Challenge

Project on Concrete Mix Design for various grades of Concrete

VVamshi Krishna Birudu9 Jun 2021

Calculate the concrete mix design for M35 grade concrete with fly ash & M50 grade without fly ash.

A: 1. Traget mean strength

F’ck=fck+1.65s

Cement content

Cement content=water/water-cement ratio

S=standard deviation

volume calculations
volume of concrete
Volume of cement=(mass of cement)/(specific gravity of cement)x1/1000
Volume of water = (mass of water)/9specfic gravity of water)x1/1000
Volume of chemical admixture=(mass of admixture)/(specfic gravity of admixture)x1/1000
Volume of all inagregate=(a-(b+c+d))
Mass of coarse aggregate=e x volume of fine aggregate x specific gravity of fine aggregate x 1000
Mass of fine aggregate =e x volume of fine aggregate x specific gravity of fine aggregate x 1000

STEP 1:Stipulatonsfor concrete design mix

Grade designation- M35

Type of cement - OPC 43 grade confirming

Type of mineral admixture - fly ash confirming

Maximum nominal size of aggregate - 20mm

Minimum cement content- 320kg/m3

Maximum water-cement ratio - 0.45

Workability - 100mm slump value

Exposure condition - severe

Method of concrete placing -pumping

Degree of supervision- good

Type of aggregate - crushed angular aggregates

Maximum cement content - 450kg/m3

Chemical admixture type - superplasticizer

Step 2: test data for materials

Cement used - OPC 43

The specific gravity of cement - 315

Fly ash - conforming to is 3812 part 1

The specific gravity of fly ash 2.2

Chemical admixture - superplasticizer

The specific gravity of course aggregate - 2.74

The specific gravity of fine aggregate - 2.74

Water absorption of course aggregate - 0.5%

Water absorption of fine aggregate - 1%

Sieve analysus coarse aggragate - conformining to table 2 of is 383

Sieve analysis fine aggregate - conformimg to grading zone 2

Step3: target strength for mix proportioning

F’ck = fck+1.65s

35+1.65 x 5=43.25 n/mm2

Traget mean strength = 43.25N/mm2

Step 4: Selection of water -cement ratio

Maximum W/C ratio = 0.45

Based on adopted water cement ratio is 0.44

0.44<0.45 hence ok

Adopted W/C ratio =0.44

Step 5: Selection of water content

Maximum water content for 20mm aggregate - 186 liters

Increasing of 3% of water content for every extra 25 slumps from 50mm slump. So 6% is to be added to it - 186+(6/100)*186= 197 liters

Estimated water content for 100mm slump

(Note: Based on trials with superplasticizers is used, the water content can be reduced up to 20% and above)

197*0.8=158 liters

Hence the arrived water content

Water required 158 liters

Step 6: calculation of cement and fly ash required

Adopted w/cratio =0.44

Total cementitous (cement + fly ash )requirement

= 158/0.44 =359kg/m3

Water content/ water cement ratio

Final cementitious content 359 x 1.10= 395kg

395kg/m3> 320 kg/m3 henceok

Cementitious material content = 395

New water cement ratio = 158/395 =118.5 kg/m3

Fly ash content is taken as 30% of thee total cementitious content = 395 x 0.3= 118.5kg/m3

Cement required 395 -118 = 277 kg/m3

Cement quantity saved due to fly ash addition 350-277 =73kg/m3

Cementitious material cement and fly ash required is 395kg/m3 and 118kg/m3 and the new water-cement ratio is 0.4 respectively.

Step 7: volume of coarse aggregate and fine aggregate content

The volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate for the water-cement ratio of 0.50 is 0.62

In the present case, water-cement is 0.44 therefore, the corrected proportion of the volume of course aggregate for the water-cement ratio of 0.44 0.632

For pumpable concrete, these values should be reduced up to 10% therefore, the volume of course aggregate 0.632 x 0.9 = 0.568

The volume of fine aggregate content 1-0.568=0.431

The volume of fine aggregate and coarse aggregate is 0.431 and 0.568

Step 8: mix calculations

The mix calculations as per unit volume of concrete shall be as follows:

the volume of concrete - 1m3

Volume of cement=(mass of cement)/(specific gravity of cement)x1/1000 (359/3.15) x (1/1000) =0.114m3

Volume of water = (mass of water)/9specfic gravity of water)x1/1000 (158/1) x (1/1000) =0.158m3

Volume of chemical admixture=(mass of admixture)/(specfic gravity of admixture)x1/1000 (7.6/1.145) x (1/1000) = 0.006m3

Volume of all in agregate=(a-(b+c+d)) (1-(0.114+0.158+0.0060)=0.755m3

Mass of coarse aggregate=e x volume of fine aggregate x specfic gravity of fine aggregate x 1000 0.722 x 0.568 x2.74 x 1000 =1123kg/m3

Mass of fine aggregate =e x volume of fine aggregate x specfic gravity of fine aggregate x 1000 0.722 x 0.431 x2.64 x 1000 =821kg/m3

Step 9: mix proportions

Cement 359kg/m3

Water 158kg/m3

Fine aggregate 821kg/m3

Coarse aggregate 20mm 1123kg/m3

Chemical agmixture 7.18kg/m3

Water cement ratio 0.44

Mix proportion by weight 1: 2.28:3.12

Result:

Mix proportion by weight of m35 grade with fly ash is 1:2.08:2.72 and its water-cement ratio was found to be 0.4

Part 2: to calculate the concrete mix design for m50 grade concrete without fly ash.

M50 grade concrete without fly ash.

Step 1: stipulations for concrete design mix

Grade designation- M50

Type of cement - OPC 53 grade confirming