Project on Concrete Mix Design for various grades of Concrete
Project
Thopic:
• Calculation of mix design for M35 grade concrete with fly ash.
• Calculation of mix design for M50 grade concrete without fly ash.
1. Calculation of mix design for M35 grade concrete with fly ash.
• Aim: calculation of mix design for M35 grade concrete with fly ash.
• Procedure:
• Data
• Grade of concrete: M35
• Type of cement = OPC 43 grade conforming to IS 8112.
• Type of minral admixture=fly ash conforming to IS 3812 (part 1).
• Maximum size of aggregate -20 mm
• Minimum cement content =320 kg/m^3. (From table 5 of IS 456: 2000 exposure condition).
• Maximum water cement ratio = 0.45. (From table 5 of IS 456:2000 condition is exposure).
• Workability=100 mm (slump for pumpeble concrete). • Exposure condition = severw for reinforcement concrete.
• Method of concrete placing= pumping.
• Degree of supervision Good.
• Maximum cement content = 450 kg/m^3.
• Admixture = superplastisizer.
• Test data for materials:
• cement OPC 43 grade conforming to IS 8112.
• Specific gravity of cement = 3.15
• Fly ash conforming to 3812 part 1.
• Specific gravity of fly ash = 2.2
• Admixture = superplastisizer conforming to 15 9103.
• Specific gravity of course aggregate and fine aggregate = 2.74.
• Water absorption
• Course aggregate = 0.5%
• Fine aggregate = 1.0 %
• Slive analysis = course aggregate conforming to table 2 of IS 383 and fine aggregate conforming to grading zone 1 of the table IS 383
• To calculate the target mean strength:
• as per IS 10262: 2009 the standard deviations = 5 N/MM^2
Fick=Fck +1.65 xS
Fck=35+ 1.65 x5= 43.25 N/MM^2
• Target mean strength = 43.25 N/mm^2
• Selection of water cement ratio:
• from table 5 IS 456 2000 for severe exosure,the maximum w/c ratio is 0.45.
Based on experience adopted water cement ratio is 0.40
0.40 <0.45 hence value is ok.
• Selection of water content:
• from table 2 of IS 10262: 2009 the maximum water content of 20 mm aggregate is 186 lit (For 25 mm to 50 mm slump).
Additional 3 % slump is increased for per 25 mm slump
• Estimated water content for 100 mm slump = 186 + 6/100 x 186 = 197 lit.
• Due to superplastisizer 30 % water content can be reduced.
Based on trial 29 % achieved with plasticizer.
Hence, arrived water content 197 x0.71 140 lit.
• Calculation of cement content:
w/c ratio = 0.40
• Cementious material (cement+ fly ash
Cement content = (140/0.4) = = 350 kg/m^3.
From table 5 of IS 456: 2000 minimum cement content for severe exposure condition is 320 kg/m^3
Now,
• Preparing the mix containing fly ash the following steps are fallow:
• Persentage of fly ash needed to be deside.
• In this design mix we increase the 10% of cementious material contant.
Cementious material content = 350×1.10-385 kg/m^3
• Water content = 140 kg/m^3.
So w/c ratio= (140/385) = 0.364.
• Fly ash @ 30% of cementious material content = 385 x 30% 115 kg/m^3.
• Cement OPC-385-115 kg/m^3.
Saving of cement with using fly ash= 350-270- 80 kg/m^3
Fly ash being utilised = 115 kg/m^3.
• Proportion of fine and course aggregate:
• fine aggregate content from table 3 of IS 10262: 2009 volume of course aggregate corresponding to 20 mm size of aggregate and fine aggregate (zone 1) from w/c ratio is 0.6.
• In present case w/c ratio = 0.4
• Therefore increase the volume of course aggregate and decrease the fine aggregate • W/c ratio is lower by 0.10 the proportion of volume of course aggregate is increasing by 0.02
• Therefore corrected volume of coarse aggregate for the w/c ratio 04-0.62
• For pumpable concrete this volume is reduced by 10% Therefore, • Volume of coarse aggregate = 0.62 x0.9 -0.56
• Volume of fine aggregate = 1-0.56= 0.44.
• Mix calculation:
• Volume of concrete:1 m^3
• Volume of cement: (mass of cement / specific gravity of cement) (1/1000).
(270/3.15)×(1/1000) = 0.0086 m^3
• Volume of cement = 0.0086 m^3
• volume of fly ash: (mass of fly ash / specific gravity of fly ash)(1/1000).
(115/22)x(1/1000) = 0.052 m^3
• Volume of fly ash: 0.052 m^3
• volume of water: (mass of water / specific gravity of water) x (1/1000)
(140/1)(1/1000) = 0.140 m^3.
• Volume of water = 0.140 m^3.
• volume of chemical admixture: (mass of chemical admixture/ specific gravity of chemical admixture)x(
1/1000).
(7.6/1.145)×(1/1000) = 0.006 m^3.
• Volume of chemical admixture: 0.006 m ^3.
• volume of aggregate (FA &CA) = (volume of concrete -(voume of cement + volume of fly ash + volume of water + volume of chemical admixture + ))
(1-(0.086+0.052+0.140+0.006))= 0.716 m^3
Volume of aggregate = 0.716 m^3
• mass of course aggregate= (volume of aggregate xvoure of coarse aggregate specific gravity of course
aggregatex 1000)
Le 0.716* 0.56* 2.74×1000 = 1098 kg/m^3
• Mass of course aggregate = 1098 kg/m^3
mass of fine aggregate : volume of aggregate x volume of fine aggregatex specific gravity of fine aggregate ×1000.
0.716×0.44×2.74×1000 = 863 kg/m^3
Mass of fine aggregate = 863 kg/m^3
• Mix proportion for trial No 1:
• Cement: 270 kg/m^3
• Fly ash: 115 kg/m^3
• Water: 140 kg/m^3
• Fine aggregate: 863 kg/m^3
• Coarse aggregate: 1098 kg/m^3 • Chemical admixture : 7 kg/m^3
• Water cement ratio: 0.364
• Results:
• Trial mix ratio = 1:2.24:1.85
2. Calculation of concrete mix design for M50 grade of concrete without fly ash.
• Data:
• Type of cement: OPC 43 grade conforming to IS 8112.
• Maximum size of aggregate: 20 mmm
• Minimum cement content: 320 kg/m^3......(From table 5 of IS 456: 2000 exposure condition)
• Maximum w/c ratio: 0.45 (From table 5 of IS 456: 2000 exposure condition).
• Workability: 100...... (Slump value of pumpable concrete).
• Exposure condition: severe for reinforced concrete. • Method of concrete placing: pumping
• Degree of supervision: Good
• Typeof aggregate: crushed angular aggregate • Maximum cement content: 450 kg/m^3
• Chemical admixture: superplastisizer.
• Test data for materials:
• Cement used: OPC 43 grade conforming to IS 8112.
• Specific gravity of cement: 3.15
• Chemical admixture : superplastisizer conforming to IS 9103
• Specific gravity of course aggregate and fine aggregate : 2.74
• Water absorption
• Course aggregate: 0.5% Fine aggregate 1.0%
• Free super moisture
• Course aggregate :Nill
• Fine aggregate: Nill
• Slive analysis: The course aggregate conforming to table 2 of IS 383 and fine aggregate conforming zone 1 of table of IS 383.
• calculation of target mean strength for mix:
• Standard deviations: 5 N/mm^2
• Fck = Fck +1.65 S
Fck 50+ 1.65×5= 58.25 N/mm^2
• Target mean strength: 58.25 N/mm^2
• Selection of water cement ratio:
• from table 5 of IS 456: 2000 for severe exposure maximum w/c ratio is 0.45
Based on experience adopted w/c ratio is 0.40.
0.40 0.45....hence ok
• From table 2 of IS 10262: 2009 maximum water content for 20 mm aggregate is 186 lit (For 25 mm to 50 mm slump range).
• Estimated water content for 100 mm slump =
186+ (6/100) × 186 = 197 mm
As per superplastisizer is used to water content can be reduced up to 20% and above based on trials water content reduced of 29 % is achieved with plasticizer.
Hence, the arrived water content: 197x 0.71 = 140 lit.
• Water cement ratio: 0.4
• Cement content: 140/0.4 = 350 kg/m^3 • From table 5 IS 456 minimum cement content for servere exposure condition is 320 kg/m^3
• 350 kg/m^3>350 kg/m^3 hence ok.
• Proportion of volume of coarse aggregate and fine aggregate content:
• from table 3 of IS 10262: 2009 volume of course aggregate corresponding to 20 mm size aggregate and fine aggregate zone I foe w/c ratio is 0.6.
• in present case w/c ratio is 0.40
• Therefore the volume of course aggregate needs to increase to decrease the fine aggregate
• As per w/c ratio is lower by the 0.10 and proportion of volume of coarse aggregate is increased by 0.02.
• Therefore corrected volume of coarse aggregate for the water cement ratio of 0.40 = 0.62
• For pumpable concrete value is reduced by 10 persent
• Volume of coarse aggregate = 0.62×0.9 -0.56
• Volume of fine aggregate= 1-0.56= 0.44.
• Mix calculation:
• volume of concrete: 1m^3
• Volume of cement: (mass of cement /specific gravity of cement)x (1/1000)
(350/3.15)x(1/1000) = 0.111m^3
• Volume of cement is 0.111 m^3
• volume of water = (mass of water / specific gravity of water ) (1/1000).
(140/1) (1/1000) = 0.140 m^3
Volume of water is 0.140 m^3
• volume of chemical admixture: (mass of chemical admixture / specific gravity of chemical admixture )x
(1/1000)
(7.6/1.145) (1/1000) = 0.006m^3 • Volume of chemical admixture is 0.006m^3
• volume of aggregate (Fa& CA): (volume of concrete (volume of cement+ volume of water + volume of chemical admixture ))
(1-(0.111+ 0.140+0.006))= 0.743 m^3
• Volume of aggregate is o.743 m^3
• mass of course aggregate : volume of aggregate xvolume of course aggregatex specific gravity of course.
aggregate x1000
0.743×0.56×2.74×1000=1140 kg/m^3
• Mass of course aggregate is 1140 kg/m^3
• mass of fine aggregate : volume of aggregatex volume of fine aggregate specific gravity of fine aggregate
1000
0.743× 0.44 x 2.74 x 1000 = 896 kg/m^3
• Mass of fine aggregate is 896 kg/m^3
• mix proportion
• Cement:350 kg/m^3
• Water: 140 kg/m^3
• Fine aggregate: 896 kg/m^3
• Coarse aggregate: 1140kg/m^3
• Chemical admixture: 7 kg/m^3
• Water cement ratio: 0.4
• Result
• Trial mix = 1:2.56: 3.26