Concrete TechnologyConcrete Technology

Ch8: Proportioning Concrete Mixes

Lecture 15

Eng: Eyad HaddadEng: Eyad Haddad

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Concrete mix design:Concrete mix design:

Concrete mix design means the process of proportioning the

components of concrete.

Determining of concrete proportions by

Empirical proportioning methods. (by experience)

Calculation methods.

Expressing proportions:

It is used to express the granular materials such as ( cement, coarse

aggregates, fine aggregates) as ratios of weigh or volume.

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Expressing proportions:Expressing proportions:

Example: when we said concrete with ratios ofCement Sand Aggregate

1 2 4

It mean that this concrete consist of 1 particle of cement , 2

particles of sand and 4 particles of aggregates.

Materials can be expressed by ratios between cement and

aggregate (cement/agg ratio)

Example: 1 : 6 it means that we use one particle of cement to 6

particles of aggregates.

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Expressing proportions:Expressing proportions:

Materials can be expressed as how it filled the unit volume

determine by weigh.

Example: cement sand aggregates

300kg 0.4m3 0.8m3

The cumulative of these quantities are 1 m3 after mixing with water.

Cement can be expressed by number of cement pockets

Example: cement sand aggregates

6pockets 0.4m3 0.8m3

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Aggregate-paste relationship:

Concrete is contain of cement paste (active) and aggregates (inert)

The strength of concrete is depended on strength of paste because of

the strength of aggregate is very high to strength of cement paste.

The crack of compressive strength test appeared on the aggregates by

using high strength concrete.

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Mix design methods:

1.1. Empirical method.Empirical method.

2.2. Trial method.Trial method.

3.3. Absolute volume method.Absolute volume method.

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1. Empirical method:

This method is determined the proportions of concrete by experience, this

method is suitable for small jobs.

It applied the components of concrete (cement, sand and aggregates) as

ratios of total weight or volume. Water content can be determined

according to workability of concrete.

Type of concretecementsandaggregatescementaggregates

High strength concrete112=13

Moderate strength concrete124=16

Low strength concrete135=18

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Disadvantages of empirical method:

1. w/c ratio is not bordered.

2. The above proportions me exceed of 1 m3 reached to 1.2 m3.

3. Sand/aggregates ratio is constant 1 : 2 .

4. The compressive strength of concrete is un accurate.

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2. Trial method:

This method is dependent of knowing the w/c ratio.

Steps of trial method:

1. Have 2.5 kg of cement (5% of the cement pocket weigh).

2. w/c ratio is determined by experience or from curves or tables.

3. Mixing the cement and water to produce cement paste.

4. Preparing quantities of sand and SSD aggregates with MSA must

be less than (b/5, 3S/4, and t/3).

5. Mixing the cement paste and aggregate with water to produce

stander concrete .

6. Weigh the quantities of retrained proportions to weigh the using

quantities.

7. Calculate the quantities by weigh and volume.

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3. Absolute volume method:

It supposed that absolute volume of concrete is equal to cumulative

absolute volume of concrete ingredients. (absolute volume of

cement, sand, aggregate and water) as following:

Where:

C = the weigh of Cement by kg used for 1 m3 of concrete.

S = the weigh of Sand by kg used for 1 m3 of concrete.

G = the weigh of aggregates by kg used for 1 m3 of concrete.

W = the weigh of Water by kg used for 1 m3 of concrete.

Gc, Gs, Gg = relative density of cement, sand, and aggregate.

Note: 1 m3 of concrete = 1000 liters.

litrersW

G

G

G

S

G

CvolumeAbsolute

gsc

10000.1

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3. Absolute volume method: (Cont.)

In this method we must determined the flowing according to the

required instruction of strength and workability for fresh concrete:

1. Water content required to 1 m3 of concrete.

2. w/c ratio (by weight) or water quantity required for 1m3 of

concrete

3. fine agg./coarse agg. Ratio.

4. Relative density of cement, fine and coarse aggregate.

Note: the above data are determining by experiences, practices, laboratory tests.

It mean that C, w/c ratio, G/S, Gc, Gs, Gg must be determined before,

then use the above equation.litrers

W

G

G

G

S

G

CvolumeAbsolute

gsc

10000.1

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3. Absolute volume method: (Cont.)

ExampleExample::

Determine the proportions of fresh concrete with the following properties.

The concrete is in the plastic state, compressive strength of hardened

concrete after 28 days is 240 kg/cm2, the percentage passing of mix

aggregates from the sieve No.3/16 is 40%.

Cement relative density Gc=3.15, aggregate relative density (Agg. and sand)

=2.65, unit volume of aggregate (sand and agg.) = 1700kg/cm2.

Solution:

a) determine fine aggregate (sand)/coarse aggregate:

Quantity passed from sieve no. 3/16 is sand and retrained quantity is

aggregate. So sand % = 40%, and aggregate% = 60%.

Note: these percentages are supposed by experience, and we can suppose

it direct as 1:2 (33%sand, 66% aggregate).

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ExampleExample Solution: (Cont.)

b) cement quantity is determined according to compressive strength

after 28 days.

cement quantity / 1 m3 = compressive strength after 28 d + (50-100)kg/cm2cement quantity / 1 m3 = compressive strength after 28 d + (50-100)kg/cm2

so cement quantity = 240 + 60 = 300 kg/cm2.so cement quantity = 240 + 60 = 300 kg/cm2.

c) water content per 1 m3 is determined according to (cement content

in concrete, MSA, and workability).

This quantity can be supposed direct by experience or from the This quantity can be supposed direct by experience or from the

following Table 5-1following Table 5-1

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ExampleExample Solution: (Cont.)

C) water content per 1 m3 is determined according to (cement content

in concrete, MSA, and workability).

This quantity can be supposed direct by experience or from the This quantity can be supposed direct by experience or from the

following Table 5-1following Table 5-1

MAS(mm)w/c ratio for cement content (kg) per one m3 of fresh concrete

200250300350400

100.700.600.500.4750.40

200.650.550.450.4250.385

400.610.480.4250.3850.370

Use w/c ratio = 0.50.

water content = cement content * w/c ratio = 300 kg * 0.5 = 150 liter.

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ExampleExample Solution: (Cont.)

d) aggrecates Weigh = 60/40 of sand weigh = 1.5 sand.

Aggregate weigh = 1.5 * S = 1.5 * 800 = 1200 kg.

litrersSS

volumeAbsolute 10000.1

150

65.2

5.1

65.215.3

300

kgSS so 800)150

15.3

3001000(

65.2

5.2

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ExampleExample (conclusion)

Concrete components ratios by weigh:

Cement sand aggregates water

300kg 800kg 1200kg 150kg

1 : 2.67 : 4 : 0.5

Concrete components ratios by volume:

Cement sand aggregates water

300kg/50 800kg/1700 1200kg/1700 150liter

6 pockets : 0.47m3 : 0.71m3 : 150liter