Project on Concrete Mix Design for various grades of Concrete
AIM: Mix Design For M35 grade concrete with fly-ash
procedure:
STEP 1: STIPULATIONS FOR PROPORTIONING
M35 grade of concrete with fly-ash - OPC 53 grade of cement ; minimum cement content is 340kg/m3; workability :100mm ; exposure condition : severe (pumping concrete)
type of aggregate is crushed angular aggregate;
max' cement concrete is 420kg/m3; superplasticizers are to be used;
STEP 2: specific gravity of coarse aggregate is:2.74 and for fine aggregate is 2.65 ;
water absorption in coarse aggregate is 0.5% and fine aggregate is 1.0%
free surface moisture is zero (should have in dry condition)
sieve analysis should be in zone1 of table 4 of IS383
STEP 3: TARGET STRENGTH FOR MIX PROPORTIONING
target strength for mix design proportion: table no 2- IS10262:2019
fck'=fck+1.65 s= 35+1.65*5 = 43.25 N/mm2
STEP 4: SELECTION OF WATER-CEMENT RATIO
the water-cement ratio for extreme conditions is 0.45
STEP 5: APPROXIMATE WATER CONTENT
water content is for 20mm aggregate = 186litre is required.
for 100 mm slump 186* 6/100 = 197 litres.
for superplasticizers 20% reduction is there is 197* 0.8 = 157.6litres appox'= 158litres =water
STEP 6: CALCULATION OF CEMENT CONTENT
cement content =158litres/0.45= 351kg/m3
cementitious content = 351*1.10= 386.1 kg/m3
water content =158litres
water cement ratio =0.40
fly-ash as 40% cementitious material content = 386* 40% = 154 kg/m3 = fly-ash
OPC cement = 386-154 =232 kg/m3= cement
as per IS456-2000 Table-5 as severe exposure condition.
hence said that 386 kg/m3 > 340 kg/m3
STEP 7: PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
the proportion of the volume of Coarse and fine aggregate content from table 5 of IS10262:2019
the volume of course aggregate to 20mm is zone 2 is 0.62.
10% reduced for pumpable concrete is 0.62*0.9= 0.56 ( Vol- of COARSE aggregate)
1-0.56= 0.44 ( Vol- of FINE aggregate)
STEP 8: Mix calculation;
a) volume of concrete is 1m3
b) volume of cement is = mass of cement /specific gravity of cement *(1/1000)= 232/3.51 *1/1000 = 0.066m3
c) volume of fly-ash = mass of fly-ash / specific gravity of fly ash *(1/1000) = 154/2.2 * 1/1000 = 0.07
d)volume of water is = mass of water / specific gravity of water *(1/1000) = 158/1 * 1/1000 = 0.158 m3
e)volume of admixture = mass of admixture = 351*0.02= 7.02
mass of admixture /specific gravity of admixture *1/1000 = 7.02/1.145 *1/1000 =0.00613m3
f) volume of aggregate (F.A and C.A) = (a-(b+c+d+e)) =(1-(0.066+0.07+0.158+0.00613)=0.699
g)mass of course aggregate =f*volume of C.A= 0.699*0.56 *2.74*1000 =1073kg/m3 = C.A
h) mass of fine aggregate = f*volume of fine aggregate* sp- the gravity of F.A =0.699*0.44*2.65*1000=815kg/m3 = F.A
cement =232kg/m3
fly-ash =154kg/m3
water =154litre
fine aggregate =815kg/m3
coarse aggregate =1073kg/m3
chemical admixture =7kg/m3
water-cement ratio =0.45
mix ratio = 1:2.11:2.77 (M35 with fly-ash)
AIM: Mix Design For M50 grade concrete without fly-ash
procedure :
Sp. Gravity
Cement = 3.15
Fine Aggregate = 2.61
Coarse Aggregate (20mm) = 2.65
Coarse Aggregate (10mm) = 2.66
Minimum Cement (As per contract) =400 kg / m3
Maximum water cement ratio (As per contract) = 0.45
Mix Calculation: –
1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 N/mm2
2. water cement ratio:-
Assume water cement ratio = 0.35
3. Calculation of water: –
Approximate water content for 20mm max. Size of aggregate = 180 kg /m3 (As per Table No. 5 , IS : 10262 ). As plasticizer is proposed we can reduce water content by 20%.
Now water content = 180 X 0.8 = 144 kg /m3
4. Calculation of cement content:-
Water cement ratio = 0.35
water content =144kg/m3
cement content = 144/0.35 =411.4 kg/m3
5. Calculation for C.A. & F.A.:
Volume of concrete = 1 m3
Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m3
Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m3
Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m3
Total weight of other materials except coarse aggregate = 0.1308 + 0.1440 +0.0043 = 0.2791 m3
Volume of coarse and fine aggregate = 1 – 0.2791 = 0.7209 m3
Volume of F.A. = 0.7209 X 0.33 = 0.2379 m3 (Assuming 33% by volume of total aggregate )
Volume of C.A. = 0.7209 – 0.2379 = 0.4830 m3
Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919 kg/ m3
Say weight of F.A. = 621 kg/ m3
Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365 kg/ m3
Say weight of C.A. = 1284 kg/ m3
Considering 20 mm: 10mm = 0.55: 0.45
20mm = 706 kg .
10mm = 578 kg .
Hence Mix details per m3
Increasing cement, water, and admixture by 2.5% for this trial
Cement = 412 X 1.025 = 422 kg
Water = 144 X 1.025 = 147.6 kg
Fine aggregate = 621 kg
Coarse aggregate 20 mm = 706 kg
Coarse aggregate 10 mm = 578 kg
Admixture = 1.2 % by weight of cement = 5.064 kg.
Cement: F.A.: C.A. = 1: 1.472: 3.043