Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Mix Design for M35 grade with fly ash
1.Stipulation for proportioning
a.Grade designation :M35
b.Type of cement :OPC 43 grade confirming to IS 8112
c.Maximum nominal size of Aggregate:20mm
d. Minimum cement content :320 kg/m3
(From table 5 of IS456:2000-Exposure condition)
e.maximum water/cement ratio:0.45
(From table 5 of IS456:2000-Exposure condition)
f.Workability :100mm(Slump value for pumpable cone)
g.Exposure condition severe (for reinforced concrete )
h.Method of concrete placinc:pumping
i.Degree of supervision:good
j.Type of aggregate:crushed angular aggregate
k.Maximum cement content:450 kg/m3
l.Chemical Admix type:superplasticizer
m.Type of mineral admix :fly ash confirming to IS 8112
2.Test data for materials
a.cement used :OPC 43 grade confirming to IS 8112
b.Specific gravity of cement :3.15
c.Specific gravity of fly ash:2.2
d.fly ash confirming to IS 3812 (part -1)
c.Chemical Admix:Superplasticizer confirming to IS 9103
d.Specific gravity of coarse aggregate:2.74
Specific gravity of Fine aggregate:2.74
e.Water Absorption
coarse aggregate:0.5%
Fine aggregate:1.0%
F.Free surface moisture (It should be user in dry condition)
coarse aggregate:NIL
Fine aggregate:NIL
g.Sieve analysis
coarse aggregate:IS383 Table-2
Fine aggregate:confirming to grading zone 1 of table 4 of IS 383
3. Target mean strength of conc(f'ck)
f'ck = fck + 1.65s
= 35 + 1.65 x 5
= 43.25 N/mm2
4.Selection of water cement ratio
Based on experience adopted water cement ratio is 0.40
5.Selection of water content
Estimated water content for 100 mm slump
= 186 + (6% x 186)
=197 litre
For superplasticizer water content reduction is 29%
Hence the arrived water content = 140 Litres
6. Calculate the cement content
W/C ratio =0.4
cement content = 140/0.4 =350 kg/m3
7.Proportion of volume of C.A and F.A
For pumpable concrete these values are reduced by 10%
The volume of C.A =0.62x 0.9 = 0.56
Volume of F.A = 1-0.56 = 0.44
Now,TO proportion a mix containing fly ash the following steps are suggested
a.The percentage of fly ash needs to decided based on project requirement and quality of materials
b.In some cases increase in cementitous materials can be warranted.
Increase in cementitous material content and its percentage may be based on experience and trails(This examples with increase the 10% cementitous material content)
cementitous material content = 350 x 1.10 = 385 kg/m3
Water content = 140 kg/m3
Water cement ratio= 140/ 385= 0.364
fly ash at 30% total cementitous material content = 385 / 30% = 115 kg/m3
Cement OPC = 385 - 115 =270 kg/m3
SAving of cement while using fly ash = 350 - 270 =80 kg/m3
Fly ash being utilized = 115 kg/m3
8.Mix calculation
a.volume of concrete = 1m3
b.volume of cement = (mass of cement/Specific gravity of cement) x (1/1000)
= 270/3.15 x 1/1000
= 0.086 m3
C.volume of fly ash= (mass of fly ash/Specific gravity of fly ash) x (1/1000)
= 115/2.2 x 1/1000
= 0.052 m3
d.volume of water= (mass of water/Specific gravity of water) x (1/1000)
= 140/1 x 1/1000
= 0.140 m3
e.volume of chemical admix= (mass of admix/Specific gravity of admix) x (1/1000)
Mass of admix = 2% of cementitous material = (350 x 2%) = 7
= 7/1.145 x 1/1000
= 0.006 m3
f.volume of aggregate (F.A & C.A) = (a-(b+c+d + e))
= (1-(0.086+0.052 +0.140 + 0.006))
= 0.716 m3
f. Mass of C.A = f x volume of C. A x S.G of C.A x 1000
= 0.716 x 0.56 x 2.74 x 1000
= 1098 kg/m3
f. Mass of F.A = f x volume of F. A x S.G of F.A x 1000
= 0.716 x 0.44 x 2.74 x 1000
= 863 kg/m3
9.Mix proportion for trial number 1:
cement = 270 kg/m3
fly ash content = 115 kg/m3
water = 140 kg/m3
F.A= 863 kg/m3
C.A= 1098 kg/m3
chemical admix = 7 kg/m3
W/C ratio = 0.364
Trail mix ratio = 1:2.24:2.85
Mix Design for M50 grade without fly ash
1.Stipulation for proportioning
a.Grade designation :M50
b.Type of cement :OPC 43 grade confirming to IS 8112
c.Maximum nominal size of Aggregate:20mm
d. Minimum cement content :320 kg/m3
(From table 5 of IS456:2000-Exposure condition)
e.maximum water/cement ratio:0.45
(From table 5 of IS456:2000-Exposure condition)
f.Workability :100mm(Slump value for pumpable cone)
g.Exposure condition severe (for reinforced concrete )
h.Method of concrete placinc:pumping
i.Degree of supervision:good
j.Type of aggregate:crushed angular aggregate
k.Maximum cement content:450 kg/m3
l.Chemical Admix type:superplasticizer
2.Test data for materials
a.cement used :OPC 43 grade confirming to IS 8112
b.Specific gravity of cement :3.15
c.Chemical Admix:Superplasticizer confirming to IS 9103
d.Specific gravity of coarse aggregate:2.74
Specific gravity of Fine aggregate:2.74
e.Water Absorption
coarse aggregate:0.5%
Fine aggregate:1.0%
F.Free surface moisture (It should be user in dry condition)
coarse aggregate:NIL
Fine aggregate:NIL
g.Sieve analysis
coarse aggregate:IS383 Table-2
Fine aggregate:confirming to grading zone 1 of table 4 of IS 383
3. Target mean strength of conc(f'ck)
f'ck = fck + 1.65s
= 50 + 1.65 x 5
= 58.25 N/mm2
4.Selection of water cement ratio
Based on experience adopted water cement ratio is 0.40
5.Selection of water content
Estimated water content for 100 mm slump
= 186 + (6% x 186)
=197 litre
For superplasticizer water content reduction is 29%
Hence the arrived water content = 140 Litres
6. Calculate the cement content
W/C ratio =0.4
cement content = 140/0.4 =350 kg/m3
7.Proportion of volume of C.A and F.A
For pumpable concrete these values are reduced by 10%
The volume of C.A =0.62x 0.9 = 0.56
Volume of F.A = 1-0.56 = 0.44
8.Mix calculation
a.volume of concrete = 1m3
b.volume of cement = (mass of cement/Specific gravity of cement) x (1/1000)
= 350/3.15 x 1/1000
= 0.111 m3
c.volume of water= (mass of water/Specific gravity of water) x (1/1000)
= 140/1 x 1/1000
= 0.140 m3
d.volume of chemical admix= (mass of admix/Specific gravity of admix) x (1/1000)
Mass of admix = 2% of cementitous material = (350 x 2%) = 7
= 7/1.145 x 1/1000
= 0.006 m3
e.volume of aggregate (F.A & C.A) = (a-(b+c+d))
= (1-(0.111+0.140 + 0.006))
= 0.743 m3
f. Mass of C.A = e x volume of C. A x S.G of C.A x 1000
= 0.743 x 0.56 x 2.74 x 1000
= 1140 kg/m3
f. Mass of F.A = e x volume of F. A x S.G of F.A x 1000
= 0.743 x 0.44 x 2.74 x 1000
= 896 kg/m3
9.Mix proportion for trial number 1:
cement = 350 kg/m3
water = 140 kg/m3
F.A= 896 kg/m3
C.A= 1140 kg/m3
chemical admix = 7 kg/m3
W/C ratio = 0.4
Trail mix ratio = 1:2.56:3.26