Project on Concrete Mix Design for various grades of Concrete
M35 with fly ash
Step1:Target mean strength
f'ck=fck+1.65S or f'ck=fck+X
f'ck=35+1.65(5) or f'ck=35+6.5
f'ck=43.25 or f'ck=41.5
f'ck=43.25
Step2:Water-Cement ratio
As per Fig1 of IS10262:2019,for OPC 53,assuming w/c ratio as 0.45
Step3:Calculation of water content
From Table 4 of IS10262:2019,for 20mm nominal size of aggregate,water content=186kg for 50mm slump.
For 75mm slump,water content=186+186(0.03)kg
=192kg
Using superplasticizer,which reduces the water content upto 20%,water content=192-0.20(192)kg
=153.6
=~154kg
Step4:Calculation of cement content
w/c=0.45
c=342.2kg
For checking the minimum cement content as per IS456:2000,minimum cement content for M35 is 340kg<343kg.
Hence,cement = 343kg
Cementitous material content=343+0.10(343)kg
= 377.3~=374kg
w/c ratio=154/374=0.4
Fly ash @ 30% of the total cementitous material=374*0.30
= 112.2kg~=112kg
Cement content=(374-112)kg
=262kg
Step5:Calculation of Coarse aggregate and Fine aggregate
From Table5 of IS10262:2019,for Zone2 fine aggregate of nominal size 20mm,for w/c ratio of 0.5 is 0.62
For 0.4 w/c ratio,coarse aggregate = 0.62+0.02
= 0.64
Volume of fine aggregate=1-0.64
=0.36
Step6:Mix Calculations
Volume of concrete=1m^3
Volume of cement=(262/3.15)*1/1000
=0.083m^3
Volume of water=154/1000
= 0.154m^3
Volume of Admixture=((2%of 262)/1.145)*1/1000
=0.004m^3
Volume of fly ash=(112/2.1)*1/1000
=0.053m^3
Volume of aggregate=1-(0.083+0.154+0.004+0.053)m^3
=0.706m^3
Mass of Coarse aggregate=0.706*2.74*0.64*1000kg
=1238kg
Mass of Fine aggregate=0.706*2.74*0.36*1000kg
= 696.4kg
Mass of Cement=262kg
Mass of FlyAsh=112kg
Mass of Chemical Admixture=2%of 262kg
=5.24kg
Trail Mix Ratio= 1:2.65:4.72
M50 Mix
Step1:Target mean strength
f'ck=fck+1.65S or f'ck=fck+X
f'ck=50+1.65(5) or f'ck=fck+6.5
f'ck=58.25 or f'ck=56.50
Therefore,f'ck=58.25N/mm^2
Step2:Calculation of w/c ratio
For f'ck=58.25N/mm^2
For OPC 53 grade of concrete,w/c=0.3 (From Fig1 of IS10262:2019)
Step3:Calculation of water content
For 20mm nominal size of aggregate,water content=186kg
For 100mm slump,water content=186+0.06(186)kg
=197.16kg~=197kg
On using admixture,20% of water content is reduced=197-0.20(197)
=157.6kg
~158kg
Step4:Cement content
w/c=0.3
c=527kg>360kg (As per Table5 of IS456:2000)
Hence,ok.
Step5:Calculation of coarse aggregate and fine aggregate
For Zone2 of fine aggregate,for 20mm nominal size of aggregate,volume of coarse aggregate per unit total aggregate=0.62 for 0.5 w/c ratio.
For 0.3w/c ratio,coarse aggregate=0.62+0.04
= 0.66
Fine aggregate=1-0.66
=0.34
Step6:Mix Calculations
Volume of concrete=1m^3
Volume of cement=527/3.15 * 1/1000
=0.167m^3
Volume of water=158/1000
=0.158m^3
Volume of chemical admixture=(2%of 527/1.145)*1/1000
=0.009m^3
Volume of aggregate=1-(0.167+0.158+0.009)
=0.67m^3
Mass of Coarse aggregate=0.67*0.66*2.74*1000
=1211.63kg
Mass of Fine aggregate=0.67*0.34*2.74*1000
=624.17kg
Mass of Cement=527kg
Mass of Water=158kg
Mass of Admixture=10.54kg
w/c ratio=0.3
Trial Mix Ratio=1:1.18:2.3