M30 Concrete Mix example: Let’s say we have to design a Reinforced Concrete Cement(RCC) with a slump of 100 using Cement (Specific Gravity 3.14), 20mm Sub Angular Coarse aggregate (Specific Gravity 2.45) and Zone 2 Sand (Specific Gravity 2.50) for a Moderate exposure condition. Plus We will be using a water reducing admixture

SuperPlasticiser 1% by weight of cement, 23% reduction in water achieved, SpGr 1.145

These are the steps you will take:

**STEP 1 : TARGET MEAN STRENGTH**

For SEVERE exposure RCC using 20mm coarse aggregate we need to atLeat use M30 Grade of concrete (As per Table 5 of IS 456 2000)

So lets design M30

Target Mean Strength Fckt = Fck + 1.65xS

Fck = 30MPa (for M30)

Standard deviation for M30= 5.0 (Table 1 IS 10262 2009)

Fckt = Fck + 1.65×5.0

Target Mean Strength= 30+ 1.65*5.0=38

**STEP 2 : CALCULATION FOR WATER CONTENT**

For 20mm Aggregates MAX Water Content =186.0 (Table 2 10262 2009)

This value is for Slump within the range of 25-50

For Every 25mm increase in slump we will increase

water by 3%

Current Slump is :100 Which is :50 mm above 50mm

Increase Water Content By :6.00%

186.0+(6.00*186.00)/100 =197.16 Kg/m^{3}

With the Admixture 23% reduction is achieved So

New Water Content = 197.16-(23/100)

Water Content = 151.81 Kg/m^{3}

**STEP 3 : SELECTION OF WATER CEMENT RATIO**

For REINFORCED Concrete and SEVERE exposure

condition MAX W:C=0.45 (As per Table 5 of IS 456 2000)

**STEP 4 : CALCULATION OF CEMENT CONTENT**

Cement=WaterContent/w:C

Cement Content=151.81/0.45 = 337.36 Kg/m^{3}

(As per Table 5 of IS 456 2000)Minimum Cement Content is 320

Since

337.36 >320 OK

**STEP 5 : CALCULATION OF RATIO OF AGGREGATES**

For 20mm Coarse and Zone 2 Fine,Volume of Coarse to Total Aggregates=0.62(Table 3 10262 2009)

This value is for w:C =0.50 our w:C=0.45

Increase Coarse Ratio by 0.01 for every decrease of 0.05 in W:C

For 0.45 there is a decrease of 0.05

So Increase Factor= 0.01

New Coarse Ratio = 0.62 + 0.01 = 0.63

Fine Agg Ratio=1-Coarse Agg Ratio=0.37

**DESIGN MIX CALCULATIONS**

Vol of Cement= WT OF CEMETN/(SPGR of

CEMENT*1000)=337.36/(3.14*1000)=0.11

Vol of Water=Wt of Water/1000

=151.81/1000

=0 .15

Vol of Admixture = Wt of ADM/(spgr*1000)=0.003

TOTAL AGGREGATE Vol(e) = 1-(Cement+Water+Admixture)=0.74

Wt of Coarse Aggregate=e*CoarseRatio*SPGR*1000

= 0.74*0.63*2.45*1000=1138.79 Kg/m^{3}

Wt of Fine Aggregate=e*FineRatio*SPGR*1000

=0.74*0.37*2.50*1000=682.47 Kg/m^{3}

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