Project on Concrete Mix Design for various grades of Concrete
Calculate the concrete mix design for M35 grade concrete with fly ash
->mean strenght : f`ck = fck + 1.65 s
f`ck = target mean strenght
fck = compressive strenght at 28 days of concrete
s = standard deviation (IS 10262:2009, table -1)
->volume of cement :mass of cement / specific gravity *1/ 1000
->volume of water : mass of water / specific of water *1/ 1000
->volume of chemical admixtures : mass of chemical admixture / specific gravity of chemical admixture * 1/1000
->volume of fly ash : mass of fly ash / specific gravity of fly ash *1 /1000
->volume of concrete :1m^3
->volume of aggregate : volume of concrete - volume of cement + volume of fly ash +volume of water + volume of admixtures
->mass of coarse aggregate volume of aggregate * volume of coarse aggregate * specific gravity of coarse aggregate *1000
->mass of fine aggregate : volume of aggregate * volume of fine aggregate * specific gravity of fine aggregate * 1000
Procedure and calculation :
->Target mean strenght
F`ck = fck + 1.65 s
fck = 35 N\mm^2
|
Sl.No |
Grade of Concrete |
Characteristic compressive strength (N/mm2) |
Assumed standard deviation (N/mm2) |
|
1. |
M10 |
10 |
3.5 |
|
2. |
M15 |
15 |
|
|
3. |
M20 |
20 |
4.0 |
|
4. |
M25 |
25 |
|
|
5. |
M30 |
30 |
5.0 |
|
6. |
M35 |
35 |
|
|
7. |
M40 |
40 |
|
|
8. |
M45 |
45 |
|
|
9. |
M50 |
50 |
|
|
10. |
M55 |
55 |
Selection of water cement ration:
->For preliminary calculation, water cement ratio as given is IS-456-Table 5 (also given below) for different environmental exposure condition, may be use
->M35
->Assume : explosure condition : extreme for reinforced concrete
Table 5 IS-456:2000
->Therefore,extreme = water/cement ratio =0.40
M35 = water/cement ratio =0.45
lower the water cement ratio better the workability
0.40<0.45
thus water cement ratio = 0.40
minimum water cement content = 360kg/m^3
->waUse table-1 for finding out water-cement ratio of plain concrete and use table -2 for finding out water-cement ratio of reinforced concrete.
|
Table -1 |
||||
| Sl.No. | Environmental Exposure Condition |
Plain Concrete |
||
| Minimum Cement Content (kg/m3) | Maximum Free Water-Cement Ratio | Minimum Grade of Concrete | ||
| 1 | Mild | 220 | 0.60 | — |
| 2 | Moderate | 240 | 0.60 | M15 |
| 3 | Severe | 250 | 0.50 | M20 |
| 4 | Very Severe | 260 | 0.45 | M20 |
| 5 | Extreme | 280 | 0.40 | M25 |
|
Reinforced Concrete |
||||
| Minimum Cement Content (kg/m3) | Maximum Free Water-Cement Ratio | Minimum Grade of Concrete | ||
| 1 | Mild | 300 | 0.55 | M20 |
| 2 | Moderate | 300 | 0.50 | M25 |
| 3 | Severe | 320 | 0.45 | M30 |
| 4 | Very Severe | 340 | 0.45 | M35 |
| 5 | Extreme | 360 | 0.40 | M40 |
Refer the table given below (As per IS-456) to choose right type of environment depending upon different exposure conditions to concrete.
Selection of water content
nominal size of aggregate = 20 mm
table 2-IS 10262:2009
|
Maximum water content per cubic meter of concrete for nominal maximum size of aggregate |
||
|
Sl.No. |
Nominal maximum size of aggregate |
Maximum water content |
|
1 |
10 |
208 |
|
2 |
20 |
186 |
|
3 |
40 |
165 |
The values given in the table shown above is applicable only for angular coarse aggregate and for a slump value in between 25 to 50mm.
Do the following adjustments if the material used differs from the specified condition.
minimum water content =186 litres or kg
slump ranges from 25 to 50 mm
as suoerplasticizer 20% reduced= 186*80/100 =148.8
water content =149 litres.
Calculation of cement content in cement + fly ash
->water cement ratio = 0.40
->water content =149 kg
->cement content =149/0.40=372.5=370 kg/m^3
->minimum cement content=360 kg/m^3
360<370 kg/m^3 its ok
increase by 10%
370 *0.10+370 = 407 kg/m^3
w=149 kg
water cement ratio =149/407 = 0.366
fly ash at 30%total cement material content=407*30/100=122 kg/m^3
ordinary portland cement= 407 - 122 = 285 kg/m^3
saving cement while using fly ash =370-285 = 85 kg/m^3
fly ash = 122 kg/m^3
Volume of coarse aggregate and fine aggregate:
fine aggregate are sieved according to zone 1 table 4 IS 383
coarse aggregate are sieved according to table 2 IS 383
according to table 3 IS 10262:2009
nominal size of aggregate =20 mm
fine aggregate zone 1
therefore volume of coarse aggregate =0.60
|
Sl.No. |
Nominal Maximum Size of Aggregate (mm) |
Volume of coarse aggregate per unit volume of total aggregate for different zones of fine aggregate |
|||
|
Zone IV |
Zone III |
Zone II |
Zone I |
||
|
1 |
10 |
0.50 |
0.48 |
0.46 |
0.44 |
|
2 |
20 |
0.66 |
0.64 |
0.62 |
0.60 |
|
3 |
40 |
0.75 |
0.73 |
0.71 |
0.69 |
water cement ratio =0.4 (+/-0.01 for every +/-0.005)
so the volume of coarse aggregate =0.62
volume of fine aggregate =1-0.62 =0.38
Mix calculation
->a)volume of concrete =1m^3
->b)volume of cement = 285/3.15 *1/1000=0.090m^3
->c)volume of fly ash =122/2.2 * 1/1000=0.055m^3
->d)volume of water = 149/1 * 1/1000 = 0.149m^3
->e)bvolume of admixture = 7.6/1.145 * 1/1000 =0.006m^3
->volume of coarse aggregate and fine aggregate = a-b+c+d+e = 1- 0.090+0.055+0.149+0.006= 0.7m^3
->mass of coarse aggregate =0.7*0.62*2.74*1000=1189.16kg/m^3
->mass of fine aggregate = 0.7*0.38*2.74*1000=728kg/m^3
Results and conclusion:
trial mix:
->cement = 285 kg/m^3
->fly ash =122 kg/m^3
->water = 149 kg or litres
->coasre aggregates = 1189 kg/m^3
->fine aggregate =728 kg/m^3
->chemical admixtures = 7.6 kg.m^3
->water cement ratio = 0.366
->trial mix ratio :285:728:1189 = 1:2.55:4.17 (c:fa:ca)
___________________________________________________________________________________________________
Calculate M50 grade without fly ash
Equations:
->target mean strength:f`ck=fck+1.65s
f`ck=target mean strength
fck=compressive strength at 28 days of concrete
s = standard deviation IS 10262:2009
->volume of cement: mass of cement/specific gravity of cement*1/1000
->volume of water: mass of water/specific gravity of water*1/1000
->volume of chemical admixture: mass of chemical admixture/specific gravity of chemical admixture*1/1000
->volume of concrete:1m^3
->volume of aggregate:volume of concrete-volume of concrete+volume of water+volume of chemical admixture
->mass of coarse aggregate:volume iof aggregate*volume of coarse aggregate*specific gravity of coasre aggregate*1000
->mass of fine aggregate:volume of aggregate*volume of fine aggregate*specific gravity of fine aggregate*1000
Pocedure and calculation:
Target mean strength
f`ck = fck + 1.65 s
fck = 50 N/mm^2 (M50)
Where,
fck’ = Target compressive strength at 28 days in N/mm2.
fck = Characteristic compressive strength at 28 days in N/mm2. (same as grade of concrete, see table below)
s = Standard deviation
Table 1-IS 10262:2009
The value of standard deviation, given in the table below, can be taken for initial calculation.
|
Sl.No |
Grade of Concrete |
Characteristic compressive strength (N/mm2) |
Assumed standard deviation (N/mm2) |
|
1. |
M10 |
10 |
3.5 |
|
2. |
M15 |
15 |
|
|
3. |
M20 |
20 |
4.0 |
|
4. |
M25 |
25 |
|
|
5. |
M30 |
30 |
5.0 |
|
6. |
M35 |
35 |
|
|
7. |
M40 |
40 |
|
|
8. |
M45 |
45 |
|
|
9. |
M50 |
50 |
|
|
10. |
M55 |
55 |
s=5 N/mm^2
f`ck = 50 + 1.65 *5 = 58.25 N/mm^2
Selection of water cement ratio:
assume :
exposure condition: extreme for reinforced concrete
Table 5- IS 456:2000
therefore, extreme = water cement ratio = 0.40
M50 = water cement ratio =0.40
lower water cement ratio better than workability= 0.40
water cement ratio=0.40
minimum cement content = 360kg/m^3
Table-5
|
Table -1 |
||||
|
Sl.No. |
Environmental Exposure Condition |
Plain Concrete |
||
|
Minimum Cement Content (kg/m3) |
Maximum Free Water-Cement Ratio |
Minimum Grade of Concrete |
||
|
1 |
Mild |
220 |
0.60 |
— |
|
2 |
Moderate |
240 |
0.60 |
M15 |
|
3 |
Severe |
250 |
0.50 |
M20 |
|
4 |
Very Severe |
260 |
0.45 |
M20 |
|
5 |
Extreme |
280 |
0.40 |
M25 |
|
Table -2 |
||||
|
Sl.No. |
Environmental Exposure Condition |
Reinforced Concrete |
||
|
Minimum Cement Content (kg/m3) |
Maximum Free Water-Cement Ratio |
Minimum Grade of Concrete |
||
|
1 |
Mild |
300 |
0.55 |
M20 |
|
2 |
Moderate |
300 |
0.50 |
M25 |
|
3 |
Severe |
320 |
0.45 |
M30 |
|
4 |
Very Severe |
340 |
0.45 |
M35 |
|
5 |
Extreme |
360 |
0.40 |
M40 |
Selection of water content
assume:Nominal size of aggregate = 20mm
Table 2 -IS 10262:2009
|
Maximum water content per cubic meter of concrete for nominal maximum size of aggregate |
||
|
Sl.No. |
Nominal maximum size of aggregate |
Maximum water content |
|
1 |
10 |
208 |
|
2 |
20 |
186 |
|
3 |
40 |
165 |
The values given in the table shown above is applicable only for angular coarse aggregate and for a slump value in between 25 to 50mm.
minimum water content = 186 litres or kg
superplasticizer 20% reuduced= 186 *80/100 =148.8
water content = 149 litres
Calculation of cement content
water cement ratio =0.40
cement content = 149/0.40 =372.5 = 373 kg/m^3
minimum content of cement =360 kg/m^3
360>373kg/m^3
Volume of coarse aggergate and fine aggregates
fine aggregate are sieved according to zone 1 of table 4 IS 383
coarse aggregate are sieved accoreding to table 2 IS 383
according,table 3 IS 10262:2009
Nominal size of aggregates= 20mm
fine aggregate zone 1
volume of coarse aggregate = 0.60
Table 3:
Volume of coarse aggregate corresponding to unit volume of total aggregate for different zones of fine aggregate is given in the following table.
|
Sl.No. |
Nominal Maximum Size of Aggregate (mm) |
Volume of coarse aggregate per unit volume of total aggregate for different zones of fine aggregate |
|||
|
Zone IV |
Zone III |
Zone II |
Zone I |
||
|
1 |
10 |
0.50 |
0.48 |
0.46 |
0.44 |
|
2 |
20 |
0.66 |
0.64 |
0.62 |
0.60 |
|
3 |
40 |
0.75 |
0.73 |
0.71 |
0.69 |
The values given in the table shown above is applicable only for a water-cement ratio of 0.4 and based on aggregates in saturated surface dry condition.
If water-cement ratio other than 0.4 is to be used then apply correction using the rule given below.
volume of coasre aggregate =0.62
volume of fine aggregate =1-0.62=0.38
Mix calculation:
->a)volume of concrete =1m^3
->b)volume of cement = 373/3.15 *1/1000=0.118m^3
->c)volume of water = 149/1 * 1/1000 = 0.149m^3
->d)bvolume of admixture = 7.6/1.145 * 1/1000 =0.006m^3
->volume of coarse aggregate and fine aggregate = a-b+c+d+e = 1- 0.118+0.149+0.006= 0.727m^3
->mass of coarse aggregate =0.727*0.62*2.74*1000=1235.02kg/m^3
->mass of fine aggregate = 0.727*0.38*2.74*1000=756.9kg/m^3
Results and conclusion:
trial mix:
->cement = 373 kg/m^3
->water = 149 kg or litres
->coasre aggregates = 1235.02 kg/m^3
->fine aggregate =756.9 kg/m^3
->chemical admixtures = 7.6 kg.m^3
->water cement ratio = 0.40
->trial mix ratio :373:756.9:1235.02 = 1:2.02:3.31 (c:fa:ca)