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AIM: To create the concrete mix design for M35 grade concrete with fly ash. INTRODUCTION: The Process of selecting suitable ingredients of concrete (cement, fine aggregates and course arrgegate mixed with water ) and determining the relative quantities with an objective of producing a concrete of required strenght , durability…
Krishna Kumar
updated on 08 Sep 2023
AIM: To create the concrete mix design for M35 grade concrete with fly ash.
INTRODUCTION:
Requirment of concrete mix design
STEP 1:
DATA REQUIRED;
a. Grade designation : M35
b. Type of cement : OPC 43 grade conforming to IS 8112
c. Type of mineral admixture: Fly ash confirming to IS 3812 ( Part 1)
d. Maximum nominal size of aggregates : 20mm
e. Minimum cement content : 320 kg/m^3 (Table :5 IS 456-2000,based on exposure condition)
f. Maximum water-cement ratio : 0.45 (Table :5 IS 456-2000,based on exposure condition)
g. Workability :100mm slump value of pumpable concrete
h. Exposure condition : Severe ( for reinforced concrete)
i. Method of concrete placing : Pumping
j. Degree of supervision : Good
k. Type of aggregate: Crushed angular aggregate
l. Maximum cement content : 450 kg/m^3.
m. Chemical admixture type : Superplaticizer.
Step :2
Test data for materials .
a. Cement used : OPC 43 grade confirming to IS 8112
b. Specific gravity of cement : 3.15
c. Fly ash : Conforming to IS 3812 part 1
d. Specific gravity of flyash : 2.2
e. Chemical admixture : Superplaticizer conforming to IS 9103
f. Specific gravity of :
Coarse aggregate : 2.74
Fine aggregate : 2.74
g. Water absorption :
Coarse aggregate : 0.5%
Fine aggregate : 1%
h. Free surface moisture
Coarse aggregate : Nil
Fine aggregate : Nil
i. Seive analysis
course aggregate;conforming to table 2 of IS 383.
Fine aggregate; conforming to grading zone 1 of table 4 for IS 383.Step :3
To find target mean strength for mix proportioning.
Target mean strength f'ck = fck +1.65s, s= 5.0 as per IS 10262-2019 table 2.
= 35 +1.65*5.0 =43.25 N/mm^2>N/mm^2
fck- Characteristic compressive strength
s- Standard deviation.
Step:4
Selection of water-cement ratio.
Maximum water-cement ratio =0.45 (for severe condition ,from table 5 of IS 456-2000)
On experience water- cement ratio =0.4
0.4 <0>
Step :5
Calculation of water content.
Maximum water content for 20mm aggregate =186 litres ( Table 4 of IS 1026:2019) for 25mm to 50mm slump range
For 100 mm slump ,= 186+(6/100)*186 =197 litres
As superplaticizer is used, the water content is reduced upto 30%
Based on trials water content reduction of 29% is achieved with plasticizer.
197-29/100*197=139.87=140 ,
Hence arrived water content = 140 litres.
Step :6
Calculation of cement content.
Water - cement ratio = 0.4.
Cementatious material ( Cement +flyash ) = 140/0.4 =350 kgm^3>kg/m^3
In table 5 of IS 456-2000, the cement content for severe exposure =320 kgm^3>kg/m^3
350 kgm^3>kg/m^3 > 320kgm^3>k/gm^3
Hence OK.
To proportion the mix containing flyash the following steps are suggested :
a. The percentage of flyash need to be decided based on project requirement and qualityof materials.
b. In some cases increase in cementatious materials can be warrned . Increase in cementatious material content and its percentage may be based on experience and trial.
In this increase of 10�mentatious material content is considered.
Cementatious material content = 350*1.1 =385 kgm3">kg/m3
Water content = 140 kgm3">kg/m3
So, water- cement ratio = 140/385 =0.364
Fly ash at 30% total cementatious material content = 35-30% =115 kgm3">kg/m3
Cement (OPC) =385-115 =270 kgm3>kg/m3
Saving of cement while using flyash = 350-270=80 kgm3">kg/m3
Fly ash beign utilized = 115 kg/m3.
Step: 7
Proportion of volume of coarse aggregate and fine aggregate content.
From table 5 of IS 1026:2019 ,volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone 1 ) for water- cement ratio =0.60
In present case the water - cement ratio is 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content .
As the water- cement ratio is lowered by 0.1 ,the proportion of volume of coarse aggregate is increased by 0.02( at the rate of +/- 0.01 for every +/- 0.05 change in water- cement ratio)
Hence the corrected value ofcoarse aggregate for the water-cemnt ratio = 0.64
( If coarse aggregate is not of angular type, then also the volume of coarse aggregate may be required to be increased suitably ,based on experience)
For pumpable concrete, these are reduced by 10%..
Volume of coarse aggregate = 0.64*0.9 = 0.576.
Volume of fine aggregate = 1-0.576 =0.424.
Step :8
Mix calculations.
Mix calculation per unit volume of concrete:
a. Volume of concrete = 1 m^3.
b. Volume of cement = (Mass of cement / Specific gravity of cement ) *1/1000
= (270/3.15 )*1/1000 =0.086 m^3>m^3
c. Volume of flyash = (Mass of flyash / Specific gravity of flyash )*1/1000
= (115/2.2) *1/1000 = 0.052m^3>m^3
d. Volume of water = ( Mass of water / Specific gravity of water)*1/1000
= (140/1)*1/1000 =0.140 m^3>m^3
e. Volume of chemical admixture = ( Mass of admixture / specific gravity of admixture)*1/1000
=( 7.6/1.14)*1/1000 =0.006 m^3>m^3
f. Volume of aggregate ( Fine and coarse) = a- (b+c+d+e)
= 1-(0.086+0.052+0.140+0.006) =0.716 m^3>m^3
g. Mass of coarse aggregate = f* Volume of coarse aggregate * Specific gravity of coarse aggregate *1000
= 0.716*0.576*2.74*1000 =1130kgm^3>kg/m^3
h. Mass of fine aggregate = f* Volume of fine aggregate* Specific gravity of fine aggreate*1000
= 0.716*0.424*2.74*1000= 831kgm^3>kg/m^3.
Step :9
Mix proportion.
Cement = 270 kg/m^3
Fly ash = 115 kg/m^3
Water = 140kg/m^3
Fine aggregate = 831 kg/m^3
Coarse aggregate = 1130 kg/m^3
Chemical admixture = 7kg/m^3
Water- cement ratio = 0.364
Mix ratio 1: 2.15: 2.94
AIM: To calculate the concrete mix design for M50 grade concrete without fly ash.
INTRODUCTION :
Requirment of concrete mix design
Guidelines for concrete mix design
IS method 10262-2019 and Recommended guidlines for concrete mix design,IS 10262-2009, British standard method - BSEN-1 and its complementary standard BS8500 part 1 & 2.
PROCEDURE :
Step 1: Data collection for concrete mix design.
Step 2 : Test data for material.
Step 3 : Target strength for mix proportioning.
ft =fck+1.65 S
Where ft = Target average compressive strength at 28 days
fck = Characteritic compressive strength at 28 days
S= Standard deviation
from Table 1 of IS 10262:2009, standard devition S = 5 N/mm^2
Therefore,Target strength ft= 50+(1.65*5)=58.25 N/mm^2.
Step 4: selection of water-cement ratio:
The water cement ratio corresponding to target strength can be assumed as 0.34 as per IS 10262-2019 Graph.
Hence 0.34<0>
The water cement ratio is 0.34
Step 5: Selection of water content
Step 6: Calculation of cement content.
water cement ratio =0.34
cement content = 136/0.34 = 400kg/m^3
from Table 5 of IS 456:2000 minimum cementitous content for severe exposure condition = 320 Kg /m^3.
Hence 400 kg/m^3 > 320 Kg /m^3
Step 7: Proportion of course aggregate & fine aggregate content.
from Table 5 of IS 10262:2019 volume of coarse aggregate corrospnding to 20mm size aggregate and fine aggregate for water cement ratio =0.6
Here water cement ratio is 0.354
Therefore the volume of course aggregate need to be increase to decrease the fine aggregate content.
Therefore the corrected volume of aggragate for the water cement ratio of 0.354=0.629=0.63m^3
Volume of fine aggregate content =1-0.63=0.37m^3.
Step 8: mix calculation
The mix calculation per unit volume of concrete shall be as follows
cement : 400 Kg/m^3
water : 136 Kg/m^3.
chemical admixture : 4 Kg/m^3.
water cement ratio : 0.34
course aggregate : 1269 Kg/m^3.
fine aggregate : 720 Kg/m^3.
Trail mix ratio =1:1.8:3.17
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