Concrete mix design is the process of finding right proportions of cement, sand and aggregates for concrete to achieve target strength in structures. So, concrete mix design can be stated as Concrete Mix = Cement:Sand:Aggregates.

The concrete mix design involves various steps, calculations and laboratory testing to find right mix proportions. This process is usually adopted for structures which requires higher grades of concrete such as M25 and above and large construction projects where quantity of concrete consumption is huge..

Benefits of concrete mix design is that it provides the right proportions of materials, thus making the concrete construction economical in achieving required strength of structural members. As, the quantity of concrete required for large constructions are huge, economy in quantity of materials such as cement makes the project construction economical.

Concrete Mix design of M20, M25, M30 and higher grade of concrete can be calculated from example below.

**Concrete Mix Design**

### Data Required for Concrete Mix Design

**(i) Concrete Mix Design Stipulation **

(a) Characteristic compressive strength required in the field at 28 days grade designation — M 25

(b) Nominal maximum size of aggregate — 20 mm

(c) Shape of CA — Angular

(d) Degree of workability required at site — 50-75 mm (slump)

(e) Degree of quality control available at site — As per IS:456

(f) Type of exposure the structure will be subjected to (as defined in IS: 456) — Mild

(g) Type of cement: PSC conforming IS:455

(h) Method of concrete placing: pump able concrete

**(ii) ****Test data of material (to be determined in the laboratory)**

(a) Specific gravity of cement — 3.15

(b) Specific gravity of FA — 2.64

(c) Specific gravity of CA — 2.84

(d) Aggregate are assumed to be in saturated surface dry condition.

(e) Fine aggregates confirm to Zone II of IS – 383

**Procedure for Concrete Mix Design of M25 Concrete**

**Step 1 — Determination Of Target Strength**

Himsworth constant for 5% risk factor is 1.65. In this case standard deviation is taken from IS:456 against M 20 is 4.0.

**f _{target} = f_{ck} + 1.65 x S**

**= 25 + 1.65 x 4.0 = 31.6 N/mm ^{2}**

Where,

S = standard deviation in N/mm^{2 }= 4 (as per table -1 of IS 10262- 2009)

**Step 2 — Selection of water / cement ratio:-**

From Table 5 of IS 456, (page no 20)

Maximum water-cement ratio for Mild exposure condition = 0.55

Based on experience, adopt water-cement ratio as 0.5.

0.5<0.55, hence OK.

**Step 3 — Selection of Water Content**

From Table 2 of IS 10262- 2009,

Maximum water content = 186 Kg (for Nominal maximum size of aggregate — 20 mm)

**Table for Correction in water content**

Parameters |
Values as per Standard reference condition |
Values as per Present Problem |
Departure |
Correction in Water Content |

Slump | 25-50 mm | 50-75 | 25 | (+3/25) x 25 = +3 |

Shape of Aggregate | Angular | Angular | Nil | – |

Total | +3 |

Estimated water content = 186+ (3/100) x 186 = 191.6 kg /m^{3}

**Step 4 — Selection of Cement Content**

Water-cement ratio = 0.5

Corrected water content = 191.6 kg /m^{3}

Cement content =

From Table 5 of IS 456,

Minimum cement Content for mild exposure condition = 300 kg/m^{3}

383.2 kg/m^{3} > 300 kg/m^{3}, hence, OK.

This value is to be checked for durability requirement from IS: 456.

In the present example against mild exposure and for the case of reinforced concrete the minimum cement content is 300 kg/m^{3} which is less than 383.2 kg/m^{3}. Hence cement content adopted = 383.2 kg/m^{3}.

As per clause 8.2.4.2 of IS: 456

Maximum cement content = 450 kg/m^{3}.

**Step 5: Estimation of Coarse Aggregate proportion:-**

From Table 3 of IS 10262- 2009,

For Nominal maximum size of aggregate = 20 mm,

Zone of fine aggregate = Zone II

And For w/c = 0.5

Volume of coarse aggregate per unit volume of total aggregate = 0.62

**Table for correction in estimation of coarse aggregate proportion **

Parameter |
Values as per Standard reference condition |
Values as per present problem |
Departure |
Correction in Coarse Aggregate proportion |
Remarks |

W/c | 0.5 | 0.5 | Nil | – | See Note 1 |

Workability | – | pump able concrete | – | -10% | See Note 2 |

Total |
-10% |

**Note 1:** For every ±0.05 change in w/c, the coarse aggregate proportion is to be changed by 0.01. If the w/c is less than 0.5 (standard value), volume of coarse aggregate is required to be increased to reduce the fine aggregate content. If the w/c is more than 0.5, volume of coarse aggregate is to be reduced to increase the fine aggregate content. If coarse aggregate is not angular, volume of coarse aggregate may be required to be increased suitably, based on experience.

**Note 2:** For pump able concrete or congested reinforcement the coarse aggregate proportion may be reduced up to 10%.

Hence,

Volume of coarse aggregate per unit volume of total aggregate = 0.62 x 90% = 0.558

Volume of fine aggregate = 1 – 0.558 = 0.442

#### Step 6: Estimation of the mix ingredients

a) Volume of concrete = 1 m^{3}

b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/100)

= (383.2/3.15) x (1/1000) = 0.122 m^{3}

c) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)

= (191.6/1) x (1/1000) = 0.1916 m^{3}

d) Volume of total aggregates = a – (b + c ) = 1 – (0.122 + 0.1916) = 0.6864 m^{3}

e) Mass of coarse aggregates = 0.6864 x 0.558 x 2.84 x 1000 = 1087.75 kg/m^{3}

f) Mass of fine aggregates = 0.6864 x 0.442 x 2.64 x 1000 = 800.94 kg/m^{3}

**Concrete Mix proportions for Trial Mix 1 **

Cement = 383.2 kg/m^{3}

Water = 191.6 kg/m^{3}

Fine aggregates = 800.94 kg/m^{3}

Coarse aggregate = 1087.75 kg/m^{3}

W/c = 0.5

For trial -1 casting of concrete in lab, to check its properties.

It will satisfy durability & economy.

For casting trial -1, mass of ingredients required will be calculated for 4 no’s cube assuming 25% wastage.

Volume of concrete required for 4 cubes = 4 x (0.15^{3 }x1.25) = 0.016878 m^{3 }

Cement = (383.2 x 0.016878) kg/m^{3} = 6.47 kg

Water = (191.6 x 0.016878) kg/m^{3} =3.23 kg

Coarse aggregate = (1087.75 x 0.016878) kg/m^{3} =18.36 kg

Fine aggregates = (800.94 x 0.016878) kg/m^{3 =} 13.52 kg

**Step 7: Correction due to absorbing / moist aggregate:-**

Since the aggregate is saturated surface dry condition hence no correction is required.

**Step 8: Concrete Trial Mixes****:-*** *

**Concrete Trial Mix 1:**** **

The mix proportion as calculated in Step 6 forms trial mix1. With this proportion, concrete is manufactured and tested for fresh concrete properties requirement i.e. workability, bleeding and finishing qualities.

In this case,

Slump value = 25 mm

Compaction Factor = 0.844

So, from slump test we can say,

Mix is cohesive, workable and had a true slump of about 25 mm and it is free from segregation and bleeding.

Desired slump = 50-75 mm

So modifications are needed in trial mix 1 to arrive at the desired workability.

**Concrete Trial Mix 2:**** **

To increase the workability from 25 mm to 50-75 mm an increase in water content by +3% is to be made.

The corrected water content = 191.6 x 1.03 = 197.4 kg.

As mentioned earlier to adjust fresh concrete properties the water cement ratio will not be changed. Hence

**Cement Content = (197.4/0.5) = 394.8 kg/m ^{3}**

Which also satisfies durability requirement.

Volume of all in aggregate = 1 – [{394.8/(3.15×1000)} + {197.4/(1 x 1000)}] = 0.6773 m^{3}

Mass of coarse aggregate = 0.6773 x 0.558 x 2.84 x 1000 = 1073.33 kg/m^{3}

Mass of fine aggregate = 0.6773 x 0.442 x 2.64 x 1000 = 790.3 kg/m^{3}

**Concrete Mix Proportions for Trial Mix 2 **

Cement = 384.8 kg/m^{3}

Water = 197.4 kg/m^{3}

Fine aggregate =790.3 kg/m^{3}

Coarse aggregate = 1073.33 kg/m^{3}

For casting trial -2, mass of ingredients required will be calculated for 4 no’s cube assuming 25% wastage.

Volume of concrete required for 4 cubes = 4 x (0.15^{3 }x1.25) = 0.016878 m^{3 }

Cement = (384.8 x 0.016878) kg/m^{3} = 6.66 kg

Water = (197.4 x 0.016878) kg/m^{3} =3.33 kg

Coarse aggregate = (1073.33 x 0.016878) kg/m^{3} =18.11 kg

Fine aggregates = (790.3 x 0.016878) kg/m^{3 =} 13.34 kg

In this case,

Slump value = 60 mm

Compaction Factor = 0.852

So, from slump test we can say,

Mix is very cohesive, workable and had a true slump of about 60 mm.

It virtually flowed during vibration but did not exhibit any segregation and bleeding.

Desired slump = 50-75 mm

So , it has achieved desired workability by satisfying the requirement of 50-75 mm slump value .

Now , we need to go for trial mix-3 .

**Concrete Trial Mix**** 3:**** **

In case of trial mix 3 water cement ratio is varied by +10% keeping water content constant. In the present example water cement ratio is raised to 0.55 from 0.5.

An increase of 0.05 in the w/c will entail a reduction in the coarse aggregate fraction by 0.01.

Hence the coarse aggregate as percentage of total aggregate = 0.558 – 0.01 = 0.548

W/c = 0.55

Water content will be kept constant.

**Cement content = (197.4/0.55) = 358.9 kg/m ^{3} **

Hence, volume of all in aggregate

= 1 – [{(358.9/(3.15 x 1000)} + (197.4/1000)] =0.688 m^{3}

Mass of coarse aggregate = 0.688 x 0.548 x 2.84 x 1000 = 1070.75 kg/m^{3}

Mass of fine aggregate = 0.688 x 0.452 x 2.64 x 1000 = 821 kg/m^{3}

**Concrete Mix Proportions of Trial Mix 3 **

Cement = 358.9 kg/m^{3}

Water = 197.4 kg/m^{3}

FA = 821 kg/m^{3}

CA = 1070.75 kg/m^{3}

For casting trial -3, mass of ingredients required will be calculated for 4 no’s cube assuming 25% wastage.

Volume of concrete required for 4 cubes = 4 x (0.15^{3 }x1.25) = 0.016878 m^{3 }

Cement = (358.9 x 0.016878) kg/m^{3} = 6.06 kg

Water = (197.4 x 0.016878) kg/m^{3} =3.33 kg

Coarse aggregate = (1070.75 x 0.016878) kg/m^{3} =18.07 kg

Fine aggregates = (821 x 0.016878) kg/m^{3 =} 13.85 kg

In this case,

Slump value = 75 mm

Compaction Factor = 0.89

So, from slump test we can say,

Mix is stable, cohesive, and workable and had a true slump of about 75 mm.

Desired slump = 50-75 mm

So , it has achieved desired workability by satisfying the requirement of 50-75 mm slump value .

Now , we need to go for trial mix-4.

**Concrete Trial Mix **4:** **

In this case water / cement ratio is decreased by 10% keeping water content constant.

W/c = 0.45

A reduction of 0.05 in w/c will entail and increase of coarse aggregate fraction by 0.01.

Coarse aggregate fraction = 0.558 +.01 =.568

W/c = 0.45 and water content = 197.4 kg/m^{3}

Cement content = (197.4/0.45) = 438.7 kg/m^{3}

Volume of all in aggregate

= 1 – [{438.7/(3.15 x 1000)} + (197.4/1000)] = 0.664 m^{3}

Mass of coarse aggregate = 0.664 x 0.568 x 2.84 x 1000 = 1071.11 kg/m^{3}

Mass of fine aggregate = 0.664 x 0.432 x 2.64 x 1000 = 757.28 kg/m^{3}

**Concrete Mix Proportions of Trial Mix 4 **

Cement = 438.7 kg/m^{3}

Water = 197.4 kg/m^{3}

FA = 757.28 kg/m^{3}

CA = 1071.11 kg/m^{3}

For casting trial -4, mass of ingredients required will be calculated for 4 no’s cube assuming 25% wastage.

Volume of concrete required for 4 cubes = 4 x (0.15^{3 }x1.25) = 0.016878 m^{3 }

Cement = (438.7 x 0.016878) kg/m^{3} = 7.4 kg

Water = (197.4 x 0.016878) kg/m^{3} =3.33 kg

Coarse aggregate = (1071.11 x 0.016878) kg/m^{3} =18.07 kg

Fine aggregates = (757.28 x 0.016878) kg/m^{3 =} 12.78 kg

A local correction due to moisture condition of aggregate is again applied on this proportions. With corrected proportions three concrete cubes are cast and tested for 28 days compressive strength.

A summary of all the trial mixes is given in the following Table.

**Recommended mix proportion of ingredients for grade of concrete M25:**

From Compressive Strength vs. c/w graph for target strength 31.6 MPa we get,

W/c = 0.44

water content = 197.4 kg/m^{3}

Cement content = (197.4/0.44) = 448.6 kg/m^{3}

Volume of all in aggregate

= 1 – [{448.6/(3.15 x 1000)} + (197.4/1000)] = 0.660 m^{3}

A reduction of 0.05 in w/c will entail and increase of coarse aggregate fraction by 0.01.

Coarse aggregate fraction = 0.558 +.01 =.568

Volume of fine aggregate = 1 – 0.568 = 0.432

Mass of coarse aggregate = 0.660 x 0.568 x 2.84 x 1000 = 1064.65 kg/m^{3}

Mass of fine aggregate = 0.660 x 0.432 x 2.64 x 1000 = 752.71 kg/m^{3}

## Kariithi Celsi

great

## Akshat Agrawal

Since, no. of cement bags = 484.6/50 = 9.692 per m3

In order to calculate the quantity of water, CA and FA per bag of cement, Should we divide their respective quantities (in per m3) by a factor of 9.692?

## vijayw

Hello

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