Project on Concrete Mix Design for various grades of Concrete

The Concrete Mix Design for M35 grade concrete with fly ash

1. Grade Designation M35
2. Type of cement OPC 53
3. Type of admixture fly ash
4. Maximum nominal size of aggregate 20mm
5. Minimum cement content 340 kg (as per IS456 table 5)
6. Maximum cement content 450kg (as per IS 456)
7. Maximum water cement ratio 0.45(IS 456 :2000 table 5 exposure condition)
8. Workability 100mm
9. Chemical admixture Superplasticizer conforming to IS 9103
10. Type of agrregate : angular crushed aggregate
11. Degree of supervision : Good

Test data for material

1. Cement used : OPC 53 grade conforming to IS 8112
2. Specific gravity 3.15
3. Chemical admixture Superplasticizer conforming to IS 9103
4. Specific gravity of

• Coarse aggregate :2.74
• Fine agregate 2.74

e. Water absorption:

• Coarse aggregate :0.5%
• Fine agregate 1.0%

f. Sieve Analysis:

• Conforming ti IS 383 table 2
• Fine aggregate zone 2 conforming to IS 383 table 4

g. Target Strength for mix proportioning:

f'ck=fck+1.65*S

f'ck= 35 + 1.65* 5 ( S=5 from table 1 IS10262)

f'ck=43.25N/mm^2

h. Selection of water cement ratio

maximum w/c ratio = 0.45 (IS456 table 5)

adopted w/c ratio = 0.45

i. Selection of water content

Maximum water content =186 L (IS 10262 table 2)

Water content for 100mm slump = 186+(.06*186)

= 197.16

Since superplastizers are used water content reduced by 20 %

therefore water adopted = 157.73 = 160 L

j. Caculation of fly ash and cement

Adopted water cement ratio = 0.45

total cementious required ( cement + fly ash)= water content/water cement ratio

= 160/0.45 = 355.55 Kg

Since fly ash is used increase cementious content by 1.1% i.e = 356*1.1 = 391.6 kg

from IS456 table 5 minimum cement content required for severe exposure is 340kg

therefore 340kg<391.6kg

adopted 391.6 kg cementious content

Fly ash content required = 391.6 * 30% (30% replacement )

= 117.48 i.e 118 kg

Cement content required = cementious content - fly ash

= 391.6-117.48

=274.12 kg

Calculation of superplastisizer at 0.6% i.e 391.6*0.6/100 = 2.35 kg

K. Coarse and fine aggregate

coarse aggregate ratio =0.62 (for W/C 0.5 table 3 IS 10262)

coarse aggregate ratio for 0.45 W/C =0.63 (adding 0.01every decrese 0.05)

final fine aggregate ratio = 1 - 0.63 = 0.37

MIX CALCULATION OF M35 GRADE CONCRETE

1. volume of concrete = 1 m^3
2. total volume of cement = cement content / specific gravity * 1000

=275/(3.15*1000)

=0.087 cu.m

C.total volume of fly ash = fly ash content / specific gravity * 1000

=118/(2.5*1000)

=0.041 cu.m

D. volume of water = water content / specific gravity * 1000

= 160/(1*1000)

=0.160 cu.m

E.volume of superplasticizer = superplasticizer content / specific gravity * 1000

= 2.35 /(1.12 *1000)

=.002 cu.m

F.Total aggregate required = A-(B+C+D+E)

= 1- (.087 + .041 + 0.16 + .002)

=0.71 cu.m

G. Coarse Aggregate = F * C.A ratio* specific gravity*1000

= 0.71 * 0.63 *2.74 *1000

= 1225.6 kg

H. Fine Aggregate = F * F.A ratio* specific gravity*1000

= 0.71*0.37*2.74*1000

= 719.8 kg

Therefore Cement = 275kg

Fly ash = 118 kg

Fine aggregate = 720kg

Coarse aggregate = 1226 kg

water = 160L

Superplastisizer = 2L

to find M35 design mix ratio divide calculated value of material by weight of cement

cement :Flyash:F.A : C.A : water

1 : 4.2: 2.6 :4.46 : 0.58