Project on Concrete Mix Design for various grades of Concrete
Aim 1 - The Concrete Mix Design for M35 grade concrete with fly ash
Introduction -
Concrete mix design is a procedure of manufacture of concrete with an ideal proportion of ingredients to fulfill the desired strength of the concrete structure. Concrete is generally a blend of cement, fine aggregate and coarse aggreate with fixed water content. Bridges, dams require an oversized amount of concrete, utilizing it in correct quantity makes the structure economical.
Procedure in detail
STIPULATIONS FOR PROPORTIONING
a) Grade designation : M35
b) Type of cement : OPC 43 grade conforming to IS 8112
c) Type of mineral admixture : Fly ash conforming to IS 3812 (Part1)
d) Maximum nominal size of aggregate : 20 mm
e) Minimum cement content and maximum
water-cement ratio to be adopted : Severe(for reinforced concrete)
and/or
Exposure conditions as per Table 3 and Table 5 of IS 456
f) Workability : 120 mm (slump)
g) Method of concrete placing : Pumping
h) Degree of supervision : Good
j) Type of aggregate : Crushed angular aggregate
k) Maximum cement content (OPC content) : As per IS 456
m) Chemical admixture type : Superplasticizer- normal
TEST DATA FOR MATERIALS
a) Cement used : OPC 43 grade conforming to IS 8112
b) Specific gravity of cement : 3.15
c) Fly ash : Conforming to IS 3812 (Part 1)
d) Specific gravity of fly ash : 2.2
e) Chemical admixture : Superplasticizer conforming to IS 9103
f) Specific gravity of
1) Coarse aggregate (at SSD condition) : 2.74
2) Fine aggregate (at SSD condition) : 2.65
3) Chemical admixture : 1.145
g) Water absorption
1) Coarse aggregate : 0.5 percent
2) Fine aggregate : 1.0 percent
h) The coarse and fine aggregates are wet and their total moisture content is 2 percent and 5 percent
respectively. Therefore, the free moisture content in coarse and fine aggregate shall be as shown in (j)
below
j) Free (surface) moisture
1) Coarse aggregate : Free moisture = Total moisture content – Water absorption
= 2.0 – 0.5 = 1.5 percent
2) Fine aggregate : Free moisture = Total moisture content – Water absorption
= 5.0 – 1.0 = 4.0 percent
k) Sieve analysis
1) Coarse aggregate : Conforming to Table 7 of IS 383
2) Fine aggregate : Conforming to grading Zone II of Table 9 of IS 383
TARGET STRENGTH FOR MIX PROPORTIONING
f’ck = fck + 1.65 S
or
f’ck = fck + X
whichever is higher.
where
f’ck = target average compressive strength at 28 days,
fck = characteristic compressive strength at 28 days
S = standard deviation, and
X = factor based on grade of concrete.
From Table 2, standard deviation, S = 5 N/mm2.
From Table 1, X = 6.5
Therefore, target strength using both equations, that is,
a) f’ck =fck+1.65 S = 35 + 1.65 × 5 = 43.25 N/mm2
b) f’ck = fck + 6.5 = 35 + 6.5 = 41.5 N/mm2
The higher value is to be adopted. Therefore, target strength will be 43.25 N/mm2 as 43.25 N/mm2 > 41.5 N/mm2.
APPROXIMATE AIR CONTENT
From Table 3, the approximate amount of entrapped
air to be expected in normal (non-air-entrained)
concrete is 1.0 percent for 20 mm nominal maximum
size of aggregate.
SELECTION OF WATER-CEMENT RATIO
From Fig. 1, the free water-cement ratio required for
the target strength of 43.25 N/mm2 is 0.38 for OPC 43
grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456.
0.38< 0.45, hence O.K.
SELECTION OF WATER CONTENT
From Table 4, water content = 186 kg (for 50 mm slump) for 20 mm aggregate.
Estimated water content for 120 mm slump ( [(70 × 3 / 25)= 8.4] i.e increasing
at the rate of 3 percent for every 25 mm slump )
= 186 + (186 × 8.4 / 100 ) = 201.62 kg
As superplasticizer is used, the water content may be
reduced.
Based on trial data, the water content reduction of
23 percent is considered while using superplasticizer
at the rate 1.0 percent by weight of cement.
Hence the arrived water content = 201.62 × 0.77 = 155.25 kg ≈ 155 kg.
CALCULATION OF CEMENT CONTENT
Water-cement ratio = 0.38
Cement content =155 / 0.38 = 407.89 kg/m3 ≈ 408 kg/m3
To proportion a mix containing fly ash the following
steps are suggested:
a) Decide the percentage fly ash to be used based
on project requirement and quality of fly ash.
b) In certain situations, increase in cementitious
material content may be warranted. The
decision on increase in cementitious material
content and its percentage may be based on
experience and trials.
NOTE — This illustrative example is with increase of 10 percent
cementitious material content.
Cementitious material content = 408 × 1.10 = 448.8 kg/m3 ≈ 449 kg/m3
Water content = 155 kg/m3
So, water-cementitious ratio = 155 / 449 = 0.345
Fly ash @ 30 percent of total cementitious material
content = 449 × 30 /100 = 134.7 ≈ 135 kg/m3
Cement (OPC) = 449 – 135 = 314 kg/m3
From Table 5 of IS 456, minimum cementitious content
for ‘severe’ exposure condition = 320 kg/m3
449 kg/m3 > 320 kg/m3, hence O.K.
PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGRAGETE CONTENT
From Table 5, volume of coarse aggregate corresponding
to 20 mm size aggregate and fine aggregate (Zone II)
for water-cement ratio of 0.50 = 0.62.
In the present case water-cementitious ratio is 0.345.
Therefore, volume of coarse aggregate is required to
be increased to decrease the fine aggregate content. As
the water-cement ratio is lower by 0.155, the proportion
of volume of coarse aggregate is increased by 0.031
(at the rate of -+ 0.01 for every ± 0.05 change in watercement
ratio) i.e[{0.155 × (0.01/0.05)}=0.031] . Therefore, corrected proportion of
volume of coarse aggregate for the water-cementitious
ratio of 0.345 = 0.62 + 0.031 = 0.651.
For pumpable concrete these values may be reduced
by up to 10 percent. (see 5.5.2). Here, 10 percent
reduction is considered.
Therefore, volume of coarse aggregate = 0.651 × 0.9 = 0.5859 , say 0.59 .
Volume of fine aggregate content = 1 – 0.59 = 0.41
MIX CALCULATIONS
The mix calculations per unit volume of concrete shall be as follows:
a) Total volume of concrete = 1 m3
b) Volume of entrapped air in wet concrete =0.01 m3 (1% of total volume of concrete)
c) Volume of cement=(Mass of cement /Specific gravity of cement)×(1/1000)= (314/ 3.15) × (1/1000) =0.09968 m3 ≈0.1 m3
d) Volume of fly ash=(Mass of fly ash /Specific gravity of fly ash)×(1/1000)= (135/2.2) × (1/1000) = 0.06136 m3 ≈ 0.0614 m3
e) Volume of water=(Mass of water /Specific gravity of water)×(1/1000)= (155/1) × (1/1000) = 0.155 m3
f) Volume of chemical admixture(superplasticizer) (@ 1.0 percent by mass of cementitious material) = (Mass of chemical admixture/Specific gravity of admixture) ×(1/1000) = (4.49/1.145)×(1/1000) = 0.0039 m3 ≈ 0.004 m3
g) Volume of all in aggregate = [(a-b)-(c+d+e+f)]= [(1-0.01) – ( 0.1 + 0.0614 + 0.155 + 0.004)]= [0.99 - 0.3204]=0.6696 m3 ≈ 0.67 m3
h) Mass of coarse aggregate= g × Volume of coarse aggregate × Specific gravity of coarse aggregate × 1 000 = 0.67 × 0.59 × 2.74 × 1000 = 1083.122 kg ≈ 1083 kg
j) Mass of fine aggregate = g × Volume of fine aggregate × Specific gravity of fine aggregate × 1 000 = 0.67 × 0.41 × 2.65 × 1000 = 727.95 kg ≈ 728 kg
MIX PROPORTIONS FOR TRIAL NUMBER 1
Cement = 314 kg/m3
Fly ash = 135 kg/m3
Water (Net mixing) = 155 kg/m3
Fine aggregate (SSD) = 728 kg/m3
Coarse aggregate (SSD) = 1083 kg/m3
Chemical admixture = 4.49 kg/m3
Free water-cementitious materials ratio = 0.345
NOTE — Aggregates shall be used in saturated surface dry
condition. If otherwise, when computing the requirement of
mixing water, allowance shall be made for the free (surface)
moisture contributed by the fine and coarse aggregates. On the
other hand, if the aggregates are dry, the amount of mixing
water shall be increased by an amount equal to the moisture
likely to be absorbed by the aggregates. Necessary adjustments
are also required to be made in mass of aggregates. The surface
water and percent water absorption of aggregates shall be
determined according to IS 2386 (Part 3).
ADJUSTMENT ON WATER, FINE AGGREGATE AND COARSE AGGREGATE (IF THE COARSE AND FINE AGGREGATE IS IN WET CONDITION)
a) Fine Aggregate ( wet )
Mass of wet fine aggregate = Mass of fine aggregate in SSD condition × {1+(Free (surface) moisture/100)}
=728 × {1+(4/100)}
=757.12 kg ≈ 757 kg
b)Coarse Aggregate ( wet )
Mass of wet coarse aggregate = Mass of coarse aggregate in SSD condition × {1+(Free (surface) moisture/100)}
=1083 × {1+(1.5/100)}
=1099.245 kg ≈ 1099 kg
The coarse and fine aggregates, being wet, contribute
water to the mix to the extent of free moisture over
SSD condition. The quantity of this water is required
to be subtracted from the calculated water content.
1) Water content contributed by wet coarse aggregate
= Mass of wet coarse aggregate – mass of SSD condition coarse aggregate
= 1099 – 1083 = 16 kg
2) Water content contributed by wet fine aggregate
= Mass of wet fine aggregate – mass of SSD condition fine aggregate
= 757 – 728 = 29 kg
The requirement for added water becomes :
= 155 - 16 - 29
= 110 kg/m3
Result
MIX PROPORTIONS AFTER ADUSTMENT FOR WET AGGREGATES
Cement = 314 kg/m3
Fly ash = 135 kg/m3
Water (to be added) = 110 kg/m3
Fine aggregate (Wet) = 757 kg/m3
Coarse aggregate (Wet) = 1099 kg/m3
Chemical admixture = 4.49 kg/m3
Free water-cementitious materials ratio = 0.345
Aim 2 - The Concrete Mix Design for M50 grade concrete without Fly ash
Procedure in detail
STIPULATIONS FOR PROPORTIONING
a) Grade designation : M50
b) Type of cement : OPC 43 grade conforming to IS 8112
c) Maximum nominal size of aggregate : 20 mm
d) Minimum cement content and
maximum water-cement ratio to be adopted and/or : Severe (for reinforced concrete)
Exposure conditions as per Table 3
and Table 5 of IS 456
e) Workability : 75 mm (slump)
f) Method of concrete placing : Chute (Non pumpable)
g) Degree of site control : Good
h) Type of aggregate : Crushed angular aggregate
j) Maximum cement content not including fly ash : 450 kg/m3
k) Chemical admixture type : Superplasticizer - normal
TEST DATA FOR MATERIALS
a) Cement used : OPC 43 grade conforming to IS 8112
b) Specific gravity of cement : 3.15
c) Chemical admixture : Superplasticizer conforming to IS 9103
d) Specific gravity of
1) Coarse aggregate [at saturated surface dry (SSD) Condtion] : 2.74
2) Fine aggregate [at saturated surface dry (SSD) Condtion] : 2.65
3) Chemical admixture : 1.145
e) Water absorption
1) Coarse aggregate : 0.5 percent
2) Fine aggregate : 1.0 percent
f) Moisture content of aggregate [As per IS 2386 (Part 3)]
1) Coarse aggregate : Nil
2) Fine aggregate : Nil
g) Sieve analysis:
1) Coarse aggregate : Conforming to Table 7 of IS 383
2) Fine aggregate : Conforming to grading Zone II of Table 9 of IS 383
TARGET STRENGTH FOR MIX PROPORTIONING
f’ck = fck + 1.65 S
or
f’ck = fck + X
whichever is higher.
where
f’ck = target average compressive strength at 28 days,
fck = characteristic compressive strength at 28 days
S = standard deviation, and
X = factor based on grade of concrete.
From Table 2, standard deviation, S = 5 N/mm2.
From Table 1, X = 6.5
Therefore, target strength using both equations, that is,
a) f’ck =fck+1.65 S = 50 + 1.65 × 5 = 58.25 N/mm2
b) f’ck = fck + 6.5 = 50 + 6.5 = 56.5 N/mm2
The higher value is to be adopted. Therefore, target strength will be 58.25 N/mm2 as 58.25 N/mm2 > 56.5 N/mm2.
APPROXIMATE AIR CONTENT
From Table 3, the approximate amount of entrapped
air to be expected in normal (non-air-entrained)
concrete is 1.0 percent for 20 mm nominal maximum
size of aggregate.
SELECTION OF WATER-CEMENT RATIO
From Fig. 1, the free water-cement ratio required for
the target strength of 58.25 N/mm2 is 0.34 for OPC 43
grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456.
0.34 < 0.45, hence O.K.
SELECTION OF WATER CONTENT
From Table 4, water content = 186 kg (for 50 mm slump) for 20 mm aggregate.
Estimated water content for 75 mm slump ( increasing
at the rate of 3 percent for every 25 mm slump )
= 186 + (186 × 3 / 100 ) = 191.58 kg
As superplasticizer is used, the water content may be
reduced.
Based on trial data, the water content reduction of
23 percent is considered while using superplasticizer
at the rate 1.0 percent by weight of cement.
Hence the arrived water content = 191.58 × 0.77 = 147.52 kg ≈ 148 kg.
SELECTION OF WATER CONTENT
Water-cement ratio = 0.34
Cement content =155 / 0.34 = 455.88 kg/m3 ≈ 456 kg/m3
From Table 5 of IS 456, minimum cementitious content
for ‘severe’ exposure condition = 320 kg/m3
456 kg/m3 > 320 kg/m3, hence O.K.
PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGRAGETE CONTENT
From Table 5, volume of coarse aggregate corresponding
to 20 mm size aggregate and fine aggregate (Zone II)
for water-cement ratio of 0.50 = 0.62.
In the present case water-cementitious ratio is 0.34.
Therefore, volume of coarse aggregate is required to
be increased to decrease the fine aggregate content. As
the water-cement ratio is lower by 0.16, the proportion
of volume of coarse aggregate is increased by 0.032
(at the rate of -+ 0.01 for every ± 0.05 change in watercement
ratio) i.e[{0.16 × (0.01/0.05)}=0.032] . Therefore, corrected proportion of
volume of coarse aggregate for the water-cementitious
ratio of 0.34 = 0.62 + 0.032 = 0.652.
Volume of fine aggregate content = 1 – 0.652 = 0.348
MIX CALCULATIONS
The mix calculations per unit volume of concrete shall be as follows:
a) Total volume of concrete = 1 m3
b) Volume of entrapped air in wet concrete =0.01 m3 (1% of total volume of concrete)
c) Volume of cement=(Mass of cement /Specific gravity of cement)×(1/1000)= (456/ 3.15) × (1/1000) =0.14476 m3 ≈0.145 m3
d) Volume of water=(Mass of water /Specific gravity of water)×(1/1000)= (148/1) × (1/1000) = 0.148 m3
e) Volume of chemical admixture(superplasticizer) (@ 1.0 percent by mass of cementitious material) = (Mass of chemical admixture/Specific gravity of admixture) ×(1/1000) = (4.560/1.145)×(1/1000) = 0.00398 m3 ≈ 0.004 m3
g) Volume of all in aggregate = [(a-b)-(c+d+e)]= [(1-0.01) – ( 0.145 + 0.148 + 0.004)]= [0.99 - 0.302]=0.688 m3
h) Mass of coarse aggregate= g × Volume of coarse aggregate × Specific gravity of coarse aggregate × 1 000 = 0.688 × 0.652 × 2.74 × 1000 = 1229.09 kg ≈ 1229 kg
j) Mass of fine aggregate = g × Volume of fine aggregate × Specific gravity of fine aggregate × 1 000 = 0.688 × 0.348 × 2.65 × 1000 = 634.47 kg ≈ 634 kg
MIX PROPORTIONS FOR TRIAL NUMBER 1
Cement = 456 kg/m3
Water = 148 kg/m3
Fine aggregate (SSD) = 634 kg/m3
Coarse aggregate (SSD) = 1229 kg/m3
Chemical admixture = 4.56 kg/m3
Free water-cementitious materials ratio = 0.34
ADJUSTMENT ON WATER, FINE AGGREGATE AND COARSE AGGREGATE (IF THE COARSE AND FINE AGGREGATE IS IN WET CONDITION)
a) Fine Aggregate ( dry )
Mass of dry fine aggregate = Mass of fine aggregate in SSD condition / {1+(water absorption/100)}
=634 / {1+(1/100)}
=627.72 kg ≈ 628 kg
b)Coarse Aggregate ( dry )
Mass of dry coarse aggregate = Mass of coarse aggregate in SSD condition / {1+(water absorption/100)}
=1229 / {1+(0.5/100)}
=1222.885 kg ≈ 1223 kg
The extra water to be added for absorption by coarse
and fine aggregate,
1) For coarse aggregate
= Mass of coarse aggregate in SSD condition – mass of coarse aggregate in dry condition
= 1 229 – 1 223 = 6 kg
2) For fine aggregate
= Mass of fine aggregate in SSD condition – mass of fine aggregate in dry condition
= 634 – 628 = 6 kg
The estimated requirement for added water, therefore, becomes= 148 + 6 + 6 = 160 kg/m3
Result
MIX PROPORTIONS AFTER ADUSTMENT FOR DRY AGGREGATES
Cement = 456 kg/m3
Water (to be added) = 160 kg/m3
Fine aggregate (Dry) = 628 kg/m3
Coarse aggregate (Dry) = 1223 kg/m3
Chemical admixture = 4.56 kg/m3
Free water-cement ratio = 0.34