Project on Concrete Mix Design for various grades of Concrete
INTRODUCTION:
Concrete Mix Design is the systematic process of choosing economical relative proportions of suitable ingredients from
available sources to produce a Concrete of pre-determined workability,compressive strength & durability.
AIM:
The primary objective of arriving at a Design Mix for Concrete is to produce Concrete of optimum Compressive
strength,workability&structural durability thus acheiving economy of scale.
Mix Design for M 35 Grade concrete with Fly ash:
Data:
1)Grade=M 35,
2)Exposure=Very severe
3)Grade of Cement=OPC 43/Maximum cement content-IS 456-2000=340 Kg/m3
4)Assumed W/C=0.4(From IS 10262-2019)/Maximum W/C-IS 456-2000=0.45
5)Admixture=Fly ash(IS 3812-Part 1),
6)Workability=75mm Slump(Pumpable concrete),
7)Super Plasticizer=2% of Cement content.
a.Target Mean Strength=35+1.65x5=43.25Nmm2
b.Water content for 75mm slump=186+186x1.03=192lts,
c.Reduce 29% for plasticizer=192x0.71=136Lts,
d.Cement&Flyash content=136/0.4=340Kg>=340 hence Ok,
e.Add 10% for admixture=340x1.10=374Kgs/m3
f.Hence W/C=136/374=0.363,
g.Fly ash @25% of cementatious material=374x0.25=93.5 say94Kg/m3
h.Volume of cement=374-94=280Kg/m3,
i.Volume of CA=0.6(Table5-IS 10262-2009),
j.Since W/C is lower by 5%,CA=0.61,
k.For pumpable concrete,CA=0.61x.9=0.55(-10%),
l.Volume of FA=1-0.55=0.45,
m.Consider Volume of Concrete=1M3(a),
n.Volume of cement=280/3.15x0.001=0.088 M3(b),
o.Volume of fly ash=94/2.2x0.001=0.042 M3(c),
p.Volume of water=136/1x0.001=0.136M3(d),
q.Volume of chemical admixture=5.4/1.145x0.001=0.004M3(e),
r.Volume of aggregate=1-(0.088+0.042+0.136+0.004)=0.73M3,
s.Mass of FA=0.73x0.45x2.74x1000=900kg/m3,
t.Mass of CA=0.73x0.55x2.74x1000=1100kg/m3,
Trail mix Ratio=1:2.4:2.94/W/C=0.363.
2.Mix Design for M 50 Grade Concrete:
Data:
1)Concrete Grade=M 50,
2)Cement Grade=53(Minimum cement content=400Kg/m3)
3)Exposure=Moderate,
4)W/C Ratio=0.34(Maximum water content=0.4),
5)Super Plasticizer=2% 0f cement content,
a.Target mean strength=50+1.65x5=58.25Nmm2,
b.Water Content=186 Lts.
c)With Superplasticizer=186x.71=132 Lts,
d)Cement content=132/0.34=388kgs<400Kgs hence assume 400kgs,
e)W/C Ratio=132/400=0.33,
f.Volume of Cement=400Kgs,
g)Volume of CA=0.6,
h.For pumpable concrete,CA=0.6x.09=0.54
i.Volume of FA=1-0.54=0.46,
j.Assume 1 M3 of Concrete(a),
k.Volume of cement=400/3.15x0.001=0.126 M3(b),
l.Volume of water=132/1x0.001=0.132 M3(c),
m.Volume of admixture=8/1.145x0.001=0.006 M3(d),
n.Volume of aggregate=1-(0.126+0.132+0.006)=0.736 M3,
o.Mass of FA=0.736x0.46x2.74x1000=927kg/M3,
p.Mass of Ca=0.736x0.54x2.74x1000=1088 kg/M3
Trial mix Ratio=1:2.31:2.72/W/C=0.33