DESIGN OF M35 GRADE CONCRETE:

Design the concrete mix by using IS Method for the following required data:

Characteristic compressive strength required at the end of 28 days=35N/mm2

Maximum site or aggregate =20mm

Shape coarse aggregate =Angular

Degree of workability = 0.75

Degree of quality control = good

Degree exposure = Moderate

Data on materials

Cement used = M35 grade

Specific gravity of cement = 3.15

Coarse aggregate=20mm and 10min mixed in the ratio 70:30

Sand=conforming zone- 2(medium sand)

Assumptions:-

Specific gravity of coarse aggregate- 2.77

Specific gravity of sand -2.42

Free surface moisture

Coarse aggregate= Nil

Sand=1.0%

Water absorption Sand = Nil

Design

Step- 1 standard deviation

Pick up the standard deviation from table- 1 of IS 10262: 2009. Standard deviation is 5.0.

Step- 2 Target strength

Target strength = 𝑓𝑐𝑘 + (1.65×standard deviation)

= 35+ (1.65× 5.0)

= 43.25 𝑵

𝒎𝒎𝟐

Step -3 water cement ratio

For the required target strength Find out water cement ratio from IS 456:2000 table-5. Assume water cement ratio based on experiences is 0.40.

Step -4 maximum water cement ratio

From the table, the maximum water cement ratio for severe exposure is 0.45 so adopt w/c of 0.40

Step-5 water content

From table -2 maximum water content= 186 liter for 20 to 50 mm slump range for 20mm Aggregate.

Required water cement is = 186 + 186×6/100

= 197.00 liters

Based on the trials with super plasticizers water content reduction achieved up to 25%.

Water content = 0.75 x 197.00

= 148.0 lit

Step-6 cement content

Determination of cement content having known the water cement ratio as 0.40 and water

contents 148.0 liters per cu. Mt.

I.e. w/c ratio= 0.40

Water content= 148.0

Cement content= 148 /0.40= 370 kg

Step-7 minimum cement content

Check for minimum cement content as per table no. 5 of IS 456:2000 for durability criteria for

severe exposure is 320 kg/m3. Hence, the worked out cement content satisfies the condition therefore adopt cement content as 370 kg.

Step -8 coarse and fine aggregate

Determination of coarse and fine aggregate content:

V=1.0

W = 148 lit

C = 370 kg

Sfa - 2.42

Sca - 2.72

fa = ? & Ca=?

From table-3 volume of C.A. corresponding 20mm size aggregate and F.A. (zone 2) for w/ c ratio of 0.5 >0.6

In our case w/c ratio =0.40 therefore C.A. volume is required to he increased or to decrease the F.A. content. As w/ c ratio lower by 0.1, the proportion of volume of C.A. is increased by 0.02.

Corrected coarse aggregate volume = .62 + .02 = .64

For pumpable concrete and R.C.C this value should be reduced by 10%

Volume of-C.A. = 0.58

Volume of F.A = 1 - .58 = 0.42

Step-9 Mix calculations

1. Volume of concrete = 1m3

2. Volume of cement = Mass of cement Sp.gr. of cement x 1000 = 370

3.15×1000 = 0.117m3

3. Volume of water = mass of water Sp.gr: of water x 1000

=148 1×1000 =0.148 m3.

4. Volume of chemical admixture # Super plasticizers @ 1 to 2% by mass of cementitious

material Therefore take 1.5% = mass of admixture/ (Sp.gr of admixture *1000)

=7.4/ (1.145×1000)

=0.006m3.

5. Volume of all aggregate = 1-(0.117-0.148-0.006) = 0.729

6. Mass of CA = 0.729× volume of CA × Sp Gr. Of CA × 1000

=0.729 × .58 × 2.72 × 1000

=1150.00 kg

7. Mass of FA = 0.729×volume of FA × Sp Gr. Of FA × 1000

=0.729×0.42×2.42×1000

=740.95 kg

Step 10: Weight proportions, ratio proportions

Total cubes: 14 volume of cube: 0.003375 m3

Total no of cylinder: 6 volume of cylinder: 0.0053 m3

Total conc. in 0.08m3

Proportion by weight: (Kg)

cement water F.A 20mm C.A 10mm C.A admixture

29.60 kg 11.84 lit 59.28 kg 64.4 kg 27.60 kg 480 ml

For CA = 20mm and 10mm ratio 70:30

Fraction I =1150 ×0.7 = 805 kg

Fraction II =1150 ×0.3 = 345 kg

Proportion by ratio:

cement water F.A 20mm C.A 10mm C.A

1 0.40 2.002 2.17 0.93