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Construction Technology_Concrete Mix Design_Project 1

CONCRETE MIX DESIGN FOR M35 GRADE CONCRETE WITH FLYASH Aim To calculate the concrete mix design for M35 grade concrete with flyash. Introduction Concrete mix design is a procedure for finding right quantities of cement, sand and aggregates to achieve the desired strength. Accurate concrete mix design makes…

Seethal Sharma
updated on 04 Jan 2021

CONCRETE MIX DESIGN FOR M35 GRADE CONCRETE WITH FLYASH

Aim

To calculate the concrete mix design for M35 grade concrete with flyash.

Introduction

Concrete mix design is a procedure for finding right quantities of cement, sand and aggregates to achieve the desired strength. Accurate concrete mix design makes construction economical.
Flyash or pulverished fuel ash is a finely divided powder, which is byprouct of thermal power plant that uses pulverised coal for raising steam in the boilers.
Since a huge amount of cement is used in concrete in mass concrete construction and the cost of flyash is negligible as compared to that of the cement, the use of flyash concrete brings about a substantial saving in cment consumption and overall construction cost.
With proper mix design the 7 an 28 days strength of flyash concret may be equal or even more than plain concrete. The 90 days strength of flyash concrete may be more than 140% of plain concrete.

Procedure

- Stipulation for proportioning.

a) Grade designation = M35

b) Type of cement = OPC 43 grade confirming to IS8112.

c) Maximum nominaal size of aggregate = 20mm

d) Minimum cement content =340kg/ $m^{3}$ $m^3$ .

e) Maximum w/c ratio = 0.45

f) Workability = 100mm

g) Exposure condition = Severe

h) Method of concret placing = Pumping

i) Degree of supervision = Good

j) Type of aggregate = Crushed angular aggregate

k) Maximum cement content = 450kg/`m^3

l) Chemical admixture used = Superplasticizers

- Test data for material

a) Cement used:OPC 43 grade confirming to IS8112

b) Specific gravity of cement = 3.15

c) Flyash = Confirming to IS3812 (part1)

d) Specific gravity of flyash = 2.2

e) Chemical admixture = Superplasticizers confirming to IS9103

f) Specific gravity of coarse aggregate = 2.74; Fine aggregate = 2.74

g) Water absorption ,Coarse aggregate = 0.5%; Fine aggregate = 1%

h) Free surface moisture, Coarse aggregate = Nil; Fine aggregate = Nil.

i) Sieve analysis, Coarse aggregate= Confirming to table 2 of IS383; Fine aggregate = Confirming to grading zone 1 of Table 4 of IS383.

- Mix proportioning

a) Target mean strength of concrete, ft = fck+1.65S = 35+1.65x5 = 43.25N/ $m m^{2}$ $mm^2$

b) Selection of w/c ratio = From table 5 of IS456:2000, the maximum w/c ratio for severe exposure condition is 0.45. Hence, adopt w/c ratio as 0.4.

c) Selection of water content = From table 2 of IS10262:2009, the maximum water content corresponding to 20mm nominal size of aggregate is 186lit for slump value of 25-50mm.

Here, slump value is 100mm, i.e an additional 50mm slump is present.

From clause 4.2 of IS10262:2009, water content may be increased by about 3% percent for every additional 25mm slump.

Hence, the water content = 186+186x(6/100) = 197 lit.

Since, Superplasticizers are used as admixture, the water content can be reduced by upto 30%.

Let, water be reduced upto 29% i.e, water content = 197x((100-29)/100) = 140 lit.

d) Calculation of cement content:

w/c ratio = 0.4

Hence the cementitious content ( cement+flyash) = 140/0.4 = 350kg/ $m^{3}$ $m^3$

From Table 5 of IS456:2000, for severe exposure condition, minimum cement is 340kg/ $m^{3}$ $m^3$ , which is less than 350kg/ $m^{3}$ $m^3$ . Hence, ok.

Let us take an increase of 10% of cementitious material content.Hence,

Cementitious material content = 350x1.1 = 385kg/ $m^{3}$ $m^3$ .
Water content = 140 lit.
w/c ratio = 140/385 = 0.36
Flyash at 30% of total cementitious material content = 385x0.03 = 115kg/ $m^{3}$ $m^3$ .
Cement content(OPC) = 385-115 = 270kg/ $m^{3}$ $m^3$
Savings in cement content by using flyash = 350-270 = 80kg/ $m^{3}$ $m^3$ .

e) Volume of Coarse aggregate and Fine aggregate:

From table 3 of IS10262:2009, volume coarse aggregate per unit volume total aggregate corresponding to Fine aggreaggte Zone I = 0.6 (corresponding to w/c ratio of 0.5).

Here, the w/c ratio is 0.4, which is lower by 0.1. Hence, coarse aggreagate needs to be increased to decrease the fine aggregate content. So increase the coarse aggregate content by 0.02% (@ of $\pm$ $+-$ 0.01% for every $\pm$ $+-$ 0.05 change in w/c ratio).

Corrected volume of coarse aggregate for w/c ratio of 0.4 = 0.62.

For pumpable concrete, values are reduced by 10%.

Hence, volume of coarse aggregate = 0.62x9 = 0.56

Volume fine aggregate = 1-0.56 = 0.44.

f) Mix calculations

Volume of concrete = 1 $m^{3}$ $m^3$
Volume of cement = (270/3.15)x(1/1000) = 0.086 $m^{3}$ $m^3$
Volume of flyash = (115/2.2)x(1/1000) = 0.052 $m^{3}$ $m^3$
Volume of water = (140/1)x(1/1000) = 0.14 $m^{3}$ $m^3$
Volume of chemical admixture = (7.6/1.145)x(1/1000) = 0.006 $m^{3}$ $m^3$
Volume of aggregate = 1-(0.086+0.052+0.14+0.006) = 0.716 $m^{3}$ $m^3$
Mass of coarse aggregate = 0.716x0.56x2.74x1000 = 1098kg/ $m^{3}$ $m^3$
Mass of fine aggregate = 0.716x0.44x2.74x1000 = 863kg/ $m^{3}$ $m^3$

g) Mix proportion:

Cement = 270kg/ $m^{3}$ $m^3$

Flyash = 115kg/ $m^{3}$ $m^3$

Water content = 140kg/ $m^{3}$ $m^3$

Coarse aggregate = 1098kg/ $m^{3}$ $m^3$

Fine aggregate = 863kg/ $m^{3}$ $m^3$

Chemical admixture = 7kg/ $m^{3}$ $m^3$

w/c ratio = 0.364

Mix ratio = 1:2.24:2.85

Result

Cement = 270kg/ $m^{3}$ $m^3$

Flyash = 115kg/ $m^{3}$ $m^3$

Water content = 140kg/ $m^{3}$ $m^3$

Coarse aggregate = 1098kg/ $m^{3}$ $m^3$

Fine aggregate = 863kg/ $m^{3}$ $m^3$

Chemical admixture = 7kg/ $m^{3}$ $m^3$

w/c ratio = 0.364

Mix ratio = 1:2.24:2.85

CONCRETE MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLYASH

Aim

To calculate the concrete mix design for M50 grade concrete without flyash.

Introduction

Concrete mix design is a procedure for finding right quantities of cement, sand and aggregates to achieve the desired strength.
It is a systematic procedure that produces a cohesive concrete of desired workability in fresh state and desired strength and durability in hardened state.
Accurate concrete mix design makes construction economical.

Procedure

- Stipulation for proportioning.

a) Grade designation = M50

b) Type of cement = OPC 43 grade confirming to IS8112.

c) Maximum nominaal size of aggregate = 20mm

d) Minimum cement content =360kg/ $m^{3}$ $m^3$ .

e) Maximum w/c ratio = 0.4

f) Workability = 100mm

g) Exposure condition = Extreme

h) Method of concret placing = Pumping

i) Degree of supervision = Good

j) Type of aggregate = Crushed angular aggregate

k) Maximum cement content = 450kg/ $m^{3}$ $m^3$

l) Chemical admixture used = Superplasticizers

- Test data for material

a) Cement used:OPC 43 grade confirming to IS8112

b) Specific gravity of cement = 3.15

c) Chemical admixture = Superplasticizers confirming to IS9103

d) Specific gravity of coarse aggregate = 2.74; Fine aggregate = 2.74

e) Water absorption ,Coarse aggregate = 0.5%; Fine aggregate = 1%

f) Free surface moisture, Coarse aggregate = Nil; Fine aggregate = Nil.

g) Sieve analysis, Coarse aggregate= Confirming to table 2 of IS383; Fine aggregate = Confirming to grading zone 1 of Table 4 of IS383.

- Mix proportioning

a) Target mean strength of concrete, ft = fck+1.65S = 50+1.65x5 = 58.25N/ $m m^{2}$ $mm^2$

b) Selection of w/c ratio = From table 5 of IS456:2000, the maximum w/c ratio for extreme exposure condition is 0.4. Hence, adopt w/c ratio as 0.35.

c) Selection of water content = From table 2 of IS10262:2009, the maximum water content corresponding to 20mm nominal size of aggregate is 186 lit for slump value of 25-50mm.

Here, slump value is 100mm, i.e an additional 50mm slump is present.

From clause 4.2 of IS10262:2009, water content may be increased by about 3% percent for every additional 25mm slump.

Hence, the water content = 186+186x(6/100) = 197 lit.

Since, Superplasticizers are used as admixture, the water content can be reduced by upto 30%.

Let, water be reduced upto 29% i.e, water content = 197x((100-29)/100) = 140 lit.

d) Calculation of cement content:

w/c ratio = 0.35

Hence the cement content= 140/0.35= 400kg/ $m^{3}$ $m^3$ .

From Table 5 of IS456:2000, for extreme exposure condition, minimum cement is 360kg/ $m^{3}$ $m^3$ , which is less than 400kg/ $m^{3}$ $m^3$ . Hence, ok.

e) Volume of Coarse aggregate and Fine aggregate:

From table 3 of IS10262:2009, volume coarse aggregate per unit volume total aggregate corresponding to Fine aggreaggte Zone I = 0.6 (corresponding to w/c ratio of 0.5).

Here, the w/c ratio is 0.35, which is lower by 0.15. Hence, coarse aggreagate needs to be increased to decrease the fine aggregate content. So increase the coarse aggregate content by 0.03% (@ of $\pm$ $+-$ 0.01% for every $\pm$ $+-$ 0.05 change in w/c ratio).

Corrected volume of coarse aggregate for w/c ratio of 0.35= 0.63.

For pumpable concrete, values are reduced by 10%.

Hence, volume of coarse aggregate = 0.63x0.9 = 0.567

Volume fine aggregate = 1-0.567 = 0.433.

f) Mix calculations