Project on Concrete Mix Design for various grades of Concrete
Question
Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Your Answers
PART 1: TO CALCULATE THE MIX DESIGN FOR M35 GRADE CONCRETE WITH FLY ASH
AIM:
To determine the concrete mix design for M35 grade concrete with fly ash.
INTRODUCTION:
Concrete mix design is a procedure of selecting the suitable ingredients of concrete and to prepare concrete of certain minimum strength, desired workability and durability as economically as possible.
Codes Used: IS 10262: 20019, IS 456: 2000.
STEP BY STEP PROCEDURE:
STEP 1: Stipulation of proportioning
STEP 2: Test Data
STEP 3: Target strength (Refer IS 10262:2019 table 1 & 2).
STEP 4: Approximate air Content (Refer IS 10262:2019 table 3)
STEP 5: Selection of Water cement ratio (Refer IS 10262:2019 Fig 1)
STEP 6: Selection of Water content (Refer IS 10262:2019 table 4)
STEP 7: Calculation of Cement content
STEP8: Proportion of Volume of Coarse aggregate (Refer IS 10262:2019 tables 5)
STEP9: Mix Calculation
STEP 10: Mix Proportion
STEP 11: Mix Proportion
STEP 1: STIPULATION PROPORTION OF CONCRETE
a) Grade designation : M35
b) Type of cement : OPC 43 grade
c) Type of mineral admixture : Fly ash
d) Max nominal size of aggregate : 20mm
e) Workability : 100 mm (slump)
f) Method of concrete placing : Chute
g) Exposure : Moderate
h) Min Cement content : 300 kg/m3
i) Maximum W/C ratio : 0.50
j) Minimum Grade : M25
k) Maximum Cement content : 450 kg/m3
l) Chemical admixture : Super plasticizer- normal
STEP 2: TEST DATA
1) Specific gravity of cement : 3.15
2) Specific gravity of Fly ash : 2.3
3) Specific gravity of Coarse aggregate : 2.77
4) Specific gravity of Fine aggregate : 2.65
5) Specific gravity of Chemical admixture : 1.1
6) Water absorption of Coarse aggregate : 0.5%
7) Water absorption o Fine aggregate : 1%
STEP 3: TARGET STRENGTH
f’ck = fck+1.65 S
IS 10262 : 2019 From Table 2, standard deviation, S = 5 N/mm2.
f’ck = fck+1.65 S
f’ck = 35+1.65 × 5
f’ck = 43.25 N/mm2
From Table 1, X = 6.5.
f’ck = fck + 6.5
f’ck = 35 + 6.5
f’ck = 41.5 N/mm2
The higher value is to be adopted. The Target Strength is = 43.25 N/mm2
STEP 4: APPROXIMATE AIR CONTENT
From Table 3, IS 10262:2019 the approximate amount of entrapped air to normal concrete is 1% of 20mm aggregate.
STEP 5: SELECTION OF WATER-CEMENT RATIO
From Table 5 IS 456:2000 ‘Moderate’ Exposure for reinforcement concrete
Maximum water cement ratio = 0.50
The target strength of 43.25 N/mm2 grade curve is assumed for the trial in IS10262:2019 Fig 1 the Water cement ratio is 0.39
0.39< 0.50, hence OK.
STEP 6: SELECTION OF WATER CONTENT
From Table 4, IS 10262:2019 water content = 186 kg for 20 mm aggregate.
Estimated water content for 100 mm slump
= 186 + [(0.06X186)]
= 197.16 Lit
The Super plasticizer to be used. The water content reduced. Based on trial data, the water content reduction of 26% is considered while using super plasticizer at the rate 1% by weight of cement.
Hence the Mass of water content
= 197.16 × 0.74
= 145.90 kg/m3
STEP 7: CALCULATION OF CEMENT CONTENT
Water-cement ratio = 0.39
Mass of Cementitious content =145.90/0.39
= 374.10 kg/m3
From Table 5 of IS 456 :2000 The minimum cement content is 300 kg/m3
300 kg/m3 < 374.10 kg/m3. Hence ok.
Since Fly ash is going to be replaced with cement, based on trail we need to add 10 %.
New Cementitious content = [Mass of Cementitious content*(100%+Assumption)]
=374.10*1.1
= 411.51 kg/ m3
New Water Cementitious ratio= [Weight of water content/Cementitious content]
=145.90/411.51
= 0.355
24% of Cementitious is going to replaced by Fly ash
= [New Cementitious content*24%]
= 411.51*0.24
Fly ash = 98.76 kg/ m3
Cement = [Cementitious content – Fly ash]
=411.51-98.76
Cement= 312.75 kg/ m3
Super Plasticizer is 1.7% of cement content = 0.017 X 312.75
Chemical admixture = 5.32 kg/ m3
STEP 8: PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGETE CONTENT
From Table 5, IS 10262:2019 20 mm size aggregate (Zone I) for water-cement ratio of 0.50 = 0.6.
Therefore, corrected proportion of volume of coarse aggregate for the water-cement ratio of
X= {[(0.5-0.355)*0.01]/0.05}
= 0.029
Proportioning Volume of coarse aggregate = 0.6 + 0.029 = 0.63.
Proportioning Volume of fine aggregate = 1 – 0.63 = 0.37
STEP 9: MIX CALCULATION
The mix calculations per unit volume of concrete shall be as follows:
Total volume = 1 m3
Volume of air entrapped =0.01 m3
1) Volume of cement = [Mass of cement /(Specific gravity of cement * 1 000)]
= [312.75 /(3.15*1000)]
= 0.099 m3
2) Volume of Fly ash = [Mass of Fly ash /(Specific gravity of cement * 1000)]
= [98.76 /(2.3*1000)]
= 0.043 m3
3) Volume of water = [Mass of water /(Specific gravity of cement * 1000)]
= [145.90 /(1*1000)]
= 0.146 m3
4) Volume of admixture = [Mass of admixture /(Specific gravity of cement * 1000)]
= [5.32 /(1.1*1000)]
= 0.048 m3
Total volume of other material expect aggregate = [Vol. cement+ Vol. Fly ash+ Vol. water+ Vol. admixture+ Vol. air entrapped]
= 0.099 +0.043 +0.146 +0.048+0.01
=0.346 m3
5) Volume of Coarse and Fine aggregate = [1- Total volume of other material expect aggregate]
=1- 0.346
=0.654 m3
Mass of coarse aggregate = [Vol. aggregate * Proportion of C. aggregate * Sp.gravity * 1000]
= 0.654*0.63*2.77*1000
= 1141.30 kg/ m3
Mass of fine aggregate = [Vol. aggregate * Proportion of F. aggregate * Sp.gravity * 1000]
= 0.654*0.37*2.65*1000
= 641.25 kg/ m3
STEP 10: MIX PROPORTION
Cement = 312.75 kg/ m3
Water = 145.90 kg/m3
Fly ash = 98.76 kg/ m3
Admixture = 5.32 kg/ m3
Fine aggregate = 641.25 kg/m3
Coarse aggregate = 1141.30 kg/ m3
Water cement ratio = 0.355
STEP 11: MIX RATIO:
The M50 Concrete mix design Mix Ratio is 1: 0.32: 2.05: 3.65: 0.355.
(Cement: Fly ash: Fine aggregate: Coarse aggregate: New W/C ratio).
PART 2: TO CALCULATE THE MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLY ASH
AIM:
To determine the concrete mix design for M50 grade concrete without fly ash
INTRODUCTION:
Concrete mix design is a procedure of selecting the suitable ingredients of concrete and to prepare concrete of certain minimum strength, desired workability and durability as economically as possible.
Codes Used: IS 10262: 20019, IS 456: 2000.
STEP BY STEP PROCEDURE:
STEP 1: Stipulation of proportioning
STEP 2: Test Data
STEP 3: Target strength (Refer IS 10262:2019 table 1 & 2).
STEP 4: Approximate air Content (Refer IS 10262:2019 table 3)
STEP 5: Selection of Water cement ratio (Refer IS 10262:2019 Fig 1)
STEP 6: Selection of Water content (Refer IS 10262:2019 table 4)
STEP 7: Calculation of Cement content
STEP8: Proportion of Volume of Coarse aggregate (Refer IS 10262:2019 table 5)
STEP9: Mix Calculation
STEP 10: Mix Proportion
STEP 11: Mix Proportion
STEP 1: STIPULATION PROPORTION OF CONCRETE
a) Grade designation : M50
b) Type of cement : OPC 43 grade
c) Type of mineral admixture : -
d) Max nominal size of aggregate : 40mm
e) Workability : 117 mm (slump)
f) Method of concrete placing : Pumping
g) Exposure : Extreme
h) Min Cement content : 360 kg/m3
i) Maximum W/C ratio : 0.40
j) Minimum Grade : M40
k) Maximum Cement content : 450 kg/m3
l) Chemical admixture : Super plasticizer- normal
STEP 2: TEST DATA
1) Specific gravity of cement : 3.18
2) Specific gravity of coarse aggregate : 2.78
3) Specific gravity of Fine aggregate : 2.68
4) Specific gravity of Chemical admixture : 1.19
5) Water absorption of coarse aggregate : 0.5%
6) Water absorption o Fine aggregate : 1%
STEP 3: TARGET STRENGTH
f’ck = fck+1.65 S
IS 10262: 2019 From Table 2, Standard deviation S = 5 N/mm2.
f’ck = fck+1.65 S
f’ck = 50+1.65 × 5
f’ck = 58.25 N/mm2
From Table 1, X = 6.5.
f’ck = fck + 6.5
f’ck = 50 + 6.5
f’ck = 56.5 N/mm2
The higher value is to be adopted. The Target Strength is = 58.25 N/mm2
STEP 4: APPROXIMATE AIR CONTENT
From Table 3, IS 10262:2019 the approximate amount of entrapped air to normal concrete is 0.8% of 40mm aggregate.
STEP 5: SELECTION OF WATER-CEMENT RATIO
From Table 5 (IS 456:2000) ‘Extreme’ Exposure for reinforcement concrete
Maximum water cement ratio = 0.40
The target strength of 58.25 N/mm2 grade curve is assumed for the trial in IS10262:2019 Fig 1 the Water cement ratio is 0.35
0.35< 0.40, hence O.K.
STEP 6: SELECTION OF WATER CONTENT
From Table 4, IS 10262:2019 water content = 165 kg for 40 mm aggregate.
Estimated water content for 117 mm slump
= 165 + [(0.080X165)]
= 178.27 Lit
The Super plasticizer to be used. The water content reduced. Based on trial data, the water content reduction of 27% is considered while using super plasticizer at the rate 1.0 percent by weight of cement.
Hence the Mass of water content
= 178.27 × 0.73
= 130.13 kg/m3
STEP 7: CALCULATION OF CEMENT CONTENT
Water-cement ratio = 0.35
Mass of Cement content =130.13/0.35
= 371.8 kg/m3
From Table 5 of IS 456, minimum cement content for ‘Extreme’ exposure condition of minimum cement content is 360 kg/m3
360 kg/m3< 371.8 kg/m3.
Super Plasticizer is 2.9% of cement content = 0.029 X 371.8
=10.78 kg/ m3
STEP 8: PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGETE CONTENT
From Table 5, IS 10262:2019 40 mm size aggregate (Zone II) for water-cement ratio of 0.50 = 0.71.
In the present case water-cement ratio is 0.35.
Therefore, corrected proportion of volume of coarse aggregate for the water-cement ratio of
X = {[(0.5-0.35)*0.01]/0.05}
= 0.03
Proportioning Volume of coarse aggregate = 0.71 + 0.03 = 0.74.
Proportioning Volume of fine aggregate = 1 – 0.74 = 0.26
STEP 9: MIX CALCULATION
The mix calculations per unit volume of concrete shall be as follows:
Total volume = 1 m3
Volume of air entrapped =0.008 m3
1) Volume of cement = [Mass of cement /(Specific gravity of cement * 1 000)]
= [371.8 /(3.18*1000)]
= 0.117 m3
2) Volume of water =Mass of water /Specific gravity of cement * 1000)]
= [130.13 /(1*1000)]
= 0.130 m3
3) Volume of admixture =Mass of admixture /Specific gravity of cement * 1000)]
= [10.78 /(1.19*1000)]
= 0.009 m3
Total volumes of other material expect aggregate = [Vol. cement+ Vol. water+ Vol. admixture+ Vol. air entrapped]
= 0.117 +0.130 +0.009 +0.008
=0.264 m3
4) Volume of Coarse and Fine aggregate = [1- Total volume of other material expect aggregate]
=1- 0.264
=0.736 m3
Mass of Coarse Aggregate = [Vol. aggregate * Proportion of C. aggregate * Sp.gravity * 1000]
= 0.736*0.74*2.78*1000
= 1514.10 kg/ m3
Mass of Fine Aggregate = [Vol. aggregate * Proportion of F. aggregate * Sp.gravity * 1000]
= 0.736*0.26*2.68*1000
= 512.84 kg/ m3
STEP 10: MIX PROPORTION
Cement = 371.8 kg/m3
Water = 130.13 kg/m3
Admixture = 10.78 kg/ m3
Fine aggregate = 512.84 kg/ m3
Coarse aggregate = 1514.10 kg/ m3
Water cement ratio = 0.35
STEP 11: MIX RATIO
The M50 Concrete mix design Mix Ratio is 1: 1.38: 4.07: 0.35
(Cement: Fine aggregate: Coarse aggregate: W/C ratio)
RESULT :
From both concrete mix designs M35 and M50, we got to know that all the concrete ingredients required to make concrete of required grade give desired properties as required in both plastic and hardened state of concrete.