Project on Concrete Mix Design for various grades of Concrete
AIM: To calculate concrete mix design for M35 grade concrete with fly ash and M50 grade concrete without fly ash
Stipulations and proportioning: (Assume & given) for M35 with fly ash
Grade designation: M35; Type of cement: OPC 43 grade conforming IS 842; Type of mineral admixture: Fly ash conforming IS 3812 (p I); Maximum nominal size of aggregate: 20 mm; Exposure condition: Extreme; Workability: 100 mm (Slump value for pumpable concrete); Minimum cement content: 360 Kg/m^3 (as per table 5 of is 456 2000); Maximum W/C ratio: 0.40 (as per IS 456 2000 table 5); Method of concreting: Pumping; Type of aggregate: Crushed angular aggregate; Maximum cement content: 450Kg/m^3; Chemical admixture: Super plasticizer; Degree of supervision: Good.
Test data for materials: Assume
Specific gravity of cement:3.15; Specific gravity of fly ash: 2.2; Specific gravity of chemical admixture: 1.145
Specific gravity and water absorption of Fine aggregate: 2.74 and 1%
Specific gravity and water absorption of coarse aggregate: 2.74 and 0.5%
sieve analysis: Fine aggregates conforming zone 1 of table 4 of IS 383 and coarse aggregate conforming IS 383 table 5
PROCEDURE:
STEP 1: Target mean strength `Fm=Fck+1.65 sigma` = 35+1.65 X 5 = 43.25 N/mm^2
the value of `sigma` = 5 as per IS 10262-2009 of table 1
Step 2:i) W/C ratio from IS 456 2000 for Extreme exposure condition is 0.40
ii) Value of W/C ratio corresponding to 28 days compressive strength of concrete from fig 46 of SP 23- 1982 for Fm =43.25 N/mm^2, W/C ratio = 0.34
W/C ratio is minimum of above two is 0.34.
Step 3: water content as per IS 10262 2009 Table 2, for 20 mm aggregate of slump value 25-50 mm is 186 kg
but assumed slump value is 100 mm so increase of 6% = 186x1.06= 197 lit
As super plasticizer used water can be reduced up to 20% and above so reduce to 29%= 197X0.71= 140 lit it is arrived water content.
Step 4: Calculation of cement content for W/C = 0.34
cementitious content (cement + Fly ash) = 140/0.34 = 412 Kg/m^3
from IS 456 2000 table 5, for extreme condition max cement content is 360 < 412 Kg/m^3 hence OK.
To proportion a mix containing fly ash, the following steps are suggested
a) percentage fly ash needs to be decided based on project requirement and quality of materials
b) In some cases, an increase in cementitious material can be warranted. Increase in cementitious material content and its percentage may be based on experience and trail.
Assuming increase of 10% cementitious material content for this case.
Cementitious material content = 412 X 1.10 = 453 Kg/m^3 and water content is 140 lit
Then new W/C ratio = 140/453 = 0.30
Fly ash @ 30% of total cementitious material content = 453 X 0.3 = 136 Kg/m^3
cement (OPC) = 453-136 = 317 Kg/m^3; saving of cement using fly ash = 412 - 317 = 95 Kg/m^3;
Fly ash utilized = 136 Kg/m^3.
Step 4: Proportion of vol of CA AND FA is as per IS 10262 - 2009 Table 3
for 20 mm zone 1 is proportion is 0.60, in this W/C ratio taken is 0.34, corrected vol of CA of W/C ratio of 0.34 is 0.63
For pumpable concrete, volume of CA and FA can be reduced by 10%
Vol of CA = 0.63 X 0.9= 0.567 m^3 and vol of FA =1-0.567 = 0.433 m^3
MIX calculation
a) Vol of concrete = 1 m^3;
b) Vol of cement = `(Mass of cement)/ (sp G of cement X 1000)`= 317`(317)/(3.15 X 1000)` =0.10 m^3
c) Vol of water = `(wt of water)/(sp G of water X 1000)`= `(140)/(1X1000)`= 0.140 m^3
d) Vol of fly ash = `(wt of fly ash)/(sp G of fly ash X 1000)`=`(136)/(2.2 X 1000)`=0.061 m^3
e) vol of Chemical admixture = `(2% of cement wt)/(sp G of chemical ad mix X1000)`= `(317 X 0.02)/(1.145 x 1000)`= 0.0055 m^3
e) Vol of total aggregates = (a -(b+c+d+e))=(1-(0.100+0.140+0.061+0.0055))=0.694 m^3
f) mass of CA = TA X SP G of CA X vol of CA X1000 = 0.694 X 2.74 X 0.567 X 1000 = 1078 Kg
g) mass of FA = TA X SP G of FA X vol of FA X 1000 = 0.694 X 2.7 X 0.433 X 1000 = 823 Kg
RESULTS:
MIX proportions for M35 with fly ash is
| ingredient | weight proportion |
| cement | 317 Kg |
| Fine aggregates | 823 Kg |
| coarse aggregates | 1078 Kg |
| chemical admixture | 6.34 Kg |
| fly ash | 136 Kg |
| W/C | 0.34 |
for M35--- C:FA:CA = 317:823:1078 I.e., C:FA:CA = 1:2.59:3.4
C:FA:CA:CHEMICAL ADMIX: FLY ASH = 317:136:894:1198:6.34 = 1:2.59:3.4:0.02:0.42
PART 2:
AIM : To calculate concrete mix design for M50 grade concrete without fly ash
Stipulations and proportioning: (Assume & given) for M50 without fly ash
Grade designation: M50; Type of cement: OPC 43 grade conforming IS 842; Type of mineral admixture: NIL Maximum nominal size of aggregate: 20 mm; Exposure condition: very severe; Workability: 100 mm (Slump value for pumpable concrete); Minimum cement content: 340 Kg/m^3 (as per table 5 of is 456 2000); Maximum W/C ratio: 0.45 (as per IS 456 2000 table 5); Method of concreting: Pumping; Type of aggregate: Crushed angular aggregate; Maximum cement content: 450Kg/m^3; Chemical admixture: Super plasticizer; Degree of supervision: Good.
Test data for materials: Assume
Specific gravity of cement:3.15; Specific gravity of chemical admixture: 1.145
Specific gravity and water absorption of Fine aggregate: 2.74 and 1%
Specific gravity and water absorption of coarse aggregate: 2.74 and 0.5%
sieve analysis: Fine aggregates conforming zone 1 of table 4 of IS 383 and coarse aggregate conforming IS 383 table 5
PROCEDURE:
STEP 1: Target mean strength `Fm=Fck+1.65 sigma` = 50+1.65 X 5 = 58.25 N/mm^2`
the value of `sigma` = 5 as per IS 10262-2009 of table 1
Step 2:i) W/C ratio from IS 456 2000 for Extreme exposure condition is 0.45
ii) Value of W/C ratio corresponding to 28 days compressive strength of concrete from fig 47 of SP 23- 1982 for Fm =58.25 N/mm^2, W/C ratio = 0.35
W/C ratio is minimum of above two is 0.35
Step 3: water content as per IS 10262 2009 Table 2, for 20 mm aggregate of slump value 25-50 mm is 186 kg
but assumed slump value is 100 mm so increase of 6% = 186x1.06= 197 lit
As super plasticizer used water can be reduced up to 20% and above so reduce to 29%= 197X0.71= 140 lit it is arrived water content.
Step 4: Calculation of cement content for W/C = 0.35
cement content = 140/0.35 = 400 Kg/m^3
from IS 456 2000 table 5, for extreme condition max cement content is 340 < 400 Kg/m^3 hence OK.
Step 4: Proportion of vol of CA AND FA is as per IS 10262 - 2009 Table 3
for 20 mm zone 1 is proportion is 0.60, in this W/C ratio taken is 0.35, corrected vol of CA of W/C ratio of 0.35 is 0.63
For pumpable concrete, volume of CA and FA can be reduced by 10%
Vol of CA = 0.63 X 0.9= 0.567 m^3 and vol of FA =1-0.567 = 0.433 m^3
MIX calculation
a) Vol of concrete = 1 m^3;
b) Vol of cement = `(Mass of cement)/ (sp G of cement X 1000)`= `(400)/(3.15 X 1000)`=0.126 m^3
c) Vol of water = `(wt of water)/(sp G of water X 1000)`= `(140)/(1X1000)`= 0.140 m^3
d) vol of Chemical admixture = `(2% of cement wt)/(sp G of chemical ad mix X1000)`= `(400 X 0.02)/(1.145 x 1000)`= 0.0069 m^3
e) Vol of total aggregates = (a -(b+c+d+e))=(1-(0.126+0.140+0.0069))=0.727 m^3
f) mass of CA = TA X SP G of CA X vol of CA X1000 = 0.727 X 2.74 X 0.63 X 1000 = 1254 Kg
g) mass of FA = TA X SP G of FA X vol of FA X 1000 = 0.727 X 2.74 X 0.47 X 1000 = 936 Kg
RESULTS:
MIX proportions for M50 with fly ash is
| ingredient | weight proportion |
| cement | 400 Kg |
| Fine aggregates | 936 Kg |
| coarse aggregates | 1254 Kg |
| chemical admixture | 8 Kg |
| W/C | 0.35 |
for M50--- C:FA:CA = 400:936:1254 I.e., C:FA:CA = 1:2.34:3.135
C:FA:CA:CHEMICAL ADMIX = 400:936:1254:8 = 1:2.34:3.135:0.02