Project on Concrete Mix Design for various grades of Concrete
# CONCRETE MIX DESIGN for M35 grade concrete with FLY ASH.
| STEP 1: STIPULATIONS FOR PROPORTIONING | |
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| STEP 2: TEST DATA FOR MATERIALS | |
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STEP 3:TARGET STRENGTH FOR MIX PROPORTIONING:
- f'ck=fck+(1.65*s)
- f'ck=35+(1.65*5)=43.25 N/mm^2
STEP 4:WATER-CEMENT RATIO
- From Table 5 of IS 456:2000;maximum water-cement ratio=0.45;
- Based on experience;water-cement ratio=0.40;
- Thus,0.40<0.45=> o.k.
STEP 5:WATER CONTENT
- Maximum water content for 20mm aggregate=186 litre(for 25-50 mm slump);
- Estimated water content for 100mm slump=[186+{(6/100)*186}]=197 litre;
- Water content to be reduced upto 29% based on trials i.e. =197*0.71=140 litre.
STEP 6:CALCULATION OF FLY ASH and CEMENT content
- Water-cement ratio=0.40;
- Cementitious material(cement+flyash)content=(140/0.40)=350 kg/m^3;
# According to Table 5 of IS 456:2000;
- Minimum cement content for "very severe" exposure condition=340 kg/m^3;
- Cementitious material content=350*1.10=385 kg/m^3;
- Water-content = 140kg/m^3;
- Water-cement ratio={(140/385)}=0.364;
- Fly ash @ 30% of total cementitious material content=385*30%=115 kg/m^3;
- Cement(OPC)=385-115=270 kg/m^3;
- Saving of cement using flyash=350-270=80 kg/m^3;
- Fly ash being utilized=115 kg/m^3.
STEP 7:VOLUME OF COARSE AGGREGATE and FINE AGGREGATE
- Volume of coarse aggregate corresponding to 20mm aggregate and fine aggregate for water-cement ratio of 0.50=0.60;
# NOTE- In present case water-cement ratio=0.40; thus volume of coarse aggregate is required to be increased to decrease the fine aggregate content;As water-cement ratio is lowered by 0.10;
Volume of coarse aggregate is increased by 0.02;
Thus Volume of coarse aggregate for water-cement ratio of 0.40=0.62;
- For pumpable concrete value is to be reduced by 0.10;
- Volume of coarse aggregate =0.62*0.9=0.56;
- Volume of fine aggregate=1-0.56=0.44.
STEP 8:MIX CALCULATIONS
- Volume of concrete=1m^3;
- Volume of cement={(270/3.15)*(1/1000)}=0.086 m^3;
- Volume of fly-ash={(115/1)*(1/1000)}=0.052 m^3;
- Volume of water={(140/1)*(1/1000)}=0.140 m^3;
- Volume of chemical admixture @ 2% by mass of cementitious material={(7/1.145)*(1/1000)}=0.007 m^3;
- Volume of all in aggregate={a-(b+c+d+e)}=0.715 m^3;
- Mass of coarse aggregate=f*volume of coarse aggregate*specific gravity of coarse aggregate*1000=1097 kg;
- Mass of fine aggregate=f*volume of fine aggregate*specific gravity of fine aggregate*1000=862 kg.
STEP 9:MIX PROPORTIONS FOR TRIAL NO. 1:
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# CONCRETE MIX DESIGN for M50 grade concrete without FLY ASH.
| STEP 1: STIPULATIONS FOR PROPORTIONING | |
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| STEP 2: TEST DATA FOR MATERIALS | |
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STEP 3:TARGET STRENGTH FOR MIX PROPORTIONING:
- f'ck=fck+(1.65*s)
- f'ck=50+(1.65*5)=58.25 N/mm^2
STEP 4:WATER-CEMENT RATIO
- From Table 5 of IS 456:2000;maximum water-cement ratio=0.45;
- Based on experience;water-cement ratio=0.40;
- Thus,0.40<0.45=> o.k.
STEP 5:WATER CONTENT
- Maximum water content for 20mm aggregate=186 litre(for 25-50 mm slump);
- Estimated water content for 100mm slump=[186+{(6/100)*186}]=197 litre;
- Water content to be reduced upto 29% based on trials i.e. =197*0.71=140 litre.
STEP 6:CALCULATION OF FLY ASH and CEMENT content
- Water-cement ratio=0.40;
# According to Table 5 of IS 456:2000;
- Minimum cement content for "very severe" exposure condition=340 kg/m^3;
- Water-content = 140kg/m^3;
STEP 7:VOLUME OF COARSE AGGREGATE and FINE AGGREGATE
- Volume of coarse aggregate corresponding to 20mm aggregate and fine aggregate for water-cement ratio of 0.50=0.60;
# NOTE- In present case water-cement ratio=0.40; thus volume of coarse aggregate is required to be increased to decrease the fine aggregate content;As water-cement ratio is lowered by 0.10;
Volume of coarse aggregate is increased by 0.02;
Thus Volume of coarse aggregate for water-cement ratio of 0.40=0.62;
- For pumpable concrete value is to be reduced by 0.10;
- Volume of coarse aggregate =0.62*0.9=0.56;
- Volume of fine aggregate=1-0.56=0.44.
STEP 8:MIX CALCULATIONS
- Volume of concrete=1m^3;
- Volume of cement={(350/3.15)*(1/1000)}=0.111 m^3;
- Volume of water={(140/1)*(1/1000)}=0.140 m^3;
- Volume of chemical admixture @ 2% by mass of cementitious material={(7/1.145)*(1/1000)}=0.007 m^3;
- Volume of all in aggregate={a-(b+c+d)}=0.743 m^3;
- Mass of coarse aggregate=e*volume of coarse aggregate*specific gravity of coarse aggregate*1000=(0.743*0.56*2.74*1000)=1140 kg;
- Mass of fine aggregate=e*volume of fine aggregate*specific gravity of fine aggregate*1000=(0.743*0.44*2.74*1000)=896 kg.
STEP 9:MIX PROPORTIONS FOR TRIAL NO. 1:
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