Project on Concrete Mix Design for various grades of Concrete

Design of M35 grade concrete with fly ash.

Step 1:Calculation of target mean strength

Assumed standard deviation(S) for M35 is- 5 N/mm2

Target mean strenth=fck+ 1.65×S

=35 + 1.65×5

=43.25 N/N/mm mm2

Step2Approximate air content

Entrapped air for nominal maximum size pf 20 mm is 1%of total volume of concrete

Step3:Calculate water cement ratio-

To calculate water cement ratio from graph from code

Compressive strength vs w/c graph need to be followed

According to graph water cement ratio for M35 grade concrete should be - 0.45

Step 4 : Selection of water content

From table-2 of IS 10262:2009

The maximum water content for 25mm to to 50 mm slump= 186 litre

But according to specification our concrete requires a slump of 100mm .Therefore to get the desired slump ,the water content must be increased.Clause 4.2 of IS 1026 recommends increasing the water content by 3% to obtain the additional 25mm slump.Therefore,to obtain a slump of 50mm in addition to our concrete, the water content must be increased by 6%.

Water -content for 100mm slump=186+(6/100)×186

=197.16

As we use superplasticizers we can reduce the water content to gain strength in the concrete. The amount 9f water -content to be reduced depends on the type of super plasticizer.

Percentage of reduction in water content =20%

Water content requirement after SP=198-(.20*198)

=157.73

Adopted water content =160 lit

Step 5:Calculation of cement content and fly ash content

It can be calculated from w/c ratio

Minimum cement content is taken from Table -5 of 456:2000

W=160kg ,w/c=0.45

C=W/0.45

=160/0.45

=355.55 kg

Since flying ash is used, the amount of cementatious content (cement+fly ash) should be increased to get good performance and strength. The percentage of increasing cementatious content depends on experience and research. Here we have increased the cementatious content by 10%

Final cementatious content(Cement +Fly Ash)=356*1.10

=391.6 kg

Minimum Cement Content required for 'severe' exposure=340 kg(table 5-IS456)

Adopted Cementatious content =391.6 kg>340 kg

Fly ash content requirement =391.6*0.30(30% replacement)

=117.48~118 kg

Cement Content requirement =(Cementatious content-Fly ash)

=391.6 -117.48

=274.12

Calculation of Super Plasticizers @0.6 percent to the mass of cementatious content =391.6*(0.6/100)

=2.35 kg

Step6 -Coarse and fine aggregate calculation

Coarse aggregate=0.62(table3-IS 10262 For W/C=0.5)

Coarse Aggregate ratio for 0.45 w/c ratio=0.63(adding 0.01 for every decrease .05 and vice versa)

Final Fine Aggregate ratio=(1-0.63)

=0.37

Step 7-Mix calculation for M35 grade concrete

a) Volume of concrete=1 cu.m

b)Toatal Volume of Cement=Cement/(S.G*1000)

=275/(3.15*1000)

=0.087 cu.m

C)Total Volume of Fly ash=Fly ash /(S.G*1000)

=118/(2.5*1000)

=0.041 cu.m

d)Volume of water=Water/(S.G*1000)

=160/(S.G*1000)

=0.160 cu.m

e) Volume of Superplasticizers =Superplasticizers/(S.G*1000)

=2.35/(1.12*1000)

=0.002 cu.m

f) Toatal aggregate requirement=A-(B+C+D+E)

=1-(0.087+0.041+0.160+0.002)

=0.71 cu.m

g)Coarse aggregate (C.A)=F*C.A ratio*S.G*1000

=0.71*0.63*2.74*1000

=1225.6 kg

h)Fine Aggregate (F.A) =F*F.A ratio*S.G*1000

=0.71*0.37*2.74*1000

=719.8 kg

Cement Flyash Sand Aggregate Water and Superplasticizer needed for M35 grade concrete

Cement-275 kg

Fly ash-118 kg

Fine aggregate -720 kg

Coarse aggregate (20mm)-1226 kg

Water-160 lit

PCE based Super Plasticizer-2L

To find the design mix ratio ,decide the calculated value of all the materials by the weight of cement.

Therefore Mix Ratio of M35 Grade concrete by weight is Cement :Flyash:F.A:C.A:Water=(1:0.4:2.6:4.46:0.58)

Ans-2 Concrete mix design for M50 grade concrete using OPC 53 grade cement(IS -12269)

Step1 Target mean strength calculation

Target mean strength=50 +(5*1.65)

=58.25 Mpa

Step2 Selection of water cement ratio:

Assume water cement ratio=0.35

Step3 Calculation of water-

Approximate water content for 20mm max .size of aggregate=180 kg/m3(As per table no 5,IS 10262)

As plasticizer is proposed we can reduce water content by 20%.

Now water content=180×.8=144 kg/m3

Calculation of cement content:

Water cement ratio=0.35

Water content per cum of concrete =144 kg

Cement content=144/0.35=411.4kg/me

Say cement content =412 kg/m3( As per contract minimum cement content 400 kg/m3)

Hence Ok

Step5 Caluclation for C.A and F.A

Volume 9f concrete= 1 m3

Volume of cement=412/(3.15×1000)=0.1308 m3

Volume of water=144/(1×1000)=0.1440 m3

Volume of admixture=4.994/(1.145×1000)

=0.0043 m3

Total weight of other materials except coarse aggregate =0.1308+0.1440+0.0043=0.2791m3

Volume of coarse and fine aggregate=1-0.2791

=0.7209m3

Volume of F.A =0.7209×0.33=0.2379 m3(Assuming 33% by volume of total aggregate)

Volume of C.A=0.7209-0.2379=0.4830 m3

Therefore weight of F.A =0.2379×2.61×1000

=620.919 kg/m3

Say weight of F.A =621 kg/m3

Therefore weight of C.A =0.4830×2.655×1000

=1282.365 kg/m3

Say weight of C.A=1284 kg/m3

Considering 20mm:10 mm=0.55:0.45

20 mm=706 kg

10 mm=578 kg

Hence Mix details per m3

Increasing cement ,water ,admixture by 2.25% for this trial

Cement =412×1.025=422 kg

Water=144×1.025=147.6 kg

Fine aggregate =621 kg

Coarse aggregate 20mm=706 kg

Coarse aggregate 10mm=578 kg

Admixture=1.2 % by weight of cement =5.064 kg

Water:Cement:F.A:C.A=0.35:1:1.472:3.043