Project on Concrete Mix Design for various grades of Concrete

Project on Concrete Mix Design for various grades of Concrete

Aim: Calculation of the Concrete Mix Design for M35 grade concrete with fly ash.

Introduction :

1.Stipulation for proportioning:
a) Grade designation : M35
b) Type of cement : OPC 43 grade conforming to IS 8112
c) Type of mineral admixture : Fly conforming to IS 3812
d) Maximum nominal size of aggregate : 20 mm
e) Minimum cement content : 340 kg /m^3 (From Table 5 of IS 456:2000)
f) Maximum water-cement ratio : 0.45 (From Table 5 of IS 456:2000)
g) Workability : 75mm (slump value for pumpable concrete)
h) Exposure condition : Severe (for reinforment concrete)
i) Method of concrete placing : Pumping
j) Degree of supervision :Good
k) Type of aggregate : Crushed angular aggregate
l) Maximum cement content : 750 kg/m^3
m) Chemical admixture type : Superplasticizer

2.Test data for materials:
a) Cement used : OPC 43 grade conforming to IS 8112
b) Specific gravity of cement : 3.15
c) Fly ash : Conforming to IS 3812
d) Specific gravity of fly ash : 2.2
e) Chemical admixture : Superplasticizer conforming to IS 9103
f) Specific gravity of :
-Coarse aggregate : 2.74
-Fine aggregate : 2.74
g) Water absorption :
-Coarse aggregate : 0.5%
-Fine aggregate : 1.0%
h) Free Surface moisture :
-Coarse aggregate : Nil
-Fine aggregate : Nil
i) Sieve Analysis:
-Coarse aggregate : Conforming to Table 2 of IS 383
-Fine aggregate : Conforming to grading Zone1 of Table 4 of IS 383

3.Target Strength for Mix proportioning :
f'ck =fck+ 1.65 S
Where, f'ck = Target average compressive strength at 28 days
fck = Characteristic compressive strength at 28 days
S = Standard deviation

f'ck = 35+(1.65x5)
= 43.25 N/mm^2

4.Selection of water cement ratio:
From table 5 IS 456:2000, For severe exposure condition maximum w/c = 0.45

From IS 10262 2019, w/c for target strength 43.25 N/mm^2 = 0.36

5.Selection of water content:
Maximum water content = 186lit
For 75mm slump, water content = 186 + ((3/100)x186) = 191
Super plasticizer reduces 29% of water content.
Hence new water content = (71/100)x191 = 135.61 ≈ 136 kg

6.Calculating cement content:

Water/cement = 0.36

Cement content = 135.61/0.36 = 375 > 340 kg/m^3

Therefore cement content = 375 kg/m^3

Cementitious material content= 375 × 1.10 = 412.5 kg/m3 ≈ 412.5 kg/m3
Water content = 136 kg/m3
So, water-cementitious ratio= 136/412.5 =0.32

Fly ash @ 30 percent of total cementitious material
content = 375 × 30 percent = 112.2 ≈ 112 kg/m3

Cement (OPC) = 357 – 112 = 245 kg/m3

From Table 5 of IS 456, minimum cementitious content
for ‘severe’ exposure condition= 340 kg/m3
375 kg/m3 > 340 kg/m3, hence O.K.

7.Proportioning of volume of coarse and fine aggregate:

Volume of coarse aggregate for w/c 0.32 = 0.654

Since the concrete is pumpable the volume of coarse aggregate is reduced by 10% = 0.654 x 0.9 = 0.589 ≈ 0.59 m^3

Volume of fine aggregate = 1-0.59 = 0.41 m^3

8. Mix Calculations:
a. Volume of concrete = 1m^3
b. Volume of entrapped air in wet concrete = 0.01 m3
c. Volume of cement = (Mass of cement/specific gravity)x(1/1000)
= (245/3.15)x(1/1000) = 0.077 m^3
d.Volume of fly ash=(Mass of fly ash/Specific gravity of fly ash)x (1/1000)
=(112/2.2)x(1/1000)=0.0509 m^3
e. Volume of water = (Mass of water/specific gravity)x(1/1000)
= (136/1)x(1/1000) = 0.136 m^3
f. Volume of chemical admixture (Super plasticizer) is upto 1% of mass of total cementitious content
= (3.57/1.145)x(1/1000) = 0.0031 m^3
g. Volume of all in aggregate= [(a-b)-(c+d+e+f)]
=[(1-0.01)(0.077+0.0509+0.136+0.0031)]
=0.2673 m^3
h.Mass of coarse aggregate = g x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
= 0.2673 x 0.59 x 2.74 x 1000
= 432.11kg ≈432kg
i.Mass of fine aggregate= g × Volume of fine aggregate × Specific
gravity of fine aggregate × 1000
= 0.2673 x 0.41 x 2.65 x 1000
= 290.42 ≈ 290 kg
9.MIX PROPORTIONS:
Cement = 245 kg/m3
Fly ash = 112 kg/m3
Water = 136 kg/m3
Fine aggregate = 290 kg/m3
Coarse aggregate = 432 kg/m3
Chemical admixture = 3.57 kg/m3
Water-cementitious ratio=0.32

Aim: Calculation of the Concrete Mix Design for M50 grade concrete without fly ash.
Introduction :

1.Stipulation for proportioning:
a)Grade designation: M50
b)Type of cement: OPC 43 grade conforming to IS 8112
c)Maximum nominal size of aggregate: 20mm
d)Exposure type: Severe
e)Minimum cement content: 360 kg/m^3
f)Workability: 75mm slump
g)Method of concrete placing: pumping
h)Type of aggregate: Crushed angular
i)Maximum cement content: 450 kg/m^3
j)Chemical admixture : Super plasticizer
k)Specific gravity of cement: 3.15
l)Specific gravity of fine aggregate: 2.61
m)Specific gravity of coarse aggregate: 2.73
n)Coarse and fine aggregate: Grade 1

2.Target Strength for Mix proportioning:
f'ck = fck + 1.65s
where, s = 5 (From table 1 of IS 10262 2019)
f'ck = 50 + (1.65 x 5) = 58.25 N/mm^2

3.Selection of water cement ratio:

From table 5 IS 456:2000, For severe exposure condition maximum w/c = 0.45
From IS 10262 2019, w/c for target strength 58.25 N/mm^2 = 0.275
0.275< 0.45, hence O.K.

4.Selection of water content:

From Table 4, water content = 186 kg (for 50 mm slump) for 20 mm aggregate.
Estimated water content for 75 mm slump = 186 + ((3/100)x186) = 191

Super plasticizer reduces 29% of water content.

Hence new water content = (77/100)x191 = 147.52 kg ≈ 148 kg

5.Calculating cement content:
Water/cement = 0.275
cement = 148/0.275 = 538.18 kg/m^3 > 360 kg/m^3

Therefore cement content = 539 kg/m^3

6.Proportioning of volume of coarse and fine aggregate:

Volume of coarse aggregate for w/c 0.275 = 0.648

Since the concrete is pumpable the volume of coarse aggregate is reduced by 10% = 0.648 x 0.9 = 0.58

Volume of fine aggregate = 1-0.58 = 0.42

7.Mix calculations:

a. Volume of concrete = 1m^3

b. Volume of cement = (Mass of cement/specific gravity)x(1/1000)
= (539/3.15)x(1/1000) = 0.171 m^3

c. Volume of water = (Mass of water/specific gravity)x(1/1000)
= (148/1)x(1/1000) = 0.148 m^3

d. Super plasticizer is upto 2% of mass of total cementitious content
= (5.39/1.145)x(1/1000) = 0.0047 m^3

e. Volume of all in aggregate = a - (b+c+d)
= 1-(0.171+0.148+0.0047) =0.676 m^3

f. Mass of coarse aggregate = e x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
= 0.676 x0.58 x 2.74 x1000 = 1074.29 kg

g. Mass of fine aggregate = e x volume of fine aggregate x specific gravity of fine aggregate x 1000
=0.676 x0.42 x2.65x 1000 =752.38 kg

8.MIX PROPORTIONS:
Cement = 539 kg/m^3
Water = 148 kg/m^3
Fine aggregate = 752.38 kg/m^3
Coarse aggregate = 1074.29kg/m^3
Chemical admixture = 5.39 kg/m^3
Water cement ratio = 0.275