How To Do Mix Design Of Concrete As Per IS Code 10262 | SSD Condition | Moisture Content Condition
The process of selecting suitable ingredients of concrete and determining their relative proportions with the object of producing concrete of certain minimum strength and durability as economically as possible is defined as MIX DESIGN.
Main Role of Mix Design
To achieve the stipulated minimum strength and durability.
To make the concrete economical.
GIVEN
Grade Of Concrete = M35
Characteristics Compressive Strength Required infield at 28 days = 35 MPA
Types Of Exposure = Moderate
Design Mix Target Slump = 150 – 175 mm
Maximum size of Coarse Aggregate = 20 mm
Fine Aggregate = ZONE – III
Specific Gravity of Cement = 2.93
Specific Gravity of Water = 1
Specific Gravity of Admixture = 1.121
Specific Gravity of Coarse Aggregate = 2.82
Specific Gravity of Fine Aggregate = 2.65
Water Absorption of Coarse Aggregate = 0.86 %
Water Absorption of Fine Aggregate = 1.25 %
1. CONDITION: SSD
TARGET MEAN STRENGTH OF CONCRETE
f´ck = f ck + 1.65 S
f´ck = 35 + 1.65 x 5
f´ck = 43.25 N/mm²
Where, f´ck = Target Mean Compressive Strength at 28 days in N/mm²
f ck = Characteristics Compressive Strength at 28 days in N/mm²
Water-Cement ratio depends on Exposure. Here Exposure is MODERATE.
As per Table 5 of IS 456: 2000 (Page-20),
Maximum Water-Cement ratio = 0.5
Based on Site Experience, Adopt water-cement ratio = 0.5-0.05 = 0.45
DETERMINATION OF WATER CONTENT
From Table 2 of IS 10262: 2009 (Page-3), We get that
Maximum water content for 20 mm coarse aggregate = 186 Litres
[NOTE: This value is for 25 mm to 50 mm slump range. Ref : IS : 10262 : 2009, Clause 4.2, Page-2]
Now, Our Targeted Slump is 150-175 mm
[NOTE: As per IS 10262: 2009, Clause 4.2, we can increase 3% for every additional 25 mm slump]
Estimated water content for 175 mm slump = 186 + (15 x 186 / 100) = 213.9 Litres
As admixture is used, the water content can be reduced by 15% and above.
The actual water to be used = 213.9 – (15 x 213.9 / 100) = 181.82 Litres
CALCULATION OF CEMENT CONTENT
As per Table 5 of IS 456: 2000, Page-20
For Moderate Exposure, Minimum Cement = 300 Kg/m³
Water Cement ratio = 0.45
Water Used = 181.82 Litres
Cement Content = Water Content / Water-Cement ratio
Cement Content = 181.82 / 0.45 = 404 Kg/m³
Which is greater than 300. Hence OK
CALCULATION OF VOLUME OF COARSE AND FINE AGGREGATE CONTENT
From Table 3 of IS 10262: 2009,
Volume of Coarse Aggregate corresponding to 20 mm size aggregate and Fine Aggregate Zone – III for Water cement ratio of 0.5 = 0.64
Now, Actual Water-Cement ratio = 0.45
It is less by (0.5 – 0.45) = 0.05
As the Water-Cement ratio is reduced, it is desirable to increase the coarse aggregate proportion to reduce the fine aggregate.
The Coarse Aggregate is increased at the rate of 0.01 for every decrease in Water-Cement ratio of 0.05
For a decrease of every 0.05 W/C ratio = Coarse Aggregate increased by 0.01
For a decrease of every 1 W/C ratio = Coarse Aggregate increased by 0.01/0.05
For decrease of every 0.05 W/C ratio = Coarse Aggregate increased by (0.01 x 0.05 / 0.05) = 0.01
Corrected Proportion of Volume of Coarse Aggregate = 0.64 + 0.01 + 0.65
Since it is angular aggregate & concrete is to be pumped, the coarse aggregate can be reduced by 10% [IS 10262: 2009, Clause 4.4.1, Page-3]
By reducing 10% we get = 0.65 – (10 x 0.65 / 100) = 0.585
So, Volume of Coarse Aggregate = 0.585
Volume of Fine Aggregate = 1 – 0.585 = 0.415
DESIGN MIX CALCULATION (Ref IS 10262: 2009)
The Mix Design Calculation per unit volume of concrete shall be as follows –
The volume of Concrete – 1 m³
The volume of Cement = Mass of Cement / (Sp. Gravity of Cement x 1000)
= 404 / (2.93 x 1000) = 0.138 m³
The volume of Water = Mass of Water / (Sp. Gravity of Water x 1000)
= 181.82 / (1 x 1000) = 0.182 m³
Volume of Admixture = Mass of Admixture / (Sp. Gravity x 1000)
Assuming a dosage of 1.2% by weight of cementitious material and assuming and Sp. Gravity as 1.121
= (1.2 x 404 / 100) / (1.121 x 1000) = 0.004 m³
The volume of Entrapped Air = 2% for 20 mm Coarse Aggregate
= 2/100 = 0.02 m³
[NOTE: As per IS 10262: 2009, the percentage of entrapped air is zero, Still 2% is considered on practical experiences]
The volume of All Aggregate (Coarse + Fine) = Volume of Concrete – (Volume of Cement + Volume of Water + Volume of Admixture + Volume of Entrapped Air)
= 1 – (0.138 + 0.182 + 0.004 + 0.02) = 0.656 m³
Mass of Coarse Aggregate = Volume of All Aggregate x Volume of Coarse Aggregate x Sp. Gravity x 1000
= 0.656 x 0.585 x 2.82 x 1000 = 1082.2 Kg
Mass of Fine Aggregate = Volume of All Aggregate x Volume of Fine Aggregate x Sp. Gravity x 1000
= 0.656 x 0.415 x 2.65 x 1000 = 721.4 Kg
PROPORTION OF TRIAL MIX : 1
Cement = 404 Kg/m³
Water = 181.82 Kg/m³
Fine Aggregate = 721.4 Kg/m³
Coarse Aggregate = 1082.2 Kg/m³ [60% of 20 mm ; 40% of 10 mm]
Admixture = 4.848 Kg/m³
Water-Cement Ratio = 0.45
2. CONDITION: MOISTURE CONTENT
Water Absorption (In SSD Condition)
Coarse Aggregate = 0.86%
Fine Aggregate = 1.25%
Due to Rain or by somehow, All aggregates mixed with water.
Free Moisture Content in Coarse Aggregate = 2%
Free Moisture Content in Fine Aggregate = 3%
Actual Moisture Content in Coarse Aggregate = 2 – 0.86 = 1.14%
Actual Moisture Content in Fine Aggregate = 3 – 1.25 = 1.75%
In SSD Condition From Mix Design, we got
Coarse Aggregate = 1082.2 Kg/m³
Fine Aggregate = 721.4 Kg/m³
Water = 181.82 Kg/m³
Corrected Mass of Coarse Aggregate = 1082.2 + (1.14 x 1082.2 / 100)
= 1082.2 + 12.34 = 1094.54 Kg
Corrected Mass of Fine Aggregate = 721.4 + (1.75 x 721.4 / 100)
= 721.4 + 12.62 = 734.02 Kg
Corrected Mass of Water = 181.82 – 12.34 – 12.62 = 156.86 Kg
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SSD CONDITION
MOISTURE CONTENT CONDITION
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Padhega India Tab Hi Badhega India | पढ़ेगाइंडियातबहीबढ़ेगाइंडिया
i have a therotical problem ,
water absorption is given forthe C.A – .5%
there is no free mosture content how can i solve that with out free moisture content.
Sir plz start series of R.R.C because apka phadne ka tarika h super se bhot upper h
Sir, please give some information to find estimation and costing
Bahut khoob sir…..
PQC Mix design ka link h to
Sir Shere kare…
Sir…plz
Long wall & short wall k bare me khuchh asani se samaj a jaaye vaise samjaaye..sir..
Pls clarify why cement specific gravity in your calculation is taken at 2.93 when in actual it is 3.15
Sir give idea about steel structure.
Sir aal interview date current month ka sharir kijiye
i have a therotical problem ,
water absorption is given forthe C.A – .5%
there is no free mosture content how can i solve that with out free moisture content.
in the data, he gave sand percentage about 35%
can I take the coarse aggregate 65% directly