Project on Concrete Mix Design for various grades of Concrete
Q.1 –
Design of concrete mix of M35 grade using fly ash –
Required Data –
- Cement –
|
Type of cement |
OPC (43 grade) |
|
Specific gravity |
3.15 |
|
Max cement content |
IS 456:2000 Cl.8.2.4.2 |
- Site Condition –
|
Exposure |
Severe |
|
Workability |
120mm |
|
Method of placing |
Pumping |
|
Degree of supervision |
Good |
- Admixture –
|
Mineral Admixture |
Fly ash (30%) |
|
Sp.gravity of mineral admixture |
2.2 |
|
Chemical admixture |
Super plasticizer |
|
Sp. Gravity of chemical admixture |
1.145 |
- Aggregates –
|
Property |
Coarse aggregate |
Fine aggregate |
|
Specific gravity |
2.74 |
2.65 |
|
Water absorption |
0.5% |
1% |
|
Moisture content |
2% |
5% |
|
Type of aggregate |
crushed angular aggregate |
|
|
Maximum Nominal Size |
20 mm |
Step 1- Target mean strength
F’ck = Fck + 1.65 x s
= 35 + 1.65 x 5
= 43.25 n/mm2
Step 2 – Approximate air content ……………cl.5.2 IS 10262:2019
Entrapped air for nominal maximum
Size of 20mm is 1% of volume of concrete.
Step 3 –Selection of water cement ratio
W/c ratio – 0.45 ………………….(IS 456-2000,table-3)
Which is 0.30>0.45 hence ok.
Step 4 – Selection of water content
= 201.624 lits or kg.
Reducing water content by 23% for 1% superplasticizer used
So, w/c = 201.624 * 0.77 = 155.250 lits or kg.
Step 5 – Calculation of cement content
Cement content = water content / w/c ratio
= 156 / 0.45 = 346.67 kg/m3 …>320kg/m3 (minimum cement content)
Cement content should be increased by 10% if we use fly ash more than 20%.
Here, we using 30% fly ash so cement content gets increased by 10%.
So, 346.67 * 1.1 = 381.337 kg/m3 = 382 kg/m3
So, Revised w/c ratio = 156/382 = 0.40
Fly ash content = 30% of 382 = 114.6 kg/m3
Cement content = 382 – 114.6 = 267.4 kg/m3 <450kg/m3 hence ok.
Step 6 – Proportion of volume of coarse aggregate & fine aggregate
Volume of coarse aggregate per unit volume =
= 0.64 %
As per cl.5.2.2 for pumping method reduce 10% of total course aggregate to make concrete more workable.
So, revised volume of course aggregate per unit volume = 0.64*0.9 = 0.576%
So, volume of fine aggregate = 1 – 0.576 = 0.424%
Step 7 – Mix calculation
Total volume = 1m3
- Volume of entrapped air = 1% of total volume = 1% *1 = 0.01m3
- Volume of water = 156 / (1* 1000) = 0.156 m3
- Volume of cement = 268 / (3.15*1000) = 0.085m3
- Volume of fly ash = 115 / (2.2*1000) = 0.052 m3
- Volume of chemical admixture = 1% of cementitious material
= 1% (268+115) =3.83 kg
So, 3.83 / (1.145*1000) = 0.00334 m3
- Total volume of aggregates = 1 – (a+b+c+d+e)
= 1 – (0.01+0.156+0.085+0.052) =0.697 m3
- Mass of course aggregate = Total volume * % of aggregate * sp.gravity *1000
= 0.697 * 0.57 * 2.74 * 1000 = 1088.57 = 1089 kg/m3
- Mass of fine aggregate = Total volume * % of aggregate * sp.gravity *1000
= 0.697 * 0.43 * 2.65 * 1000 = 794.23 = 795 kg/m3.
Required quantities –
|
Cement |
268 kg/m3 |
|
Water |
156 lits /m3 |
|
w/c ratio |
0.40 |
|
Fly ash |
115 kg/m3 |
|
Superplasticizer |
3.83 kg/m3 |
|
Coarse aggregate |
1089 kg/m3 |
|
Fine aggregate |
795 kg/m3 |
Design mix = cement: Fine aggregate: Course aggregate
= 383: 795: 1089
= 1: 2.07: 2.84
So, Design mix of concrete grade M35 is 1:2.07:2.84
Q.2 –
Concrete mix design for M50 grade concrete -
- Required data –
- Grade designation – M50
- Type and grade of cement – OPC 43 grade
iii. Maximum nominal size of aggregate – 20mm
- Exposure condition – Severe …………….( IS-456, Table-30)
- Minimum cement content – 320kg/m3…..(IS-456, Table-5)
- Workability in terms of slump – 100 mm..(IS456, Cl.7.1)
vii. Transportation time – 30 min
viii. Method of placing – Pumping
- Standard deviation – 5.0 N/mm2 …………(IS-10262: Table-2)DS
- Type of coarse aggregate – Crushed angular aggregate.
- Water cement ratio – 0.30 …………..(IS Graph)
xii. Fine aggregate – Zone-II
Admixture type – Super plasticizer (PC based - 30 % reduction)
- Calculation –
- Target strength –
F’ck = fck + 1.65 *S
= 50 + 1.65 * 0.5 (S- IS10262, Table1, Page3)
= 58.25 N/mm2
- Water cement ratio –
Exposure condition – severe
So,
Water/cement ratio – 0.45 (IS-456-2000, table-3 & p.no.20)
Which is 0.30 < 0.45 hence ok
It can be reduced since we use admixture
W/c ratio = 0.30 – 0.05 = 0.25
- Water content –
20 mm aggregate – 186 kg (for 50mm slump)
So for 100 mm, add 3% for every 25mm (IS-10262-cl.5.3)
Therefore,
186+6% of 186 = 197 kg.
For super plasticizer reduce 30% (197 – 30% of 197) = 137.9 kg.
Water content = 137.9 kg
- Calculation of cement content –
Water cement ratio = water content / cement content
But,
Cement content = water content / w/c ratio
= 137.9/0.25
= 551.6 kg/m3 > 320 kg/m3
So, Cement content = 552 kg/m3
- Aggregate proportion between C.A & F.A –
Zone-II – 0.62 (W/c – 0.5) (IS-10262, table-5)
Every 0.05 decrease increase 0.01, (w/c – 0.45-0.25 = 0.2)
0.62+0.04 = 0.66
For pumpable concrete C.A can be reduced up to 10%
So, volume of C.A = 0.66 – (10% of 0.66) = 0.594
Volume of F.A = 1- 0.594 = 0.406
- Mix calculation –
- a) Volume of concrete – 1m3
- b) Volume of cement = ( Mass / Sp.gravity) x (1 / 1000)
= 552 / (3.16 x 1000) = 0.174 m3
- c) Volume of water = ( Mass / sp.gravity) x (1 / 1000)
=137.9 / (1.0 x 1000) = 0.137 m3
- d) Volume of admixture = ( Mass / Sp.gravity) x (1 / 1000)
Mass – 1% of cement content
(1.1 / 100) x 552 / (1.12 x 1000) = 0.00521 m3
- e) Volume of all in aggregate
1- (b+c+d) = 1- (0.174+0.137+0.00521) = 0.684
- f) Mass of coarse aggregate =
Vol. of all in aggregate x Vol. of C.A x Sp. Gravity of C.A x 1000
0.684 x 0.594 x 2.73 x 1000
=1109 kg
- g) Mass of fine aggregate –Vol. of all in aggregate x Vol. of F.A x Sp. Gravity of F.A x 1000
0.684 x 0.406 x 2.46 x 1000
=688.15 kg.
Summary –
Cement – 552 kg/m3
Water – 137.9 kg/m3
Fine agg. – 689 kg/m3
Coarse agg. – 1109 kg/m3
W/c ratio – 0.25
Mix design for M50 concrete –
Cement: F.A agg. : C.A agg.
552 : 689 : 1109
1 : 1.2 : 2