Project on Concrete Mix Design for various grades of Concrete

Calculate the Concrete Mix Design for M35 grade concrete with fly ash 

a) Grade designation : M35
b) Type of cement : OPC 43 grade conforming to IS 269
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 : 100 mm (slump)
g) Method of concrete placing : Pumping
h) Degree of supervision : Good
j) Type of aggregate : Crushed angular aggregate (no correction required)
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 269
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) : 2.75
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

Target strength

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 = 5.0 N/mm2
.
Therefore, target strength using both equations, that is,
a) f’ck = fck+1.65 S
= 35 +1.65 × 5.0 = 43.25 N/mm2

b) f’ck = fck + 8.0 (The value of X for M 35 grade as per Table 1 is 6.5 N/mm2)
= 35 + 6.5 = 41.5 N/mm2

Therefore the target strength is 43.25 N/mm2,

as 43.25 N/mm2>41.5 N/mm2

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. (For PPC, the strength corresponding to OPC 43 grade curve is assumed for the trial).

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 100 mm slump

=50slump+extra25slump+extra25slump=186+186x3/100+186x3/100

= 186 +5.58+5.58

= 197.16 kg
As superplasticizer is used, the water content may be further reduced from 20 to 30% considering 23% as water reduction percentage due to super plasticizers.

=197.16x(100-23)=151.81kg of water

CALCULATION OF CEMENT CONTENT
Water-cement ratio = 0.38
Cement content =151.81/0.38
=399.51 kg/m3 ≈ 400kg/m3 of cement

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 8 percent cementitious material content.
Cementitious material content
= 400 × 1.08 = 432 kg/m3 

Water content = 151.81 kg/m

So, water-cementitious ratio
= 151.81/ 432 =0.351 <0.45

Fly ash @ 30 percent of total cementitious material content = 432× 30 percent = 432x30/100=129.6 kg/m3
Cement (OPC) = 432-129.6 = 302.4 kg/m3

From Table 5 of IS 456, minimum cementitious content
for ‘severe’ exposure condition= 320 kg/m3
432 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 0.351.

The water to cementitious material ratio is lower by 0.5-0.351=0.149

Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content.

As the water-cement ratio is lower by 0.149, the proportion of volume of coarse aggregate has to be increased

Aggregate to increase           water cement ratio

0.01                                             0.05

x                                                  0.149

Therfore x=0.01x0.149/0.05= 0.0298
(at the rate of -+ 0.01 for every ± 0.05 change in water cement ratio).

Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.351 = 0.62 + 0.0298 = 0.6498.

For pumpable concrete these values may be reduced by up to 10 percent. (see 5.5.2) IS 10262. Here, 10 percent reduction is considered.
Therefore, volume of coarse aggregate = 0.6498 ×0.9 = 0.5848, say 0.59 m3.

Volume of fine aggregate content = 1 – 0.59 = 0.41 m3

MIX CALCULATIONS
The mix calculations per unit volume of concrete shall be as follows:
a) Total volume = 1 m3
b) Volume of entrapped air in wet concrete = 0.01 m3
c) Volume of cement = Mass of cement/(Specificgravity of cementx 1000 )= 302.4/(3.15x1000)= 0.096m3
d) Volume of fly ash= Mass of fly ash x(Specificgravity of fly ash x1000 )= 129.6/(2.2 x1000)= 0.0583 m3
e) Volume of water= Mass of water (Specific gravity of waterx 1000)= 151.81/1x 1000)= 0.1518 m3
f) Volume of chemical admixture(superplasticizer) (@ 1.0 percent by mass of cementitious material)

=1/100 x432=4.32kg.m3
= Mass of chemical admixture /(Specific gravity of admixture x1000)= 4.32/(1.145x1000)= 0.00377 m3

g) Volume of all in aggregate= [(a-b)-(c+d+e+f)}]
= (1-0.01)-(0.096 + 0.0583 + 0.1518+0.00377)= 0.680m3

h) Mass of coarse aggregate= Vol.in all × volume of coarse aggregate × Sp gr of coarse aggregate × 1 000
= 0.680 × 0.59 × 2.74 × 1000= 1 099.28 kg ≈ 1 099 kg

j) Mass of fine aggregate=Vol.in all × Volume of fine aggregate × Sp gr of fine aggregate × 1 000
= 0.680 × (1-0.59=0.41) × 2.65 × 1 000
= 738.82 ≈ 739kg

MIX PROPORTIONS FOR TRIAL 
Cement = 302.4 kg/m3
Fly ash = 129.6kg/m3
Water (Net mixing) = 151.81 kg/m3
Fine aggregate (SSD) = 739 kg/m3
Coarse aggregate (SSD) = 1 099 kg/m3
Chemical admixture = 4.32 kg/m3
Free water-cementitious materials ratio= 0.351

ADJUSTMENT ON WATER, FINE AGGREGATE AND COARSE AGGREGATE (IFTHE 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))

= 739 × (1+4/100)= 768.56 kg/m3 ≈ 769 kg/m3

b) Coarse aggregate (Wet)

Mass of wet coarse aggregate=mass of coarse aggregate in SSD conditionx(1+(Free surface moisture/100))

= 1 099 × (1+(1.5/100))= 1 115.48 kg/m3 ≈ 1 116kg/m3

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 coars aggregate
= Mass of wet coarse aggregate – mass of SSD
condition coarse aggregate
= 1116– 1 099 = 17 kg
2) Water content contributed by wet fine aggregate
= Mass of wet fine aggregate – mass of SSD condition fine aggregate
= 769 – 739 = 30 kg

The requirement for added water becomes :
= 151.81 - 17 - 30= 104.81 kg/m3

MIX PROPORTIONS AFTER ADUSTMENT FOR WET AGGREGATES

Cement = 302.4 kg/m3
Fly ash = 129.6kg/m3
Water (Net mixing) = 104.81 kg/m3
Fine aggregate (SSD) = 769 kg/m3
Coarse aggregate (SSD) = 1116 kg/m3
Chemical admixture = 4.32 kg/m3
Free water-cementitious materials ratio= 0.351

Cement     Flyash          FA            CA            water    

302.4         129.6          769          1116         104.81      

1                  0.428         2.54        3.69        0.346

2. M50 grade concrete without Fly ash

ASSUMPTIONS FOR PROPORTIONING
a) Grade designation : M 50
b) Type of cement : OPC 53 grade conforming to IS 269
c) Silica fume : none
d) Maximum nominal size of aggregate : 20 mm
e) Exposure conditions as per Table 3 and Table 5 of IS 456 : Severe (for reinforced concrete)
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 (OPC) content : 450 kg/m3
m) Chemical admixture type : Superplasticizer (Polycarboxylate ether based)

TEST DATA FOR MATERIALS
a) Cement used : OPC 53 Grade conforming to IS 269
b) Specific gravity of cement : 3.15
c) Specific gravity of
1) Coarse aggregate (at SSD condition) : 2.74
2) Fine aggregate (at SSD condition) : 2.65
3) Fly ash : -
4) Silica fume : -
5) Chemical admixture : 1.08
d) Water absorption
1) Coarse aggregate : 0.5 percent
2) Fine aggregate : 1.0 percent
f) Moisture content
1) Coarse aggregate : Nil
2) Fine aggregate : Nil
g) Sieve analysis

2) Fine aggregate : Conforming to grading Zone II of Table 9 of IS 383

An example illustrating the mix proportioning for a concrete of M50 grade using silica fume and fly ash is
given below. Use of silica fume is generally

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 = 5.0 N/mm2

Therefore, target strength using both equations, that is,
a) f’ck = fck+1.65 S
= 50 +1.65 × 5.0 = 58.25 N/mm2

b) f’ck = fck + 6.5 (The value of X for M 50 grade as per Table 1 is 8.0 N/mm2)
= 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 M50 for 20 mm aggrgate is 1.0% of volume of concrete.

SELECTION OF WATER-CEMENT RATIO

From the figure for 53 grade choosig curve3 and the W/C ratio is 0.33 (strength point of view)

Our exposure condition is severe therefore the minimum cement content= 320kg/m3

and W/C max= 0.45 (durability point of view)

Here we are providing 0.33 < 0.45 therefore ok

SELECTION OF WATER CONTENT from IS 10262

From Table 4, water content for 20 mm aggregate = 186 kg/m3 (for 50 mm slump without using superplasticiser).

Estimated water content for 120 mm slump

= 186 +186x8.4/100=201.624 ≈ 202 kg/m3

As superplasticizer (Polycarboxylate ether based) isused, the water content can be reduced by 30 percent.
Hence, the reduced water content

= 202 × 0.70 = 141.4 kg/m3 ≈ 141 kg/m3

CALCULATION OF CEMENT CONTENT
Water–cement ratio = 0.33
Water content = 141 kg/m3
Cement content = 141 / 0.33= 427.27 ≈ 428 kg/m3

Check for minimum cementitious materials content,
320 kg/m3 < 428 kg/m3

Check for maximum cement(OPC) content, 450 kg/m3> 428 kg/m3. Hence OK.

PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATECONTENT

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 0.33.

The water to cementitious material ratio is lower by 0.5-0.33=0.17

Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content.

As the water-cement ratio is lower by 0.17, the proportion of volume of coarse aggregate has to be increased

Aggregate to increase           water cement ratio

0.01                                             0.05

x                                                  0.17

Therfore x=0.01x0.17/0.05= 0.034
(at the rate of -+ 0.01 for every ± 0.05 change in water cement ratio).

Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.33 = 0.62 + 0.034 = 0.654.

From Table 5, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate grading Zone II = 0.62 per unit volume of total aggregate. This is valid for water-cementitious materials ratio of 0.50. After correction we got the coarse aggregate ratio as 0.654

Fine aggregate ratio= 1 – 0.654=0.346

MIX CALCULATIONS

The mix calculations per unit volume of concrete shall be as follows:
a) Total volume = 1 m3
b) Volume of entrapped air in wet concrete = 0.01 m3
c) Volume of cement = Mass of cement/(Specificgravity of cementx 1000 )= 428/(3.15x1000)= 0.136m3
d) Volume of water= Mass of water (Specific gravity of waterx 1000)= 141/1x 1000)= 0.141 m3
e) Volume of chemical admixture(superplasticizer) (@ 1.0 percent by mass of cementitious material)

=1/100 x428=4.28kg.m3
= Mass of chemical admixture /(Specific gravity of admixture x1000)= 4.28/(1.145x1000)= 0.00374 m3

F) Volume of all in aggregate= [(a-b)-(c+d+e)}]
= (1-0.01)-(0.136 + 0.141 + 0.00374)= 0.709m3

g) Mass of coarse aggregate= Vol.in all × volume of coarse aggregate × Sp gr of coarse aggregate × 1 000
= 0.709 × 0.654 × 2.74 × 1000= 1270.5 kg ≈ 1271 kg

h) Mass of fine aggregate=Vol.in all × Volume of fine aggregate × Sp gr of fine aggregate × 1 000
= 0.709 × (1-0.654=0.346) × 2.65 × 1 000 = 650.08≈ 650kg

MIX PROPORTIONS FOR TRIAL 
Cement =428 kg/m3
Water (Net mixing) = 141 kg/m3
Fine aggregate (SSD) = 650 kg/m3
Coarse aggregate (SSD) = 1271 kg/m3
Chemical admixture = 4.28 kg/m3
Free water-cementitious materials ratio= 0.33

MIX PROPORTIONS AFTER ADUSTMENT FOR ABSORBTION OF AGGREGATES

(WHICH MEANS THE AGGREGATES ARE DRY)

ADJUSTMENT ON WATER, FINE AGGREGATE AND COARSE AGGREGATE 

Water content for aborption

=141+0.5/100=141.05

Fine aggregate (SSD) = 650 kg/m3 -1/100x650=650-6.5=643.5kg/m3

Coarse aggregate (SSD) = 1271 kg/m3-0.5/100x1271=1264.65kg/m3

MIX PROPORTIONS AFTER ADUSTMENT FOR ABSORBTION OF AGGREGATES

Cement =428 kg/m3
Water (Net mixing) = 141.05 kg/m3
Fine aggregate (SSD) = 643.5 kg/m3
Coarse aggregate (SSD) = 1264.65 kg/m3
Chemical admixture = 4.28 kg/m3
Free water-cementitious materials ratio= 0.33

Cement         FA            CA            water    

428            643.5        1264.65        41.05      

1                 1.5             2.95             0.329