Question

Problem with different moisture conditions: 5. Calculate the quantities of cement, water, fine aggregate, and coarse aggregate per trial mix of 1 m3 for the following specifications. Target mean compressive strength-30 MPa at 7 days; Cement strength class 42.5; Slump required 100 mm; Max. Aggregate size 20 mm; specific gravity of aggregates 2.65% Coarse aggregate UNCRUSHED (10, 20mm), fine aggregate CRUSHED (50% pass 600 micros); Maximum allowable free- water/cement ratio 0.40; Minimum allowable cement content 275 kg/m3; Absorption of fine aggregate : 1.0%; Absorption of coarse aggregate 1.0%; Total Moisture content of coarse aggregate 3.0%; Total Moisture content of fine aggregate-30% Chapter 5 Concrote Mix Design Calculations Total 60

Answer 1

Step 1:- Target Strength For Mix Proportioning

ck
= $_{f}{ck}^{}$ +1.65 S [Refer IS 10262:2009]

where,

= target average compressive strength at 28 days.

ck

$_{f}{ck}^{}$ = characteristic compressive strength at 28 days

S = standard deviation depending upon the grade of concrete

Here in the question, the characteristic strength is given for 7 days. So we have two options to proceed with. Either proceed with 7 days strength and take data accordingly from IS 10262:2009 or increase the 7 day strength by about 60 % to get 28 day compressive strength and take data accordingly from IS 10262:2009. Lets move with 7 day strength.

$_{f}{ck}^{}$= 30 MPa (M30)

S= 5 MPa {As per $_{f}{ck}^{}$ from Table 1 of IS 10262:2009)

So, = 30 + (1.65 x 5) =38.25 MPa

ck

Therefore, Target Strength is 38.25 MPa.

Step 2:- Selection of Water-Cement Ratio

Now from Table No 5 of IS 456:2000, As per M30 grade of concrete and for severe exposure

Maximum Water-Cement ratio = 0.45

Adopt Water-Cement ratio = 0.40 < 0.45, Hence OK.

Step 3:- Selection of Water Content(as per Clause 4.2 of IS 10262:2009)

Slump Required = 100 mm

maximum aggregate size = 20 mm

so, from Tabe No 2 of IS 10262:2009,

maximum water content for 20 mm aggregate = 186 litre (for 25 to 50 mm slump range)

so estimated water content for 100 mm slump = 186 + [(2 x 3%)(186)] = 197 litre

Step 4:- Calculation of Cement Content

Minimum allowable cement content = 275 Kg/m³

Since Water-Cement ratio = 0.40

Hence, Cement Content = (197)/(0.40) =492.5 Kg/m³  > 275 Kg/m³ hence OK.

Step 5:- Proportion of Volume of Coarse and Fine Aggregate

Grading Zone of Fine aggregate as per Table No 4 of IS 383:1970 = Zone II

From Table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 =0.62.

In our case water-cement ratio is 0.40. Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.12, the proportion of volume of coarse aggregate is increased by 0.024 (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 of 0.40 = 0.62 + 0.024 = 0.644

For pumpable concrete these values should be reduced by 10 percent.

Therefore, volume of coarse aggregate = 0.644 x 0.9 = 0.5796.

Volume of fine aggregate content =1 - 0.5796 =0.4204

Step 6:- Mix Calculation for Trial mix of 1 m³

1. Volume of Concrete = 1 m³
2. Volume of Cement =[(mass of cement)/(specific gravity of cement)] x [1/1000] = [492.5/3.15] x 1/1000] = 0.156  m³
3. Volume of water = [(mass of water)/(specific gravity of water)] x [1/1000] =[197/1] x [1/1000] = 0.197 m³  
4. Mass of Coarse aggregate = [1-(0.156+0.197)] x [volume of coarse aggregate] x [specific gravity of coarse aggregate] x 1000 = 0.647 x 0.5796 x 2.65 x 1000 =993.753 Kg
5. Mass of Fine aggregate = [1-(0.156+0.197)] x [volume of fine aggregate] x [specific gravity of fine aggregate] x 1000 = 0.647 x 0.4204 x 2.65 x 1000 =720.796 Kg

Therefore for 1 cubic metre of trail mix, proportions are as following:

Cement = 492.5 Kg/m3

Water = 197 Kg/m3

Fine Aggregate = 720.796 Kg/m3

Coarse Aggregate = 993.753 Kg/m3

water-cement ratio = 0.40