Question:

Published on: 29 June, 2022

**Design concrete mixes of M30 to suit the following data per IS:10262-1982 :**

**Characteristic cube strength = M-30, type of cement - ordinary Portland, fine aggregate natural river sand conforming to grading zone II of table-4 of IS:383-1970 coarse aggregate-crushed (angular) coarse aggregate of 20 mm maximum size conforming to IS:383 code requirements, specific gravities of cement, sand and coarse aggregate are 3·14, 2·63 and 2·61 respectively. Type of exposure - mild, Degree of quality control - very good. Degree of workability = 0·08. use IS:10262-1982. **

Answer:

Grade = M 30

F.A. = grade zone II

C.A. = 20 mm.

Sp. Gr. Of cement = 3.14

Sp. Gr. Of sand = 2.63

Sp. Gr. Of C.A.= 2.61

Expose condition – mild;

Degree of quality control – very good;

Workability = 0.08 = 80mm;

Use code IS: 10262 – 2009 and IS: 456 – 2000;

From IS: 10262 – 2009:- f’ck = fck + 1.65 S

= 30 + 1.65 x 5 = 38.25 N/mm^{2}

S = 5 [ from table 1 of IS: 10262 – 2009].

**Selection of W/C ratio:-** from table 5 of IS: 456 – 2000 maximum W/C ratio = 0.45

For trial 1 adopt W/C ratio as 0.40;

0.4 < 0.45; hence ok.

**Selection of water content:-** from table 2 maximum water content for 20 mm aggregate = 186 lit. [ for25 to 50 mm slump range.

Here 80 mm slump;

As per 4.2 of code the required water content may be increased by 3% for every additional 25 mm slump. = 186 + (3/100) x 186 = 192 lit.

**Calculation of cement content :-** W/C = 0.4

Cement content = 192 / 0.4 = 480 kg/m^{3}.

From table 5 of IS 456, minimum cement content for mild exposure condition = 300 kg/m^{3}.

480 kg/m^{3} > 300 kg/m^{3}.

Proportion of volume of C.A. and F.A. content:- from table 3 volume of coarse aggregate corresponding to 20 mm size agg. And F.A. (zone II) for W/C ratio of 0.5 = 0.62.

In the present case W/C ratio is 0.4. as the w/c ratio is lowered by 0.1, the prop. Of volume of coarse agg. Is increased by 0.02, .i.e. ( -/+ 0.01 for every +/- 0.05 change in w/c ratio).

Proportion of C.A. for w/c ratio of 0.4 = 0.62+0.02 = 0.64

For pumpable concrete this values are reduced by 10%.

Volume of C.A. = 0.64 x 0.9 = 0.58

Volume of F.A. = 1-0.58 = 0.42

**Mix calculation:-**

- Volume of concrete = 1 m
^{3} -
Volume of cement = (mass of cement/sp.gr. of cement) x(1/1000)

= 0.153 m

^{3} -
Volume of water = (mass of water/sp. Gr. Of water)x (1/1000)

= 0.192 m

^{3} - Volume of all in aggregate = 1-(0.153+0.192) = 0.66 m
^{3} - Mass of C.A. = 0.66 x 2.61 x 0.58 x 1000 = 999.11 kg.
- Mass of F.A. = 0.66 x 2.63 x 0.42 x 1000 = 729 kg.

**Mix proportion:-** cement = 480 kg/m^{3}.

water = 192 lit.

F.A= 729 kg/m^{3}.

C.A=999.11 kg /m^{3}.

w/c ratio = 0.4

**proportion:-** 1 : 0.4 : 1.52 : 2.08

two more trials with w/c ratio 0.35 and 0.45 are done.

Subjects

Trending

**What is an air entraining agents? Give some examples. What is efflorescence in concrete structure?**

View : 929

30 June, 2022

Random questions

**Give the physical characteristics of 53 grade OPC.**

29 June, 2022