Project on Concrete Mix Design for various grades of Concrete
Ans 1) Equations used:
1. Target means strength
f'ck = fck + 1.65 $
where, f'ck = target avg compressive strength at 28 days
fck = characteristics compressive strength at 28 days
$ = Standard deviation (table 1 of 1S 10262 = 2000)
2. Cement content
Cement content = water content / w/c ratio
3. Volume calculations
a. Volume of concrete
b. Volume of cement=Mass of cement/sp.gs of cement * 1/1000
c. Volume of water = Mass of water/sp.gs of water * 1/1000
d. Volume of chemical admixture= Mass of admixture/sp.gs of admixture * 1/1000
e. Volume of all aggregate = (a-(b+c+d))
f. Mass of coarse agg = ex vol. of CA*Sp.gs of FA *1000
g. Mass of fine agg = ex vol of FA * Sp.gs of FA * 1000
Part : 1
To calculate the concrete mix design for M35 grade concrete with fly ash
Step 1 : Stipulations for concrete design mix
Grade designation : M35
Type of cement : OPC 43 grade confirming to 1S 8112
Type of mineral admixture : Fly ash confirming to 1S 3812 (part 1)
Max. nominal size of agg : 20mm
Min cement content : 320kg/m3 (table 5 of 15 456:2000-exposure condition)
Max w/c ratio : 0.45 (Table 5 of 1S 456:2000-exposure condition)
Workability : 100mm(Slump value for pumpable concerete)
Exposure condition : Severe (for re forced concrete)
Method of concrete placing : Pumping
Degree of supervision : Good
Type of aggregate : Crushed angular aggregates
Max. cement content : 450kg/m3
Chemical admixture type : Super plasticizer
Step 2:
Cement used : opc 43 grade confirming 1S 8112
Specific gravity of cement : 3.15
fly ash : Confirming to 1S 3812(part 1)
Sp.gs of fly ash : 2.2
Chemical admixture : Super plasticizer confirming to 1S 9103
Sp.gs of CA : 2.74
Sp.gs of FA : 2.74
Water absorption moisture of CA : 0.5%
Water absorption of FA : 1.0%
Free surface moisture of CA : Nill
Free surface moisture of FA : Nill
Sieve analysis : CA confirming to table 2 of 1S 3*3
Sieve analysis : FA confirming to grading zone 11 of table 4 of 1S 383
Step 3 : Target strength for mix proportioning
f'ck = fck + 1.65 $ fck = 3.5
35 + 1.65*5 Therefore, 5=5
f'ck = 43.25N/nm2
Step 4: Selection of w/c ratio
From table 5 of 1S 456:2000 for fever exposure max w/c ratio =0.45
adopted max w/c ratio is 0.44
0.44<0.45, hence OK.
Step 5 : Selection of water content
From table 2 of 1S 10262:2009, max water content for 20mm aggregate.
186 lts (for 25 to 50mm slump range)
Clause 4.2 suggest increasing of 3% of water content for every extra 25mm slump from 50mm slump . So a 6% is to added to it.
186 + (6/100) * 186 = 197lts.
If super plasticizer is used, the water content can be reduced up to 20% & above)
Based on trials with super plasticizer water content reduction 20% has been achieved.
Hence, the arrived water content
197 * 0.8= 158lts.
water required = 158lts.
Step 6 : Calculation of cement & Fly ash required.
adopted w/c ratio = 0.44
Total cementitious (cement +fly ash) requirements.
water content / w/c ratio
= 158/0.44 = 359kg/m3
Hence, we are increasing cementitious content by 10%
Final cementitious content = 359*1.10=395kg.
From table 5 of 1S 456, min cement content for 'severe' exposure condition = 320kg/m3
395kg/m3 > 320kg/m3, hence ok.
cementitious material content = 395.
New water cement ratio = 158/395 = 0.4
Fly ash content is taken as 30% of the total cementitious content
= 395*0.3 = 118.5kg/m3
cement required
= 395-118.5=277kg/m3
Cement content required saved due to fly ash addition
= 350-277=73kg/m3
Cementitious material, cement & fly ash required is 395kg/m3, 277kg/m3 & 118kg/m3 & the new w/c ratio is 0.4 respectively.
Step 7: Volume of CA & FA content
From table 3 (1S 10262:2009) vol of CA corresponding to 20mm size agg & FA (zone 11) for w/c ratio of 0.50 is 0.62.
In the present case w/c ratio is 0.44.
Therefore, as the w/c ratio is lower by 0.06.
The proportion of vol of CA is increased by 0.02 (at the rate of -/+ 0.01 for every +- 0.05 change in water cement ratio).
Therefore, corrected proportion of vol. of CA for the w/c ratio is 0.44=0.632.
Therefore, vol of CA = 0.632*0.9
= 0.568
vol of FA = 1 - 0.568
= 0.431
Therefore, vol of CA & FA = 0.568 & 0.431.
Step 8: Mix calculation
The mix calculations as per unit vol of concrete shall be as follows.
a) Volume of concrete = 1m3
b) Volume of cement = Mass of cement/Sp.gs of cement * 1/1000
= 277/3.15 * 1/1000
= 0.087m3.
c) Volume of fly ash = Mass of fly ash/Sp.gs of fly ash * 1/1000
= 118/2.2 * 1/1000
= 0.053m3
d) Vol of water = Mass of water/Sp.gs of water * 1/1000
= 158/1 * 1/1000
= 0.158m3
e) Vol of chemical admixture = Mass of admixture/Sp.gs of admixture * 1/1000
= 7.6/1.145 * 1/1000
= 0.006m3.
f) Vol of all in aggregate = a-(b+c+d+e)
= 1-(0.087+0.053+0.158 + 0.006)
= 0.696m3
g) Mass of CA = f * vol of CA * Sp.gs of FA * 1000
= 0.696 * 0.568 * 2.74 * 1000
=1077kg/m3
h) Mass of FA = f * vol of FA * Sp.gs of FA * 1000
= 0.696 * 0.431 * 2.74 *1000
= 822kg/m3
Step 9: Mix proportion
Cement - 277kg/m3
fly ash - 118kg/m3
water - 158 lts/m3
FA - 822kg/m3
CA -20mm - 1077kg/m3
w/c ratio - 0.4
Mix proportion by weight = 1: 2.08 : 2.72.
PART 2 :
To calculate the concrete mix deisgn for M50 grade concrete without fly ash.
Step 1 : Stipulation for conceret design mix
Grade designation : M50
Type of cement : OPC 53 grade confirming to 1S B1S 12269-1987.
Max nominal size of agg : 20mm
Min cement content : 320kg/m3 (table 5 of 1S 456:2000- exposure condition)
Max w/c ratio : 0.45( Table 5 of 1S 456:2000 exposure condition)
Workability : 100mm (slump value for pumpable concrete)
Exposure condition : Sever (For reinfused conceret)
Method of concrete placing : Pumping
Degree of supervision : Good
Type of aggregate : Crushed angular agg
Max cement content : 450kg/m3
Chemical admixture type : Super plasticizer
Step 2 : Test data for materials
Cement used : OPC 53 grade conforming 1S 12269-1987
Sp.gs of cement : 3.15
Chemical admixture : Super plasticizer conforming 1S 3103
Sp.gs of CA : 2.74
Sp.gs of FA : 2.64
Water absorption of CA : 0.5%
Water absorption of FA : 1.0%
Free surafce moisture of CA : Nil
Free surface moisture of FA : Nil
Seive analysis : CA : Conforming to table 2 of 1S 383
Seive analysis : FA : Conforming to grading zone 11 of table 4 of 1S 383
Step 3 : Target strength for mix proportions
f'ck = fck + 1.65 $ fck = 50
= 50 + 1.65 * 5 S = 5
f'ck = 58.25N/mm2.
Step 4 : Selection of water cement ratio
The max w/c ratio to design M50 grade
Table 5 of 1S 456-2000 for max w/c ratio = 0.45 sever exposure.
adopted water cement ratio = 0.44
0.44 < 0.45 hence ok
Step 5 : Selection of water content
From table 2 of 1S 10262:2009, max water content for 20mm agg
186 lts (for 25 to 50mm slump range)
Clause 4.2 suggest incraesing 3% of water content for every extra 25mm slump from 50mm slumpp. So a 6% is to be added to it.
186+ (6/100) * 186 = 197lts.
(If super plasticizer is used, the water content can reduction of 20% & above )
Based on trials with super plasticizer water content reduction of 20% has been achieved. Hence, the arrived water content.
197 * 0.8 = 158lts.
Therefore, water required 158lts.
Step 6 : Calculation of cement required
adopted w/c ratio = 0.44
cement content = 158/0.44 = 359kg/m3
From table 5 of 1S 456, min cement content for ' severe' exposure condition = 320kg/m3
359kg/m3 > 320kg/m3. Hence ok,
Cement required is 359kg/m3.
Step 7 : Volume of CA & FA content
From table 3 of (IS 10262:2009) vol of CA corresponding to 20mm size agg & FA (zone 11) for water cement ratio of 0.50=0.62.
The present w/c ratio is 0.44. The vol. of CA is required to be increased to decreased the FA content. As the w/c ratio lower by 0.06. The proportion of CA is increased by 0.02 (at the rate of -/+ 0.01 for every +- 0.05 change in w/c ratio).
Therefore, corrected proportion of volume of CA for the w/c ratio of 0.44=0.632.
For pumpable concrete these values should be reduced up to 10%. Therefore,
Vol. of CA = 0.632*0.9 = 0.568
Vol. of FA = 1 - 0.568 = 0.431
Therefore, Vol. of CA & FA is 0.568 & 0.431 respectively.
Step 8 : Mix calculations
a) Vol. of concrete = 1m3
b) Vol.of cement = mass of cement/Sp.gs of cement * 1/1000
= 359/3.15 * 1/1000
=0.114m3.
c) Vol. of water = mass of water/sp.gs of water * 1/1000
= 158/1 * 1/1000
= 0.158m3.
d) Vol. of chemical admixture = mass of admixture/sp.gs of admixture * 1/1000
= 7.6/1.145 * 1/1000
= 0.006m3.
e) Vol of all in aggregate = a-(b+c+d)
= 1-(0.114+0.158+0.006)
= 0.722m3.
f) Mass of cA = exvol of CA xsp.gs of FA *1000
= 0.722*0.568*2.74*1000
= 1123kg/m3.
g) Mass of FA = ex vol of fA * sp.gs of FA *1000
= 0.722*0.431*2.64*1000
= 821kg/m3.
Step 9: Mix proportions
Cement - 359kg/m3
Water - 158l/m3
FA - 821kg/m3
CA-20mm - 1123kg/m3
Chemical admixture - 7.18kg/m3
w/c ratio - 0.44
Mix proportions by weight = 1:2:28:3.12
RESULTS:
1) Mix proportions by weight of M35 grade of concrete with fly ash is 1:2.08:2.72 & w/c ratio is 0.4.
2) Mix proportions by weight of M50 grade of concrete without fly ash is 1:2:28:3.12 & w/c ratio is 0.44.