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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