Project on Concrete Mix Design for various grades of Concrete

Project on concrete mix design for various grades of concrete

1.Calculate the concrete mix design for M35 grade concrete with fly ash?

Given data grade of concrete M35.

Assumption are

Exposure = very severe

Flyash = 15-30% from the IS-3812 pat-1:2003

Coarse aggregate = 20mm

Slump value = 100mm

Sp. Gravity of coarse and fine aggregate = 2.74

Sp. Gravity of cement = 3.15

Sp. Gravity of admixture = 1.145

Sp. Gravity of fly ash = 2.25

1. Target mean strength of concrete

F’ck=Fck+1.65S

Here Fck = characteristic compressive strength

S= standard deviation from table-1 IS-10262-2009

Therefore F’ck= 35+1.65X5

F’ck = 43.25N/mm2

2. water cement ratio

Exposure condition is very severe from IS-456-2000 table-5, the water cement ratio = 0.45

For the nominal size of aggregate W/C=0.45

3. water content of concrete

As per IS-10262-2009 table-2 for nominal size aggregate 20mm

Water W=186 L

Assumed slump value is 100mm. As per code we need to increase the water content to get the required workability an increase of 3% of water content for every increase of 25mm slump

Therefore for slump value 100mm 6% to be increased

W= 186+186X6/100

= 186+11.16 = 197.16 = 197 say

Now by adding super plasticizer the water content should reduces around 20% as per IS-10262-2009

W = 197-197X20/100

= 197-39.4 = 157.6 = 160 L say

4. cement content of concrete

From the water cement ratio and water content, cement weight can be calculated

W/C=0.45

C = 160/0.45= 355 = 360kgs

From IS-456-2000 table-5 the maximum cement for very severe is 340kg.

Hence C=360 kg ok

From the cement fly ash added is 30% = 360X30/100 = 108kg

Cement content C =360-108 = 252 kgs

5. Volume of coarse aggregate

As per IS-10262-2009 table-3 volume of coarse aggregate = 0.62 for zone-2 of 0.5 water cement ratio

But water cement ratio is 0.45 (at rate of increase of 0.01 for decrease of 0.05 change in water cement ratio)

Hence volume of coarse aggregate = 0.61

Volume of fine aggregate = 1-0.61= 0.39

Calculation

Now consider a. volume of concrete = 1 cum

1. b Volume of cement = mass of cement/sp. Gravity of cement X(1/1000)

= 252/3.15X(1/1000)

=0.08 cum

1. c.Volume of fly ash = Mass of fly ash/sp. Gravity of fly ash X(1/1000)

= 108/2.25X(1/1000)

= 0.048

1. d.Volume of water = Mass of water/ sp. Gravity of water X(1/1000)

= 160/1 X(1/1000)

= 0.16 cum

1. e.Volume super plasticize = Mass of admixture/ sp. Gravity of admixture X(1/1000)

(2% of mass of

cementious material)

= 7.2/1.145 x(1/1000)

= 0.0062 kg/m3

1. f.Volume of aggregate = (a-(b+c+d+e)

= (1-(0.08+0.048+0.16+0.0062)

= (1-0.2942)

= 0.705 kg/ m3

1. g.Mass of coarse aggregate = fX volume of coarse aggregate X sp. Gravity of CA

X1000

= 0.705X0.61X2.74X1000

= 1178 kg/m3

1. h.Mass of fine aggregate = fX volume of fine aggregate X sp. Gravity of FA

X1000

= 0.705X0.39X2.74X1000

= 753 kg/m3

Mix proportion of concrete listed

Cement = 252 kg/m3

Fly ash = 108 kg/m3

Coarse aggregate = 1178 kg/m3

Fine aggregate = 753 kg/m3

Super plasticizes = 7.2 kg/m3

Water cement = 0.45

Mix ratio = 1: 0.42: 2.9 :4.67 Cement: flyash: fine agg: coarse agg

2.Calculate the concrete mix design for M50 grade concrete with fly ash?

Given data grade of concrete M50.

Assumption are

Exposure = extreme

Coarse aggregate = 20mm

Slump value = 75mm

1.Target mean strength of concrete

F’ck=Fck+1.65S

Here Fck = characteristic compressive strength

S= standard deviation from table-1 IS-10262-2009

Therefore F’ck= 50+1.65X5

F’ck = 58.25N/mm2

2. water cement ratio

Exposure condition is extreme from IS-456-2000 table-5, the water cement ratio = 0.40

For the nominal size of aggregate W/C=0.40

3.water content of concrete

As per IS-10262-2009 table-2 for nominal size aggregate 20mm

Water W=186 L

Assumed slump value is 75mm. As per code we need to increase the water content to get the required workability an increase of 3% of water content for every increase of 25mm slump

Therefore for slump value 75mm 3% to be increased

W= 186+186X3/100

= 186+5.58 = 191.58= 192 say

Now by adding super plasticizer the water content should reduces around 20% as per IS-10262-2009

W = 192-192X20/100

= 192-38.4= 153.5 = 155 L say

4. cement content of concrete

From the water cement ratio and water content, cement weight can be calculated

W/C=0.40

C = 155/0.40= 387.5 = 390kgs

From IS-456-2000 table-5 the minimum cement for extreme is 360kg.

Hence C=390 kg ok

5. Volume of coarse aggregate

As per IS-10262-2009 table-3 volume of coarse aggregate = 0.60 for zone-1 of 0.5 water cement ratio

But water cement ratio is 0.40 (at rate of increase of 0.01 for decrease of 0.05 change in water cement ratio).

The concrete is pumpable as per IS-10262-2009 10% reducing of aggregate

Hence volume of coarse aggregate = 0.62-0.62X10/100= 0.558 =0.56 say

Volume of fine aggregate = 1-0.56= 0.44

Calculation

Now consider a.volume of concrete = 1 cum

b.Volume of cement = mass of cement/sp. Gravity of cement X(1/1000)

= 390/3.15X(1/1000)

=0.123 cum

c.Volume of water = Mass of water/ sp. Gravity of water X(1/1000)

= 155/1 X(1/1000)

= 0.155 cum

d.Volume super plasticize = Mass of admixture/ sp. Gravity of admixture X(1/1000)

(2% of mass of

cementitious material)

= 7.8/1.145 x(1/1000)

= 0.0068 kg/m3

e.Volume of aggregate = (a-(b+c+d)

= (1-(0.123+0.155+0.0068)

= (1-0.2848)

= 0.715 kg/ m3

f. Mass of coarse aggregate = e X volume of coarse aggregate X sp. Gravity of CA

X1000

= 0.715X0.56X2.74X1000

= 1097 kg/m3

g.Mass of fine aggregate = e X volume of fine aggregate X sp. Gravity of FA

X1000

= 0.715X0.44X2.74X1000

= 862 kg/m3

Mix proportion of concrete listed

Cement = 390 kg/m3

Coarse aggregate = 1097 kg/m3

Fine aggregate = 862 kg/m3

Super plasticizes = 7.8 kg/m3

Water cement = 0.40

Mix ratio = 1: 2.2: 2.81 Cement: fine agg: coarse agg