Construction Technology Concrete Mix Design Project 1
Mix design for M35 grade of concrete with flyash
Essential data:
- Grade :M35
- Type of cement :OPC 53 grade
- Type of mineral admixture :Flyash
- Maximum nominal size of aggregate :20mm
- workability :100mm (slump)
- Exposure condition :very severe
- Method of concrete placing :pumping
- Chemical admixture type :Superplasticizer
- Specific gravity of cement :3.15
- Specific gravity of flyash :2.2
- Specific gravity of aggregate(CA+FA):2.74
- Specific gravity of admixture :1.145
- Fine aggregate confining to grading zone I
Step 1: Calculation of target mean strength
Target mean strength :fck'=fck+1.65S
where
fck=characteristic compressive strength
S=Standard deviation taken from IS 10262:2009 Table 1
fck'=35+1.65*5=43.25 N/mm^2
Step 2: Calculation of water cement ratio
From IS 10262:1982
Fig 1: 28 days compressive strength of concrete in N/mm^2 VS water cement ratio.
Water-cement ratio is 0.35
From IS 10262:1982
Fig 2:Concrete strength for different grade of cement VS water cement ratio.
Water-cement ratio is 0.40
From IS 456:2000
Table 5:Maximum water-cement ratio and minimum grade of concrete for different exposures.
Water-cement ratio is 0.45
From trail or error or by experience the water-cement ratio is taken as 0.4<0.45 Hence OK
Step 3: Calculation of water content
From IS 10262:2009
Table 2:Maximum water content per cubic metre of concrete for nominal maximum size of aggregate
For 20mm size of aggregate water-content is 186 kg for 25mm-50mm slump
For every 25mm slump extra 3% is added so for 100mm slump 6% is added
Water-content=186+(186*(6/100))=197 kg
As superplasticizer is used so water-content is reduced upto 30%
Therefore from trail and error
water-content is 197*0.71=140 (29% is reduced)
Step 4: Calculation of cementitious content
a) Calculation of cement content.
w/c=0.4
Cement-content= 140/0.4=350 kg
From IS 456:2000 Table 5
For very severe exposure condition minimum cement-content should be 340kg/m^3
since 350kg/m^3 > 320kg/m^3 Hence OK
b) Calulation of fly ash content
In certain situation an increase in the cemetitious material is done and this is done by experience or trail and error method.
cementitious content=350*1.10=385kg/m^3 (10% is increased)
water-content= 140kg/m^3
then, w/c=(140/385)=0.3636
Fly ash content= 385*0.3=115 kg/m^3 (30% of total cementitious material)
Cement used=385-115= 270kg/m^3.
Cement saved=350-270= 80kg/m^3
Step 5: Calculationing the proportion of volume of coarse and fine aggegate content
From IS 10262:2009
Table 5: volume of coarse aggregate per unit volume of total aggregate for different zones of fine aggregate
For Zone I, 20mm nominal maximum size of aggregates
Volume of coarse aggregate=0.60 for w/c=0.5
For every 0.05 reduction in w/c volume of CA is increased by 0.01
Therefore, for w/c=0.4 the volume of CA=0.62
We are using a pumpable concrete so 10% reduction in volume of CA is done.
Volume of CA=0.62*0.9=0.56
Volume of FA=(1-0.56) =0.44
Step 6: Mix calculations
a) Volume of concrete=1m^3
b) Volume of cement=(Mass of cement/Specific gravity of cement)*(1/1000)
=(270/3.15)*(1/1000)=0.086 m^3
c) Volume of flyash =(Mass of flyash/Specific gravity of flyash)*(1/1000)
=(115/2.2)*(1/1000)=0.052m^3
d) Volume of water =(Mass of water/Specific gravity of water)*(1/1000)
=(140/1000)=0.140m^3
e) Volume of Admixture=(Mass of admixture/Specific gravity of admixture)*(1/1000)
=(7/1.145)*(1/1000)=0.006m^3
(Note: volume of admixture is 2% of the cement content= 350*0.02=7)
f) Volume of Aggregates= a-(b+c+d+e)
= 1-(0.086+0.052+0.140+0.006)= 0.716 m^3
g) Volume of coarse aggregate= e*volume of aggregate*Specific gravity of aggregate*1000
=0.56*0.716*2.74*1000=1099 kg ( e=proportion of coarse aggregate)
h) Volume of fine aggregate= (1-e)*volume of aggregate*specific gravity of fine aggregate*1000
=(1-0.56)*0.716*2.74*1000
=0.44*0.716*2.74*1000= 863 kg
Step 7: Mix proportion
Cement : 270 kg/m^3
flyash : 115kg/m^3
water : 140kg/m^3
Fine aggregate : 863kg/m^3
coarse aggrgate : 1099kg/m^3
Admixture : 7kg/m^3
Water-cement ratio :0.36
Step 8: Mix proportion for 50 kg of cement ( 1bag of cement)
C: F: FA : CA: Adm
1: 0.43: 3.19 : 4.067: 0.02
Mix design for M50 grade of concrete.
Essential data:
- Grade :M50
- Type of cement :OPC 53 grade
- Maximum nominal size of aggregate :40mm
- workability :125mm (slump)
- Exposure condition :Extreme
- Method of concrete placing :pumping
- Chemical admixture type :Superplasticizer
- Specific gravity of cement :3.15
- Specific gravity of aggregates(CA+FA) :2.74
- Specific gravity of admixture :1.145
- Fine aggregate confining to grading zone IV
Step 1: Calculation of target mean strength
Target mean strength :fck'=fck+1.65S
where
fck=characteristic compressive strength
S=Standard deviation taken from IS 10262:2009 Table 1
fck'=50+1.65*5=58.25 N/mm^2
Step 2: Calculation of water cement ratio
From IS 456:2000
Table 5:Maximum water-cement ratio and minimum grade of concrete for different exposures.
Water-cement ratio is 0.40
From trail or error or by experience the water-cement ratio is taken as 0.35<0.4 Hence OK
Step 3: Calculation of water content
From IS 10262:2009
Table 2:Maximum water content per cubic metre of concrete for nominal maximum size of aggregate
For 40mm size of aggregate water-content is 165 kg for 25mm-50mm slump
For every 25mm slump extra 3% is added so for 125mm slump 9% is added
Water-content=165+(165*(9/100))=180 kg
As superplasticizer is used so water-content is reduced upto 30%
Therefore from trail and error
water-content is 180*0.75=135 (25% is reduced)
Step 4: Calculation of cement content
w/c=0.35
Cement-content= 135/0.35=386 kg
From IS 456:2000 Table 5
For extreme exposure condition minimum cement-content should be 400kg/m^3
From IS 456:2000
Table 6:Adjustments to minimum cement contents for aggregates other than 20mm nominal maximum size.
=400-30= 370 (-30 kg/m^3 for 40mm maximum aggregate size)
since 386kg/m^3 > 370kg/m^3 Hence OK
Step 5: Calculationing the proportion of volume of coarse and fine aggegate content
From IS 10262:2009
Table 5: volume of coarse aggregate per unit volume of total aggregate for different zones of fine aggregate
For Zone IV, 40mm nominal maximum size of aggregates
Volume of coarse aggregate=0.75 for w/c=0.5
For every 0.05 reduction in w/c volume of CA is increased by 0.01
Therefore, for w/c=0.35 the volume of CA=0.78
We are using a pumpable concrete so 10% reduction in volume of CA is done.
Volume of CA=0.78*0.9=0.70
Volume of FA=(1-0.70) =0.30
Step 6: Mix calculations
a) Volume of concrete=1m^3
b) Volume of cement=(Mass of cement/Specific gravity of cement)*(1/1000)
=(386/3.15)*(1/1000)=0.123 m^3
c) Volume of water =(Mass of water/Specific gravity of water)*(1/1000)
=(135/1000)=0.135m^3
d) Volume of Admixture=(Mass of admixture/Specific gravity of admixture)*(1/1000)
=(7.72/1.145)*(1/1000)=0.007m^3
(Note: volume of admixture is 2% of the cement content= 386*0.02=7.72)
e) Volume of Aggregates= a-(b+c+d)
= 1-(0.123+0.135+0.007)= 0.735 m^3
f) Volume of coarse aggregate= e*volume of aggregate*Specific gravity of aggregate*1000
=0.70*0.735*2.74*1000=1410 kg ( e=proportion of coarse aggregate)
g) Volume of fine aggregate= (1-e)*volume of aggregate*specific gravity of fine aggregate*1000
=(1-0.70)*0.735*2.74*1000
=0.30*0.735*2.74*1000= 605 kg
Step 7: Mix proportion
Cement : 386 kg/m^3
water : 135kg/m^3
Fine aggregate : 605kg/m^3
coarse aggrgate : 1410kg/m^3
Admixture : 7.72kg/m^3
Water-cement ratio :0.35
Step 8: Mix proportion for 50 kg of cement ( 1bag of cement)
C: W: FA: CA: Adm
1: 0.35: 1.57 : 3.65: 0.02