# M25 -Concrete Mix Design Example

In a previous article we Designed M20 so let’s design M25 in this one. 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. These are the steps you will take:

STEP 1 : TARGET MEAN STRENGTH

For MODERATE exposure RCC using 20mm coarse aggregate we need to atLeat use M25 Grade of concrete (As per Table 5 of IS 456 2000)

So lets design M25

Target Mean Strength Fckt = Fck + 1.65xS

Fck = 25MPa (for M25)

Standard deviation for M20-M25 = 4.0 (Table 1 IS 10262 2009)

Fckt = Fck + 1.65×4

Target Mean Strength= 25+ 1.65*4.0=31.60

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/m3

STEP 3 : SELECTION OF WATER CEMENT RATIO

For • REINFORCED Concrete and MODERATE exposure

condition MAX W:C=0 50 (As per Table 5 of IS 456 2000)

STEP 4 : CALCULATION OF CEMENT CONTENT

Cement=WaterContent/w:C

Cement Content=197.16/0.50 = 394.32 Kg/m3

(As per Table 5 of IS 456 2000)Minimum Cement Content is 300

Since

394.32>300 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=0.50 OK

Fine Agg Ratio=1-Coarse Agg Ratio=0.38

DESIGN MIX CALCULATIONS

Vol of Cement= WT OF CEMETN/(SPGR of

CEMENT*1000)=394.32/(3.14*1000)=0.13

Vol of Water=Wt of Water/1000

=197.16/1000

=0 .20

TOTAL AGGREGATE Vol(e) = 1-(Cement+Water)=0.68

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

= 0.68*0.62*2.45*1000=1028.76 Kg/m3

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

=0.68*0.38*2.50*1000=643.40 Kg/m3

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