0 0.0
5
10
15 12 t/sq.m. 20
25
30
35 23
40
45
20.0 40.0 60.0 80.0 100.0 120.0 140.0 160.0 180.0 200.0 220.0 240.0 260.0 280.0
Now if one want to decided the modulus of subgrade reaction (slope of the initial tangent line) then it is quite clear that the values are very different and the modulus calculated from plate load test can not be used for footing /raft of width 3m since the value willl be quite low for the actual footing. Evanthough the applicability of plate load test is very low for actual conditions since the pressure bulb for actual situations will be quite large compared to the bulb formed by plate which is well documented and i need not elaborate it here.
0.6M WIDE PLATE 3m wide footing
0 0.0
5
10
15 12 t/sq.m. 20
25
40.0 60.0 80.0 100.0
140.0 160.0 180.0 200.0 220.0 240.0 260.0 280.0
35 23
40
45
0.6M WIDE PLATE 3m wide footing
20.0
120.0
30
The curve on which one should work is calculated here. There are two way to look at which this graph. One way is to draw tangent lines which was drawn in earlier slide and second is to directly go to the allowable settlement and read the allowable bearing pressure. The later makes more sense since you might have diffrent allowable settlement limits based on type of sthe structure you have and catagory of the structure. With first approach, for FOS = 2.5 the allowable bearing pressure is 9 what we got in earlier slide however as per second one we get 12 t/sq.m. i.e. some improved one. The first approach does not say anything about allowable settlement limit where as in second approach one has flexibility over selection of the settlement limit.
0.6M WIDE PLATE 0
5
10
15
20
25
30
35
40
45
0.000 0.6M WIDE PLATE
5.000
10.000
Setlement (mm)
15.000
20.000
25.000
30.000
35.000
40.000
45.000
50.000
It is usual trend that the safe bearing pressure is taken as 22.5 t/sq.m. as illustrated in the plot which Mr. Picajol tried also. Some engineers of the opinion that FOS shall be applied to it so if we take FOS = 2.5 then net safe bearing pressure will be = 22.5/2.5 = 9 t/sq.m. I just want to say that the procedure is not correct. one should predict the load settlement curve for the footing widht and then try to pedict the allowable bearing pressure for acceptable settlement.
Though as per code procedure, numerically correct however, the soil does not look to me so bad. For sand the acceptable settlement value is 40mm (min) and its just scale at which plot is drawn which allows you to draw two tengent lines. Based on scaling relation given i have tried to predict the load settlement curve for the 3m wide footing based on 0.6m wide plate load test. Applied pressure (t/sq.m.)
0
5
10
15
20
25
0.0
50.0
100.0
150.0
0.6M WIDE PLATE
1.5m wide footing
200.0
2m wide footing
3m wide footing
250.0
3.5m wide footing
4m wide footing
5m wide footing
6m wide footing
300.0
350.0
30
35
40
45
Width of Plate
Load ( t/m2)
0.6 m
Settlement (mm) 0 0.000 5 2.360 10 5.393 20 13.399 25 18.708 30 25.028 35 33.202 40 45.169
For Sandy Stratum
1.5 0 11.65 26.63 66.17 92.38 123.60 163.96 223.05
2 0 12.69 29.00 72.05 100.59 134.58 178.53 242.87
If stratum is clay then it is directly proportional to the footing size.
13.38 30.57 75.96 106.05 141.88 188.22 256.06
Settlement of Footing various wi 2.5 3 0 0 13.87 31.70 78.74 109.95 147.09 195.13 265.45
nt of Footing various widths (m) 3.5 0 14.23 32.54 80.83 112.86 150.99 200.30 272.48
4 0 14.52 33.19 82.45 115.12 154.01 204.31 277.94
5 0 14.93 34.13 84.80 118.40 158.40 210.13 285.87
6 0 15.22 34.79 86.42 120.67 161.43 214.15 291.34