 11.9).
Since Table 11.9 is based on certain specific conditions, adjustments as per Table 11.10 have
to be made for any deviations in these values as given below:
Thus after incorporating the above adjustments the sand content = 35–3.5 = 31.5
and the water content = 186 + 1.5 × 186/100 = 188.8 lit.
Step 4: Determination of cement contents:
Water-cement ratio = 0.40
Quantity of water after adjustment = 188.8 litres
Therefore, cement content = 188.8/0.40 = 472 kg
This cement content has now to be checked against the minimum cement content required for mild exposure conditions in reinforced concrete. The minimum cement content specified in Table 11.1 is 320 kg/m3 from durability considerations. Therefore the value of 472 kg/m3 is acceptable.
Step 5: Now quantities of coarse and fine aggregates are worked out per m3 of concrete as
given below.
The volume of concrete = 1– 0.02 (entrapped air)
= 0.98 m3
= 980 litres.
980 = 188.8 + 742./0.315 + fa/(0.315×2.66)
fa=537.39 kg
Similarly,
980 = 188.8 + 742./0.315 + Ca/(0.315×2.66)
Ca = 1195 kg
Therefore, the mix proportion becomes:
Cement Water Sand Coarse aggregate
472 188.8 537.39 1195
or
1 0.40 1.138 2.531
Adjusting required for water absorption:
Water absorbed by coarse aggregates
= (1195 x 0.5))/100 = 5.98
Free water in fine aggregate
= (537.30 x 2) /100
= 10.75 lit
Therefore, the actual quantity of water required
= 188.8 + 5.98 – 10.75
=184.03 lit
Actual quantity of coarse aggregate = 1195 – 5.98
= 1189.02
and sand = 537.39 + 10.75
= 548.14
Therefore, the actual quantities of materials required are:
Cement Water Sand Coarse
472 kg 184.03 lit 548.14 kg 1189.02 kg
Step 6: The fraction I and II of the coarse aggregates are to be now combined to give a combined grading in accordance with IS:383–1970 of 20 mm maximum size aggregate. To obtain a combined grading, IS: 383–1970 recommends that the fraction passing the 10 mm sieve shall be in the range of 25 to 55 per cent, or on an average of 40 per cent. Trials are then made to combine the fraction I and II in the proportion of, say 40:60 to see whether the combined grading is obtained. This is given below.
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