Project on Concrete Mix Design for various grades of Concrete
1. Concrete Mix Design for M35grade concrete with fly ash .
Aim :
To Calculate the concrete mix design for M35 grade concrete with Fly ash as given below.
Introduction :
Concrete mix design is the process of finding right proportions of cement , sand and aggregate for concrete to achieve target strentgh in strutures .
Concrete mix = cement:sand:aggregate
The concrete mix design involves various steps , calculations and laboratory testing to find right mix proportions . this process is usally to adopted for structures which requires higher grades of concrete.
Given :
Grade = M35
admixture = fly ash.
cement = 0pc 53
maximum nominal size of aggrregate = 20mm
minimum cement content = 340 kg ( table 5 IS456)
maximum cement content = 450 kg as per code
partial replacement of cement by fly ash = 30% as per requirement
maximum water cement ratio = .45 ( table 5 IS456)
workability = 100mm
chemical admixture = pcs type
Procedure :
Test for materials.
a. cement used opc53 (confirming to IS8112)
b. specific gravity of cement = 3.15 (test IS:2720)
c. mineral mixture = fly ash (confirming to IS 12089)
d. specific gravity of fly ash = 2.90
e. chemical admixture @0.6% (IS9103)
f. * specific gravity of CA =2.74
* specific gravity of FA = 2.74
* admixture pce based = 1.12
g. water absorption (test IS 2386)
CA = .5%
FA = 1%
h. sieve analysis (test IS 2386)
CA = confirming to table2 IS383
FA = Zone II confirming to table IS 383
Target strength:
Fck = f'ck + 1.65s ( where s from table 1 IS10262)
= 35 + 1.65 * 5
= 43.25 N/mm^2
Selection of water cement ratio:
Maximum water cement ratio = 0.45 (table 5 IS456)
adopted w/c ratio = 0.45
Selection of water content:
maximum water content = 186L ( from table 2 IS 1026)
adopted water content is for 25 to 50 mm slump , but we need 100 mm slump , clause of 4.2 of IS 1026 suggests to increase 3% of water content to get an extra 25mm slump. hence 6% to get additional 50mm slump of our concrete.
Water content for 100 mm slump = 186 + ( 0.06 + 186)
= 197.16 L
Since we are use super plasticizers :
% of reduction in water content = 20%
water content required after sp = 198 - ( 0.20 * 198)
= 157.73
adopted water content = 160 L
Calculation of cement and fly ash content :
adopted water cement ratio = 0.45
Total cementitious (cement+fly ash) = water content / water cement ratio
= 160/0.45
= 355.55 kg.
here we have increased the cementitious content by 10%
final cementitious content(cement+fly ash) = 356*1.10 = 391.6kg
minimum cement content for severe exposure = 340 kg (table 5 IS 456)
adopted cementitious content = 391.6kg > 340kg
fly ash = 391.6 * 0.30 = 117.48~118kg
cement content required = cementitious content - fly ash
= 391.6 - 117.48 = 274.12
calculation of super plasticizer @ 0.6% to the mass of cementitiuos content = 391.6*(0.6/100) = 2.35kg
Coarse and fine aggregate :
Coarse aggregate ratio = 0.62 (from table 3 IS10262 , for w/c ratii = 0.5)
coarse aggregate ratio for 0.45w/c = 0.63 ( adding 0.01 for every decrease .05 and viceversa)
final fine aggregate ratio = 0.37 ( 1-0.63 = 0.37)
Mix calculation for m35 grade concrete:
a. volume of concrete = 1 cu.m
b. total volume of cement = (cement/ sg * 1000) = 275(3.15*1000) = 0.087 cu.m
c. total volume of fly ash = flyash/(sg*1000) = 118/(2.5*1000) = 0.041 cu.m
d. volume of water = water/(sg*1000) = 160/(1*1000) = 0.160 cu.m
e. volume of super plasticizers = superplasticizers/(sg*1000) = 235/(1.12*1000) = 0.002 cu.m
f. total aggregates requirement = a-(b+c+d+e) = 1-(0.087+0.047+0.160+0.002) = 0.71 cu.m
g. coarse aggregate (CA) = F * CA ratio * sg*1000 = 0.71*0.63*2.74*1000 = 1225.6kg
h. Fine aggregate (FA) = F * FA ratio*sg*1000 = 0.71*0.37*2.74*1000 = 719.8 kg
Ingredients needed for m35 grade concrete :
1. cement = 275 kg
2. flyash = 118 kg
3. FA = 720 kg
4. CA = 1226 kg
5. water = 160 L
6. PCE based super plasticizer = 2L
RESULT :
The mix ratio of M35 Grade concrete by weigth is cement:flyasf:CA:FA:Water =
1 : 0.4 : 2.6 : 4.46 : 0.58 .
2.Concrete Mix Design for M50 grade without fly ash:
Aim:
To calculate concrete mix design for M50 grade of concrete without flyash.
Introduction:
Concrete mix design is the process of finding right proportions of cement , sand and aggregate for concrete to achieve target strentgh in strutures .
Concrete mix = cement:sand:aggregate
The concrete mix design involves various steps , calculations and laboratory testing to find right mix proportions . this process is usally to adopted for structures which requires higher grades of concrete.
Given :
Grade = M50
cement = 0pc 53
maximum nominal size of aggrregate = 20mm
minimum cement content = 360 kg ( table 5 IS456)
maximum cement content = 450 kg as per code
maximum water cement ratio = .45 ( table 5 IS456)
workability = 50 to 75 mm slump
exposure condition : extreme
chemical admixture type : super plasticizer.
degree of supervision : good
Procedure:
Test data for Materials:
a. cement used opc53 (confirming to IS8112)
b. specific gravity of cement = 3.15 (test IS:2720)
c. chemical admixture @0.6% (IS9103)
d. * specific gravity of CA =2.74
* specific gravity of FA = 2.74
* admixture pce based = 1.12
e. water absorption (test IS 2386)
CA = .5%
FA = 1%
f. sieve analysis (test IS 2386)
CA = confirming to table2 IS383
FA = Zone II confirming to table IS 383
Target strength:
Fck = f'ck + 1.65s ( where s from table 1 IS10262)
= 50 + 1.65 * 5
= 58.25 N/mm^2
Selection of water cement ratio:
Maximum water cement ratio = 0.35 (table 5 IS456)
adopted w/c ratio = 0.35
Selection of water content:
maximum water content = 186L ( from table 2 IS 1026)
adopted water content is for 25 to 50 mm slump , but we need 100 mm slump , clause of 4.2 of IS 1026 suggests to increase 3% of water content to get an extra 25mm slump. hence 6% to get additional 50mm slump of our concrete.
Water content for 100 mm slump = 186 + ( 0.06 + 186)
= 197.16 L
based on trails with super plasticizer water content reduction of 27% has been achieved, hence the arrived water content = 197-[197*(27/100)] = 143L
Calculation of cement content:
water - cement ratio =0.35
cement content = 143/0.35 = 410 kg/m^3
from table 5 of IS 456, minimum cement content for extreme exposure condition = 360 kg/m^3
hence , 410 kg/m^3 > 369 kg/m^3 ok.
Proportion of volume of CA & FA content:
Coarse aggregate ratio = 0.62 (from table 3 IS10262 , for w/c ratio = 0.5)
coarse aggregate ratio for 0.35w/c = 0.65 ( adding 0.01 for every decrease .05 and viceversa)
for pumpable concrete , theses values are reduced by 10%.
therefore, volume of coarse aggregate = 0.65*0.9 = 0.585
volume of fine aggregate = 1 - 0.585 = 0.415.
Mix calculation for m50 grade concrete:
a. volume of concrete = 1m^3
b. volume of cement = (mass of cement/SG) *(1/1000) = (410/3.12)*(1/1000) = 0.131 m^3
c. volume of water = (mass of water/(SG*1000)) = (143/(1*1000) = 0.143 m^3
d. volume of super plasticizer = (mass of superplasticizer / (SG*1000)) = (8.2/(1.145*1000)) = 0.007m^3
e. volume of aggregate = (a - ( b + c + d )) = 1 - (0.131 + 0.143 + 0.007) = 0.719 m^3
f. mass of coarse aggregate = volume of all aggregate * volume of CA * SG of CA *1000
= 0.719 * 0.585 * 2.74 * 1000 = 1152.48 kg
g. mass of FA = volume of all aggregate * volume of FA * SG of FA * 1000
= 0.719 * 0.415 * 2.56 * 1000 = 763.86kg
RESULT :
Concrete mix proportion :
cement = 410 kg/ m^3
water = 143 kg/m^3
super plasticizer = 7 kg/m^3
FA = 763.865 kg/m^3
CA = 1152.48 kg/m^3
w/c ratio = 0.35.