Project on Concrete Mix Design for various grades of Concrete

Ans 1.

calculate the concrete mix design for M35 grade concrete with fly ash & M50 grade concrete without fly ash

MIX DESIGH M35 GRADE WITH FLY ASH

step 1. stipulations for proportioning

• grade designation : M35
• type of cement : opc 43
• minimum cement content : 320 kg /m^3 ( table 5 is 456 ;2000 based on exposure condition )
• maximum water cement ratio ; 0.45
• minimum nominal size of aggregate : 10 mm
• maximum nominal size of aggregate : 20 mm
• workability : High
• exposure condition : severe
• method of concrete placing : pumping
• degree of supervision : good
• type of aggregate : crushed
• maximum cement content : 450 kg /m^3 ( clouse 8.2.4.2 is 456 : 2000 )
• chemical admixture type : super plasticizers
• minerals admixture : fly ash

step 2. test data for materials

• cement used : opc 43
• specific gravity of cement : 3.14
• specific gravity of chemical admixture ; super plasticizers ( 1.14 according to lab )
• specific gravity of fly ash :2.2 ( is 3812 part 1)
• specific gravity of coarse aggregate :- 2.74
• specific gravity of fine aggregate : 2.6
• water absorption of coarse aggregate : 0.45 %
• water absorption of fine aggregate : 1%
• fine aggregate : zone 2

step -3. target mean strength for mix proportioning

f'ck = fck + 1.65 s

f'ck = 35 +1.65 *5

= 43.25 n/mm^2

step 4. selection of water cement ratio

• from table 5 of is 456;2000 maximum water content : 0.45
• from graph water cement ratio : 0.39
• finalised water cement ratio ( lesser of two ) ; 0.39

step : 5 selection of water content

• from table 2 max water content for 20 mm aggregate = 186 litr ( for 50 mm slump )
• estimated water content for 100 mm slump = 186 + (6/100 ) *186 + 197 litr.
• estimated reduction in water content due to use of super plasticizers = 20 % ( decrease 20% water USE MRWR admixture )
• computed reduction in water content after parcticular trials = 158 litr.

step 6. calculation of cement content

• water cement ratio = 0.39
• cement content = 405.12 kg/m^3
• exposure condition = severe
• from table 5 is 456 minimum cement content for the given exposure condition =320 kg/m^3 ( ok pass )

add fly ash = 10 % ( increase )

= 405 + ( 10/100 ) *405

total = 446 kg

will add fly ash to 30 % of the total quantity of cementatious material = ( 30/100 ) * 446

= 134 kg ( fly ash )

cement = 446-134

=312 kg

step 7. proportion of volume of coarse aggregate and fine aggregate

• maximum aggregate size = 20 mm
• max water cement ratio = 0.45
• from table 5 volume of CA corresponding to above
• MAS and zone of FA as above and for w/c as above = 0.62
• finalized water cement ratio = 0.35 ( cement content = 312, fly ash =134 ) = 446 kg w/c ratio = 158/446 = 0.35
• difference between max to finalized w/c ratio = 0.10

• the rate of change of aggregate content is +- 0.01 for every +- 0.05 change in water cement ratio
• correceted proportion of volume of coarse aggregate for the finalized water cement ratio = 0.02
• concrete is pumpable = yes
• therefore finalized volume of CA = 0.64
• volume of FA = 0.36

step -8. mix calculations

a. volume of concrete = 1 cum

b. volume of cement = ( mass of cement / spg. og cement ) *1/1000

= ( 312/3.14 ) *1/1000

= 0.10 m^3

c. volume of water = mass of water / spg of water *1/1000

=( 158/1)* 1/1000 = 0.158 m^3

d. volume of chemical admixture = ( mass of admixture / spg of admixture ) *1/1000

312*(0.6 /100) = 1.87 kg ( admixture )

( 1.87/1.14)*1/1000 = 0.0016 m^3

e. volume of fly ash = ( mass of fly ash / spg of fly ash ) *1/1000

= ( 134/2.2) *1/1000 = 0.06 m^3

f. volume of all in aggregate : 1-( 0.10+0.158+0.0016+0.06 ) = 0.68 m^3

G. Mass of coarse aggregate = f* volume of CA* sp. gravity of CA*1000

= 0.68*0.64*2.74*1000

=1192 kg

h. mass of fine aggregate = f*volume of FA * sp gravity of FA * 1000

= o.68*0.36*2.6*1000

= 636 kg

since it is angular aggregate & concrete is to be pumped the coarse aggregate ( 20mm size ) can be reduced by 10% and increase 10 % fine aggregate

coarse aggregate : 20mm, 1192*0.6 = 715.2 kg ( 10 % reduce )

10mm, 1192*0.4 = 476.8 kg

final content 20 mm aggregate = 715.2 - ( 10/100 ) * 715.2 = 644 kg

10 mm aggregate = 477 kg

fine aggregate = 636+ ( 10/100 )* 636 = 700kg ( increase 10 % )

step. 9 mix proportion for trial

• cement = 312 kg/m^3
• water = 158 kg / m^3
• fine aggregate = 700 kg /m^3
• coarse aggregate = 1121 kg /m^3
• fly ash = 134 kg / m^3
• chemical admixture = 1.87 kg / m^3
• water cement ratio = 0.35

MIX RATIO :- 1:2.25:3.50

MIX DESIGN OF M50 GRADE CONCRETE WITHOUT FLY ASH

STEP 1. STIPULATIONSFOR PROPORTIONING

• grade designation = m50
• type of cement = opc 43
• max. nominal size of aggregate = 20mm
• workability = high 100 mm slump
• exposure condition = severe
• type of aggregate = crushed
• chemical admixture = super plasticizers

STEP 2. TEST DATA FOR MATERIALS

• specific gravity of cement = 3.15
• sp . gravity of CA = 2.74
• sp. gravity of FA = 2.70
• water absorption of CA = 0.5%
• water absorption of FA = 1 %
• seive analysis FA conforming to grading zone 2

STEP 3. TO FIND FOR MEAN STRENGTH FOR PROPORTIONING

TARGET MEAN STRENGTH, f'ck =fck=1.65 s

= 50+1.65*5 = 58.25 N/mm^2

STEP-4. SELECTION OF WATER CEMENT RATIO

• from table 5 of is 456 maximum water cement ratio = 0.45
• from graph w/c ratio = 0.40
• finalised water cement ratio ( lasser of two ) =0.40

STEP -5. SELECTION OF WATER CONTENT

• from table 4 maximum water content for 20 mm aggregate = 186 litr
• estimated water content for 100 mm slump = 186+ ( 6/100) *186 = 197 litr
• estimated reduction in water content due to use of super plasticizers = 20% ( decrase 20 % water )
• computed reduction in water content after partical trials = 158 litr.

STEP 6. CACLULATION OF CEMENT CONTENT

• w/c ratio = 0.40
• cement content = 158/0.40 = 395 kg/ m^3
• exposure condition = severe
• from table 5 of is 456:2000 min. cement content for the given exposure codition = 320 kg/m^3 ( ok pass )

STEP- 7. PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE

• MAS = 20mm
• zone of FA = 2
• max water cement ratio = 0.45
• from table 5 volume of CA corresponding to above MAS and zone of FA as above and for w/c as above o.62
• finalized water cement ratio = 0.40
• diffrence between max to finalized w/c ratio = 0.05

the rate of change of aggregate content is +- 0.01 for every +- 0.05 change in water ratio

• correceted proportion of volume of CA for the finalized water cement ratio = o.o5
• therefore finalized volume of CA = 0.62
• volume of CA = 0.38

STEP-8. MIX CALCULATIONS

a. colume of concrete = 1 cum

b. volume of cement = ( 395/3.15)*1/1000 = 0.1253 m^3

c. volume of water = ( 158/1)*1/1000 = 0.158 m^3

d. volume of chemical admixture = ( mass od admixture / sp. gravity of admixture )*1/1000 { 395*(0.6/100) } = 2.37 kg

= ( 2.37 /1.14)*1/1000

= 0.0020

e. volume oa all in aggregate = [a-(b+c+d) ]

= [ 1-( 0.1253+0.158+0.0020)]

= 0.714

f. mass of CA = e( volume of CA* sp. gravity of CA*1000)

= 0.714*0.62*2.74*1000

=1213 kg

g. mass of fine aggregate = 0.714*0.38*2.70*1000

= 733 kg

since it is angular aggregate & concrete is to be pumped, the coarse aggregate can be reduced bt 10% and increase 10 % fine aggregate

CA : 20 mm, 1213*0.6 = 728 kg

10 mm. 1213*0.4 = 485 kg

final content 20 mm aggregate = 728- (10/100) = 655 kg ( reduce 10 %)

10 mm aggregate = 485 kg

FA aggregate = 733+(10/100)*733 = 806 kg ( increase 10 %)

STEP-9. MIX PROPORTION FOR TRIAL

• CEMENT = 395 KG/M^3
• WATER = 158 KG/M^3
• FA = 806 KG/M^3
• CA = 1140 KG/M^3
• CHEMICAL ADMIXTURE = 2.37 KG /M^3
• WATER CEMENT RATIO = 0.40

MIX RATIO = 1:2.04:2.88