Project on Concrete Mix Design for various grades of Concrete
AIM
- To calculate the concrete mix design for M35 grade concrete with fly ash
INTRODUCTION
- Concrete mix design is the method to finding the acurate proportion of cement, fine aggregates and coarse aggregates for concrete to acive the strength in structure.
- Therefore the concrete mix design is stated as concrete mix terms Cement : Fine aggregates : Course aggregates
- The design is done according to the requirements concrete strength so we can achive the desired state of concrete are in fresh stage and harden stage.
OBJECTIVE
- This project to produce the mix design for M35 grade concrete containg fly ash
- So the mix will be prepared as per IS 10262:2009
PROCEDURE
a) Grade of concrete : M35
b) Type of cement : OPC 43 grade confirming IS 8112
c) Type of mineral admixture : Fly ash conforming IS 3812 (Part 1)
d) Maximun noninal size of aggregates : 20mm
e) Minimun cement content : 320 kg/m^3
f) Maximun water-cement ratio : 0.45
g) Workability : 100mm
h) Exposure condition : Severe(for reinforced concrete)
i) Method of concrete placing : Pumping
j) Degree of supervision : Good
k)Type of aggregates : Crushed angular aggregates
l) Maximun cement content : 450 kg/m^3
m) Chemical admixture : Superplasticizer
Data for material tests
a) Cement used : OPC 43 Grade IS 8112
b) Specific gravity of cement : 3.15
c) Fly ash : Confitming IS 3812(Part 1)
d) Specific gravity of fly ash : 2.2
e) Chemical admixture : Super plasticizier IS 9103
f) Specific gravity of CA : 2.74
g) Specific gravity of FA : 2.74
h) Water absorption of CA : 0.5%
i) Water absorption of FA : 1.0%
j) Free moisture of coaese aggregates : Nil
k) Free moisture of fine aggregates : Nil
l) Sieve analysis of coarse aggregates : Confirming table 2 of IS 383
m) Fine aggregates of fine aggregates : Confirming to grading Zone 1 table 4 of IS 383
1.Target strength for mix proportions
-> f'ck= fck+1.65 S
f'ck = Target mean strength for 28 days
fck = comcpressive strength at 28 days
S = Standard deviation (From table 1 in IS 10262 : 2009)
f'ck = 35+1.65 x 5
= 43.25 N/mm^2
2.Selection of water-cement ratio
Maximun water-cement ratio = 0.45 (From table of IS 456 : 2000)
Based on experience adopt water cemrnt ratio as 0.40
0.40 < 0.45 So it is ok
3.Selection of water content
Maximun water-cement ratio for 20mm aggregates = 186 litre (From table 2 IS 10262 : 2009)
Estimated ewater content for 100 mm slup = 186 + (6%) x 186
= 197.16 litre
As superpasticizer is used the water content can c=reduced upto 30%
So based on trails with super plasticizer water concent reductionupto 28%
so Water content = 197.16 x 0.72
= 139.68 litres
4.Calculation of cement and fly ash content
Water-cement ratio = 0.40
Cementitious material (cement+flyash) = 139.68/0.40
= 349.20 kg/m^3
349.20kg/m^3 > 320kg/m^3 so it ok (From table 5 of IS 456:2000 minimun cement content for severe exposure condition)
So the illustrative example is with increase of 10% cementitious material content
cementitious material content = 350 x 1.10 = 385 kg/m^3
Water content = 139.68 liters
So water-cement ratio = 139.68/385
= 0.362
Fly ash @25% of total cementitious material content = 385 x 25%
= 96.25 kg/m^3
Cement(OPC) = 385-96.25
= 288.75 kg/m^3
Saving cement while using fly ash = 350 - 288.75
= 61.25 kg/m^3
Fly ash being utilized = 96.25 kg/m^3
5.Proportion of volume of corse aggregates and fine aggregates
Volume of coarse aggregates corresponding to 20 mm size aggregates and fine aggregates(Zone 1) for water-cement ratio of 0.50=0.60 (From table 3)
In the present case water-cement ratio is 0.40
Therefore volume of aggregates is required to be increased to decrease the fine aggregates content.
As the water-cement ratio is lower by 0.10 the proportion of volume of coarse aggregates is increased by 0.02(at rate of -/+ 0.01 for every -/+ 0.05 change in water-cement ratio
There fore corrected proportion of volume of coarse aggregates for the water-cement ratio of 0.40=0.64
For the pumpable concrete we need reduce by 10%
Volume of coarse aggregates = 0.64 x 0.9 = 0.576
Volume of fine aggregates content = 1 - 0.576 = 0.424
6.Mix calculation
The mix calculation per unit volume of concrete
a) Volume of concrete = 1m^3
b) Volume of cement = (Mass of cement/Specific gravity of cement) x (1/1000)
= (288.75/3.15) x (1/1000)
= 0.0916 m^3
c) Volume of fly ash = (Mass of fly ash /Specific gravity of fly ash) x 1/1000
= (96.25/2.2) x (1/1000)
= 0.04375 m^3
d) Volume of water = (Mass of water/Sepecfic gravity of water) x (1/1000)
= (139.68/1) x (1/1000)
= 0.13968 m^3
e) Volume of chemical admixture = (Mass of admixture/specific gravity of admixture) x (1/1000)
= (7/1.145) x (1/1000)
= 0.0061 m^3
f) Volume of all aggregates = [a - (b+c+d+e)]
= [1 - (0.0916 + 0.04375 + 0.1396 + 0.0061)]
= [1 - 0.28113]
= 0.718 m^3
g) Mass of coarse aggregates = f x volume of coarse aggregates x specific gravity of coarse aggregates x 1000
= 0.718 x 0.576 x 2.74 x 1000
= 1133.176 kg
h) Mass of fine aggregates = f x volume of fine aggregates x specific gravity of fine aggregates x 1000
= 0.718 x 0.424 x 2.74 x 1000
= 834.14 kg
7.Mix proportion for trail 1
Cement = 288.75 kg/m^3
Fly ash = 96. 25 kg/m^3
Water = 139.68 Kg/m^3
Fine aggregates = 834.14 Kg/m^3
Coarse aggregates = 1133.176 kg/^3
Chemical admixture = 7kg/m^3
RESULT
The mix proportion ratio for M35 grade with fly ash is 1 : 2.88 : 3.92 : 0.33 is Cement : FA : CA : FLYASH
First stage : So the slump shall be measured and water content and doasge of admixture shall be adjusted for acheiving the required slump basked on trail if required the mix proportions shall be reworked for actual water content and checked for durability requirements.
Second stage : Two or more trails having variation of +/- 10% of water-cement ratio in stage 1 shall be caried out and afraph between three water-cement ratio and their correspomding strength shall be plotted to work out the mix proportion for the given target strength for field trails.
So durability required shall be met.
AIM
- To calculate the concrete mix design for M50 grade concrete without fly ash
INTRODUCTION
- Concrete mix design is the method to finding the acurate proportion of cement, fine aggregates and coarse aggregates for concrete to acive the strength in structure.
- Therefore the concrete mix design is stated as concrete mix terms Cement : Fine aggregates : Course aggregates
- The design is done according to the requirements concrete strength so we can achive the desired state of concrete are in fresh stage and harden stage.
OBJECTIVE
- This project to produce the mix design for M50 grade concrete
- So the mix will be prepared as per IS 10262:2009
PROCEDURE
a) Grade of concrete : M50
b) Type of cement : OPC 43 grade confirming IS 8112
c) Maximun noninal size of aggregates : 20mm
d) Minimun cement content : 320 kg/m^3
e) Maximun water-cement ratio : 0.45
f) Workability : 100mm
g) Exposure condition : Severe (For reinforced concrete)
h) Method of concrete placing : Pumping
i) Degree of supervision : Good
j)Type of aggregates : Crushed angular aggregates
k) Maximun cement content : 450 kg/m^3
l) Chemical admixture : Superplasticizer
Data for material tests
a) Cement used : OPC 43 Grade IS 8112
b) Specific gravity of cement : 3.15
c) Chemical admixture : Super plasticizier IS 9103
d) Specific gravity of CA : 2.74
e) Specific gravity of FA : 2.74
f) Water absorption of CA : 0.5%
g) Water absorption of FA : 1.0%
h) Free moisture of coaese aggregates : Nil
i) Free moisture of fine aggregates : Nil
j) Sieve analysis of coarse aggregates : Confirming table 2 of IS 383
k) Fine aggregates of fine aggregates : Confirming of grading Zone 1 table 4 of IS 383
1.Target strength for mix proportions
-> f'ck= fck+1.65 S
f'ck = Target mean strength for 28 days
fck = comcpressive strength at 28 days
S = Standard deviation (From table 1 in IS 10262 : 2009)
f'ck = 50+1.65 x 5
= 58.25 N/mm^2
2.Selection of water-cement ratio
Maximun water-cement ratio = 0.45 (From table of IS 456 : 2000)
Based on experience adopt water cemrnt ratio as 0.40
0.40 < 0.45 So it is ok
3.Selection of water content
Maximun water-cement ratio for 20mm aggregates = 186 litre (From table 2 IS 10262 : 2009)
Estimated ewater content for 100 mm slup = 186 + (6%) x 186
= 197.16 litre
As superpasticizer is used the water content can c=reduced upto 30%
So based on trails with super plasticizer water concent reductionupto 29%
so Water content = 197.16 x 0.71
= 139.98 litr
4.Calculation of cement content
Water-cement ratio = 0.40
Cement content = 139.98/0.40
= 349.45 kg/m^3
349.45kg/m^3 > 320kg/m^3 so it ok (From table 5 of IS 456:2000 minimun cement content for severe exposure condition)
5.Proportion of volume of corse aggregates and fine aggregates
Volume of coarse aggregates corresponding to 20 mm size aggregates and fine aggregates(Zone 1) for water-cement ratio of 0.50=0.60 (From table 3)
In the present case water-cement ratio is 0.40
Therefore volume of aggregates is required to be increased to decrease the fine aggregates content.
As the water-cement ratio is lower by 0.10 the proportion of volume of coarse aggregates is increased by 0.02(at rate of -/+ 0.01 for every -/+ 0.05 change in water-cement ratio
There fore corrected proportion of volume of coarse aggregates for the water-cement ratio of 0.40=0.64
For the pumpable concrete we need reduce by 10%
Volume of coarse aggregates = 0.64 x 0.9 = 0.576
Volume of fine aggregates content = 1 - 0.576 = 0.424
6.Mix calculation
The mix calculation per unit volume of concrete
a) Volume of concrete = 1m^3
b) Volume of cement = (Mass of cement/Specific gravity of cement) x (1/1000)
= (349.45/3.15) x (1/1000)
= 0.11093 m^3
c) Volume of water = (Mass of water/Sepecfic gravity of water) x (1/1000)
= (139.98/1) x (1/1000)
= 0.13998 m^3
d) Volume of chemical admixture = (Mass of admixture/specific gravity of admixture) x (1/1000)
= (7/1.145) x (1/1000)
= 0.0061 m^3
e) Volume of all aggregates = [a - (b+c+d)]
= [1 - (0.11093 + 0.1396 + 0.0061)]
= [1 - 0.256]
= 0.744 m^3
f) Mass of coarse aggregates = e x volume of coarse aggregates x specific gravity of coarse aggregates x 1000
= 0.744 x 0.576 x 2.74 x 1000
= 1174.210 kg
g) Mass of fine aggregates = e x volume of fine aggregates x specific gravity of fine aggregates x 1000
= 0.744 x 0.424 x 2.74 x 1000
= 864.34 kg
7.Mix proportion for trail 1
Cement = 349.45 kg/m^3
Water = 139.98 Kg/m^3
Fine aggregates = 864.34 Kg/m^3
Coarse aggregates = 1174.21 kg/^3
Chemical admixture = 7kg/m^3
RESULT
The mix proportion ratio for M50 grade is 1: 2.47 : 3.36 is Cement : FA : CA
First stage : So the slump shall be measured and water content and doasge of admixture shall be adjusted for acheiving the required slump basked on trail if required the mix proportions shall be reworked for actual water content and checked for durability requirements.
Second stage : Two or more trails having variation of +/- 10% of water-cement ratio in stage 1 shall be caried out and afraph between three water-cement ratio and their correspomding strength shall be plotted to work out the mix proportion for the given target strength for field trails.
So durability required shall be met.