Simplified Procedure for Calculating
Bearing Capacity of Cohesionless Soil
using Standard Penetration Test as an
alternative to the IS 6403 Method
Ravi Kant Mittal
Associate Professor, Department of Civil Engineering,
Birla Institute of Technology & Science, Pilani, Rajasthan - 333031
(India)
Phone: +91-9887692025,
e-mail: ravimittal@pilani.bits-pilani.ac.in
Piyush Bansal
Undergraduate Student, Department of Civil Engineering, Birla Institute
of Technology & Science, Pilani, Rajasthan - 333031 (India)
Phone: +91-9468708130,
e-mail: piyushb3593@gmail.com
ABSTRACT
This paper presents a simple and general procedure for calculating the bearing capacity of
shallow foundations valid for strip, square, rectangular and circular footings; in
cohesionless soils. As per Indian Standard IS:6403, 1981, the method of analysis (general
shear failure; local shear failure; or interpolation between the two) is decided on the basis
of relative density of soil Dr for computing the bearing capacity. This is time consuming
and requires precision at every step; more so if Dr is between 20% to 70% and
interpolation is needed. This paper gives a direct correlation between SPT N1)60 and the
bearing capacity factors ( N q )int erpolated , Nγ )int erpolated , N q and Nγ ) by generating
regression models using MATLAB®. By using the regression models, the need for
determining Dr, deciding a method of analysis as well as carrying out interpolation
between two methods of analysis is eliminated. A comparison is drawn between the
bearing capacity values obtained from the detailed procedure outlined in IS:6403 (1981)
and the proposed simplified procedure. The percentage variation is found to be within
±2.2%.
KEYWORDS:
®
Bearing capacity; SPT; Shallow foundation; Regression; MS
®
Excel ; MATLAB .
- 10929 -
Vol. 15 [2011], Bund. J
10930
INTRODUCTION
The load per unit area of the foundation at which shear failure in the soil occurs is called
ultimate bearing capacity. Several theories for estimating the ultimate bearing capacity of shallow
foundations have been proposed and summarized by Das (2009). The bearing capacity of soil is
needed for dimensioning the foundation for any structure. Vesic (1973) identified three types of
bearing capacity failures occurring in shallow foundations as shown in Figure 1: a) General shear
failure, b) Local shear failure, c) Punching shear failure.
Figure 1: Modes of foundation failure at different relative depths
Df
B
in sand, taken
from IS: 6403 (1971) after Vesic (1973)
The failure mode significantly affects the bearing capacity calculation in shallow foundations.
As per IS:6403 (1981) the mode of failure of soil support beneath the shallow foundation can be
determined by Dr as shown in Table 1. Knowing the mode of failure, appropriate equations can
be used to determine the bearing capacity. If the relative density lies between certain limits, as
mentioned in Table 1, one has to interpolate between two methods of bearing capacity
calculation. This makes the process time consuming and very cumbersome.
Table 1: Method of analysis based on relative density Dr
% Relative Density
Greater than 70%
Less than 20%
20% to 70%
Condition
Dense
Loose
Medium
Method of Analysis
General Shear
Local shear (as well as punching shear)
Interpolate between 1 and 2
Vol. 15 [2011], Bund. J
10931
In order to make the process easier for the designer, an effort has been made to directly
correlate SPT N1)60 and the bearing capacity factors ( N q )int erpolated , Nγ )int erpolated , N q and Nγ ) for
SPT N1)60 ranging from 10-30 and 31-50. This reduces the number of steps involved and the
chance of committing calculation errors. It automatically takes care of Dr computation and
interpolations required thereby eliminating the need of computing Dr, mode of failure of soil and
the steps of interpolation. A parametric study is carried out by changing SPT N1)60 from 10-50 and
a comparison is drawn between the bearing capacity values obtained from both the methods. The
results indicate that the regression models are able to accurately predict the bearing capacity and
outperforms the existing method.
IS 6403 METHOD
IS:6403 (1981) suggests that bearing capacity calculations shall be made on the basis of shear
strength parameters - angle of internal friction φ and cohesion c, obtained from appropriate shear
tests (IS:2720 (Parts XI and XIII), 1993 and 1986); or from plate load test results as given in
IS:1888 (1982); or from static cone penetration resistance qc obtained from static cone penetration
test as given in IS:4968 (Part III, 1976).
For cohesionless soils it is difficult to get an undisturbed sample for laboratory testing and
therefore, one has to rely on the SPT as per IS:6403 (1981) to get φ , from the chart given in
Figure 2.
Angle of internal friction ϕ (degrees)
46
44
42
40
38
36
34
32
30
28
0
10
20
30
40
SPT N1)60
50
60
70
80
Figure 2: Correlation between SPT N1)60 and angle of internal friction φ adapted from IS
6403 (1981) after Peck, et al. (1974)
In the case of cohesionless soils, IS:6403 (1981) suggests using Dr to determine the method
of analysis for bearing capacity calculation (Table 1).
Vol. 15 [2011], Bund. J
10932
The net ultimate bearing capacity originally given by Terzaghi (1943) and then modified to
take into account the shape of the footing, inclination of loading, depth of embedment and effect
of water table (Meyerhof, 1953; 1963; Brinch Hansen, 1961; 1970; Vesic, 1973; 1975) and
adopted by IS:6403 (1981) is:
a) In the case of general shear failure:
1
qn=
cN c sc d c ic + q ( N q − 1) sq d q iq + B 'γ Nγ sγ dγ iγ W '
.u
2
(1)
b) In the case of local shear failure:
=
qn' .u
2 ' '
1
cN c sc d c ic + q ( N q' − 1) sq d q' iq + B 'γ Nγ' sγ dγ' iγ' W '
3
2
(2)
where;
net ultimate bearing capacity/kN m-2
cohesion of soil/kN m-2
unit weight of soil/kN m-3
effective surcharge/kN m-2
effective width of footing considering the effect of load eccentricity/m
shape factors
qn.u
c
γ
q
B'
sc , sq , sγ
ic , iq , iγ
load inclination factors
d c , d q , dγ
depth factors
W'
correction factor for the location of water table
bearing capacity factors that are non-dimensional and are only functions
of angle of internal friction of soil
N c , N q , Nγ , N c' , N q' , Nγ'
Table 2: Shape factors adapted from (IS:6403, 1981)
Shape of Base
sq
sγ
sc
Continuous strip
Rectangle
Square
Circle
1.0
1 + 0.2B/L
1.3
1.3
1.0
1 + 0.2B/L
1.2
1.2
1.0
1 – 0.4B/L
0.8
0.6
Note: Use B as the diameter in the bearing capacity formula for circular footing.
The depth factors as given by Meyerhof (1963) also adopted by IS:6403 (1981) are:
Df
Nf
B
d=
d=
1 for f < 10o
γ
q
d c = 1 + 0.2
D
dq =
dγ =
1 + 0.1 f Nff
for > 10o
B
π φ 
+ 
 4 2
where,
=
Nφ tan 2 
(3)
(4)
(5)
Vol. 15 [2011], Bund. J
10933
The load inclination factors (Meyerhof, 1953; Brinch Hansen, 1970) adopted by IS:6403
(1981) are:
ic= iq= (1 − α / 90o ) 2
(6)
iγ= (1 − α / φ ) 2
(7)
where, α is the angle with the vertical
The bearing capacity factors adopted by IS:6403 (1981) are:
a) N c and N q are given by NCHRP Report 651 (2010) after Prandtl (1920) & Reissner
(1924):
=
N c ( N q − 1) cot φ
(8)
φ
=
N q exp(p tan φ ) tan 2 (45o + )
2
(9)
b) Nγ is given by Vesic (1973):
=
Nγ 2( N q + 1) tan φ
(10)
'
'
'
'
'
'
'
For obtaining the values of N c , N q , Nγ , d c , d q , dγ and iγ calculate φ ' = tan −1 (0.67 tan φ ) .
Use the above mentioned formulas with φ ' .
Water table correction:
=
W ' 0.5(1 +
Z
)
B
(11)
Z=depth of the water table below the base of footing; (0.5≤W '≤1).
When Dr is higher than 70% or lower than 20% one can use Equation (1) and Equation (2)
respectively to calculate the bearing capacity. But when Dr is between 20% to 70% one has to
calculate bearing capacity by both the general and local shear formulae, and then interpolate
between them on the basis of Dr.
PROPOSED METHOD
IS:6403 (1981) recommends the usage of Figure 2 for obtaining φ from SPT N1)60. As an
alternative, Figure 2 can be curve fitted and an equation is generated which relates SPT N1)60 to φ
(degrees):
=
φ 26.625 + 0.3486 N1)60 − 0.001( N1)60 ) − 7 ×10−6 ( N1)60 )
2
3
(12)
R =1
2
Relative density can also be calculated to a good approximation by using the correlation with
SPT N1)60 (Skempton, 1986; Eurocode 7, 2007):
Vol. 15 [2011], Bund. J
10934
N1)60
Dr =×100 ; for N1)60 > 7.35
60
(13)
A spreadsheet is prepared where SPT N1)60 is taken as input and gives φ and Dr by using
Equations (12) and (13), following which φ ' can also be calculated. After obtaining φ and φ ' , the
bearing capacity factors N q , Nγ , N q' , Nγ' can be calculated using Equations (9) and (10). When
20% ≤ Dr ≤ 70% (SPT N1)60 lies between 10-30) instead of calculating bearing capacity by both
the methods i.e. general shear failure and local shear failure and then interpolating (as suggested
by IS:6403 (1981)), it was thought appropriate to use the interpolated bearing capacity factors in
calculating the bearing capacity. That can save a lot of time and greatly reduce the chances of
committing calculation errors.
'
'
For 20% ≤ Dr ≤ 70%, interpolation is done between N q − N q and Nγ − Nγ on the basis of Dr
by the spreadsheet, as shown in Table 3. The rules for interpolation are kept the same as
recommended by IS:6403 (1981) for the method of analysis (Table 1).For SPT N1)60 > 30, Dr
exceeds 70%, for which only general shear failure governs. As a result, no interpolation is
required.
Table 3: Nq)interpolated and Nγ)interpolated for SPT N1)60 = 10-30; Nq and Nγ for SPT N1)60 =
31-50
Nγ
Nγ'
Nγ )int erpolated
11.82
22.42
6.24
12.98
12.68
23.52
6.47
14.25
7.48
13.58
24.68
6.70
15.59
20.57
7.67
14.52
25.89
6.94
17.00
48.3%
21.33
7.87
15.49
27.15
7.19
18.49
22.30
50.0%
22.13
8.07
16.51
28.47
7.44
20.06
31.92
22.55
51.6%
22.95
8.27
17.56
29.84
7.70
21.71
17
32.23
22.80
53.2%
23.81
8.48
18.67
31.28
7.97
23.46
18
32.53
23.04
54.8%
24.69
8.70
19.82
32.77
8.25
25.30
19
32.84
23.28
56.3%
25.60
8.92
21.02
34.33
8.53
27.25
20
33.14
23.52
57.7%
26.54
9.14
22.27
35.96
8.83
29.30
21
33.44
23.76
59.2%
27.51
9.37
23.57
37.65
9.13
31.47
22
33.74
24.00
60.6%
28.51
9.60
24.94
39.41
9.44
33.75
23
34.03
24.24
61.9%
29.54
9.84
26.36
41.25
9.76
36.16
24
34.32
24.47
63.2%
30.61
10.09
27.84
43.15
10.09
38.69
25
34.61
24.70
64.5%
31.71
10.33
29.38
45.14
10.43
41.35
26
34.89
24.93
65.8%
32.84
10.59
30.99
47.20
10.77
44.16
27
35.17
25.16
67.1%
34.01
10.85
32.66
49.34
11.13
47.11
28
35.45
25.39
68.3%
35.21
11.11
34.40
51.56
11.50
50.21
29
35.72
25.62
69.5%
36.45
11.38
36.21
53.87
11.87
53.47
SPT
N1)60
φ
φ'
Dr
Nq
N q'
N q )int erpolated
10
30.00
21.05
40.8%
18.41
7.11
11
30.33
21.31
42.8%
19.10
7.29
12
30.65
21.56
44.7%
19.82
13
30.97
21.81
46.5%
14
31.29
22.06
15
31.61
16
Vol. 15 [2011], Bund. J
10935
30
35.99
25.84
70.7%
37.72
39.03
11.65
-
38.09
-
31
36.26
71.9%
73.0%
40.38
-
-
56.26
58.73
12.25
-
56.88
-
61.30
-
-
32
36.53
-
33
36.79
-
74.2%
41.76
-
-
63.95
-
-
34
37.05
-
75.3%
43.18
-
-
66.69
-
-
35
37.30
-
76.4%
44.63
-
-
69.53
-
-
36
37.55
-
77.5%
46.12
-
-
72.46
-
-
37
37.80
-
78.5%
47.65
-
-
75.48
-
-
38
38.04
-
79.6%
49.22
-
-
78.59
-
-
39
38.28
-
80.6%
50.82
-
-
81.80
-
-
40
38.52
-
81.6%
52.45
-
-
85.10
-
-
41
38.75
-
82.7%
54.13
-
-
88.50
-
-
42
38.98
-
83.7%
55.83
-
-
91.99
-
-
43
39.21
-
84.7%
57.57
-
-
95.57
-
-
44
39.43
-
85.6%
59.35
-
-
99.25
-
-
45
39.65
-
86.6%
61.15
-
-
103.01
-
-
46
39.86
-
87.6%
62.99
-
-
106.87
-
-
47
40.07
-
88.5%
64.85
-
-
110.80
-
-
48
40.28
-
89.4%
66.75
-
-
114.83
-
-
49
40.48
-
90.4%
68.67
-
-
118.93
-
-
50
40.68
-
91.3%
70.62
-
-
123.11
-
-
where ‘–‘ means Not Applicable.
An effort has been made to get reliable correlations with N q )int erpolated , Nγ )int erpolated , N q and
Nγ directly from SPT N1)60 without the need for calculating other parameters. Regression models
are generated for that purpose as a function of SPT N1)60 alone, and for both the ranges of SPT
N1)60 i.e. 10-30 and 31-50. MATLAB® (MathWorks, 2012) is used to obtain the regression
models. The equations generated are as follows:
a) For SPT N1)60 10-30
=
N q )interpolated 0.06087( N1)60 )1.823 + 7.928
R 2 = 0.9999
=
Nγ )interpolated 0.0453( N1)60 ) 2.051 + 8.198
R 2 = 0.9999
(14)
(15)
b) For SPT N1)60 31-50
=
N q 0.04255( N1)60 )1.828 + 16.36
R2 = 1
=
Nγ 0.02808( N1)60 ) 2.095 + 21.29
R2 = 1
(16)
(17)
Vol. 15 [2011], Bund. J
10936
where, R2 = coefficient of determination
These bearing capacity factors (Equations (14), (15), (16) and (17)) can be directly used in
calculating bearing capacity without calculating Dr, choosing a method of analysis or performing
any interpolation. Parametric study is carried out for different values of SPT N1)60 and presented
in Table 4 and Table 5. Details of foundation and the soil on which it rests are given in Annexure.
The percentage variation is found to be within ±2.2%.
On calculating the bearing capacity by both the methods (Annexure), it is evident that the
IS:6403 (1981) method takes longer than the proposed simplified procedure. The simplified
procedure takes approximately one third the number of steps required in IS:6403 (1981) for SPT
N1)60 10-30.
Table 4: Comparison of bearing capacity values by using author’s Equations (14) and
(15) versus IS:6403 (1981) method and corresponding % error between the two for SPT
N1)60 = 10-30
SPT
N1)60
Net Ultimate Bearing Capacity by using
N q )int erpolated and Nγ )int erpolated (in kN m-2)
Net Ultimate Bearing Capacity as
per IS:6403 (in kN m-2)
%
error
10
10.5
11
11.5
12
12.5
13
13.5
14
14.5
15
15.5
16
16.5
17
17.5
18
18.5
19
19.5
20
20.5
21
21.5
22
22.5
23
297.92
308.75
320.07
331.85
344.12
356.86
370.07
383.76
397.91
412.53
427.62
443.18
459.20
475.69
492.64
510.06
527.93
546.27
565.06
584.32
604.03
624.20
644.83
665.91
687.44
709.44
731.88
291.48
304.05
316.88
330.00
343.41
357.13
371.16
385.52
400.22
415.27
430.67
446.45
462.60
479.14
496.08
513.43
531.19
549.39
568.02
587.10
606.64
626.65
647.13
668.11
689.58
711.57
734.07
2.2%
1.5%
1.0%
0.6%
0.2%
-0.1%
-0.3%
-0.5%
-0.6%
-0.7%
-0.7%
-0.7%
-0.7%
-0.7%
-0.7%
-0.7%
-0.6%
-0.6%
-0.5%
-0.5%
-0.4%
-0.4%
-0.4%
-0.3%
-0.3%
-0.3%
-0.3%
Vol. 15 [2011], Bund. J
23.5
24
24.5
25
25.5
26
26.5
27
27.5
28
28.5
29
29.5
30
754.78
778.13
801.93
826.18
850.88
876.03
901.64
927.69
954.18
981.13
1008.52
1036.36
1064.65
1093.38
10937
757.10
780.68
804.80
829.48
854.74
880.58
907.00
934.03
961.68
989.94
1018.84
1048.37
1076.67
1097.83
-0.3%
-0.3%
-0.4%
-0.4%
-0.5%
-0.5%
-0.6%
-0.7%
-0.8%
-0.9%
-1.0%
-1.1%
-1.1%
-0.4%
Table 5: Comparison of bearing capacity values by using author’s Equations (16) and
(17) versus IS:6403 (1981) method and corresponding % error between the two for SPT
N1)60 = 31-50
SPT
N1)60
Net Ultimate Bearing Capacity by using
N q and Nγ (in kN m-2)
Net Ultimate Bearing Capacity as per
IS:6403 (in kN m-2)
%
error
31
31.5
32
32.5
33
33.5
34
34.5
35
35.5
36
36.5
37
37.5
38
38.5
39
39.5
40
40.5
41
41.5
1127.89
1149.64
1171.72
1194.12
1216.86
1239.93
1263.33
1287.05
1311.10
1335.48
1360.19
1385.23
1410.60
1436.29
1462.31
1488.66
1515.33
1542.33
1569.66
1597.32
1625.30
1653.61
1141.18
1163.36
1185.90
1208.78
1232.01
1255.59
1279.53
1303.82
1328.46
1353.46
1378.82
1404.53
1430.61
1457.04
1483.83
1510.98
1538.49
1566.35
1594.57
1623.14
1652.06
1681.34
-1.2%
-1.2%
-1.2%
-1.2%
-1.2%
-1.2%
-1.3%
-1.3%
-1.3%
-1.3%
-1.4%
-1.4%
-1.4%
-1.4%
-1.5%
-1.5%
-1.5%
-1.5%
-1.6%
-1.6%
-1.6%
-1.6%
Vol. 15 [2011], Bund. J
42
42.5
43
43.5
44
44.5
45
45.5
46
46.5
47
47.5
48
48.5
49
49.5
50
10938
1682.25
1711.21
1740.50
1770.12
1800.06
1830.33
1860.92
1891.84
1923.09
1954.66
1986.56
2018.78
2051.33
2084.20
2117.40
2150.92
2184.77
1710.97
1740.94
1771.25
1801.91
1832.90
1864.23
1895.88
1927.86
1960.16
1992.78
2025.70
2058.93
2092.46
2126.27
2160.37
2194.75
2229.39
-1.7%
-1.7%
-1.7%
-1.8%
-1.8%
-1.8%
-1.8%
-1.9%
-1.9%
-1.9%
-1.9%
-2.0%
-2.0%
-2.0%
-2.0%
-2.0%
-2.0%
DISCUSSION
The summary of both the methods is presented by Figure 3. It clearly shows that the proposed
method takes lesser number of steps. Moreover, the designer is not required to determine the
mode of failure and perform cumbersome calculations.
Vol. 15 [2011], Bund. J
10939
Figure 3: Step by step procedure of calculating bearing capacity by IS:6403
(1981) method and proposed method
CONCLUSIONS
In the present study a regression equation is developed to compute φ from SPT N1)60 which
eliminates reading error from the graph. Also, regression models are developed for the bearing
capacity factors as a direct function of design SPT N1)60 value from 10 to 30 and 31 to 50. These
simple equations automatically take care of appropriate method of analysis to be adopted, and
therefore, one does not need to check the mode of failure. The proposed equations also eliminate
the need of calculating Dr and interpolation between general shear failure and local shear failure
methods of analysis. Number of steps required to calculate the bearing capacity are greatly
reduced by using the proposed simplified procedure. The percentage variation lies within ±2.2%.
ANNEXURE
Example 1) A rectangular footing whose length is 3m and breadth is 1.5m is founded on a
cohesion-less soil at a depth of 0.75m from the ground surface. The ground water table is present
at a depth of 1m from the ground surface. The bulk density of soil is 20 kN/m3. The design SPT
N1)60 is 20.
Solution: By (IS:6403, 1981) method:
Vol. 15 [2011], Bund. J
10940
From SPT N1)60 = 20, get Dr and φ .
Dr =
N1)60
60
× 100 =
20
= 58% (Interpolation between general shear failure and local
60
shear failure required)
φ (deg) ≈ 33.14o (From Fig. 2)
General Shear Failure:
φ
33.14
2
φ ) tan 2 (45o + ) exp(p tan(33.14)) tan
(45 +
) 26.54
=
N q exp(p tan=
=
2
2
Nγ =2( N q + 1) tan φ =2(26.54 + 1) tan(33.14) =35.96
Shape factors are:
Sq =
1 + 0.2
B
1.5
=
1 + 0.2*
=
1.10
L
3
Sγ =
1 − 0.4
B
1.5
1 − 0.4*
0.8
=
=
L
3
Depth factors are:
D
1 + 0.1 f Nff
dq =
dγ =
for > 10o
B
dq =
dγ =
1 + 0.1*
0.75
33.14
tan(45 +
)=
1.09
1.5
2
Load inclination factors are:
iq= iγ= 1.0 (Because there is no load inclination)
Water table correction:
W ' = 0.5(1 +
Z
0.25
) = 0.5(1 +
) = 0.5833
B
1.5
N c , s c , d c and ic are not required because c = 0.
qn=
cN c sc d c ic + q ( N q − 1) sq d q iq +
.u
1 '
B γ Nγ sγ dγ iγ W '
2
qn.u = (20*0.75)(26.54 − 1) *1.10*1.09*1 +
qn.u = 735.21 kN/m 2
Local Shear Failure:
1
*1.5* 20*35.96*0.8*1.09*1*0.5833
2
Vol. 15 [2011], Bund. J
10941
−1
=
φ ' tan
=
(0.67 tan(33.14)) 23.52o
φ'
23.52
2
=
N q' exp(p tan φ=
=
') tan 2 (45o + ) exp(p tan(23.52)) tan
(45 +
) 9.14
2
2
Nγ' =
2( N q' + 1) tan φ ' =
2(9.14 + 1) tan 23.52 =
8.83
Shape factors are:
Sq =
1 + 0.2
B
1.5
=
1 + 0.2*
=
1.10
L
3
Sγ =
1 − 0.4
B
1.5
1 − 0.4*
0.8
=
=
L
3
Depth factors are:
Df
1 + 0.1
dq =
dγ =
Nff
for ' > 10o
B
1 + 0.1*
d q' =
dγ' =
0.75
23.52
tan(45 +
)=
1.08
1.5
2
Load inclination factors are:
iq= iγ= 1.0 (Because there is no load inclination)
Water table correction:
W ' = 0.5(1 +
Z
0.25
) = 0.5(1 +
) = 0.5833
B
1.5
N c' , s c , d c' and ic are not required because c = 0.
'n.u
q=
2 ' '
1
cN c sc d c ic + q ( N q' − 1) sq d q' iq + B 'γ Nγ' sγ dγ' iγ' W '
3
2
=
q 'n.u (20*0.75)(9.14 − 1) *1.10*1.08*1 +
1
*1.5* 20*8.83*0.8*1.08*1*0.5833
2
q 'n.u = 211.06 kN/m 2
Table A.1: Net ultimate bearing capacity qn.u corresponding to Dr at which it is
calculated
qn.u (kN/m2)
735.21
211.06
?
Dr
70%
20%
58%
Interpolating between general shear failure and local shear failure on the basis of %Dr.
Vol. 15 [2011], Bund. J
10942
735.21 − 211.06 735.21 − qn.u
=
70% − 20%
70% − 58%
Solving this, qn.u at 58% relative density = 606.64 kN/m2
By the simplified procedure proposed by the paper:
From SPT N1)60 = 20, get N q )int erpolated and Nγ )int erpolated from author’s Equations (14) and (15)
For SPT N1)60 10-30
N
=
0.06087( N1)60 )1.823
=
+ 7.928 0.06087 *(20)1.823
=
+ 7.928 22.26
q )interpolated
N=
0.0453( N1)60 ) 2.051=
+ 8.198 0.0453*(20) 2.051=
+ 8.198 29.31
γ )interpolated
Shape factors are:
Sq =
1 + 0.2
B
1.5
=
=
1 + 0.2*
1.10
L
3
Sγ =
1 − 0.4
B
1.5
1 − 0.4*
0.8
=
=
L
3
Depth factors are:
(From Equation (12)) -
=
φ 26.625 + 0.3486 N1)60 − 0.001( N1)60 ) − 7 ×10−6 ( N1)60 )
2
= 26.625 + 0.3486* 20 − 0.001( 20 ) − 7 ×10−6 ( 20 )
2
3
3
= 33.14o
D
dq =
dγ =
1 + 0.1 f Nff
for > 10o
B
1 + 0.1*
dq =
dγ =
0.75
33.14
tan(45 +
)=
1.09
1.5
2
Load inclination factors are:
iq= iγ= 1.0 (Because there is no load inclination)
Water table correction:
W ' = 0.5(1 +
Z
0.25
) = 0.5(1 +
) = 0.5833
B
1.5
N c , s c , d c and ic are not required because c = 0.
qn=
cN c sc d c ic + q ( N q − 1) sq d q iq +
.u
1 '
B γ Nγ sγ dγ iγ W '
2
Vol. 15 [2011], Bund. J
qn.u = (20*0.75)(22.26 − 1) *1.10*1.09*1 +
10943
1
*1.5* 20* 29.31*0.8*1.09*1*0.5833
2
qn.u = 604.03 kN/m 2
604.03 − 606.64
) *100 = −0.4%
606.64
% error = (
Example 2) A rectangular footing whose length is 3m and breadth is 1.5m is founded on a
cohesion-less soil at a depth of 0.75m from the ground surface. The ground water table is present
at a depth of 1m from the ground surface. The bulk density of soil is 20 kN/m3. The design SPT
N1)60 is 40.
Solution: By (IS:6403, 1981) method:
From SPT N1)60 = 40, get Dr and φ .
Dr =
N1)60
60
×100 =
40
= 82% (General shear failure governs)
60
φ (deg) ≈ 38.52o (From Fig. 2)
General Shear Failure:
φ
38.52
2
=
N q exp(p tan=
φ ) tan 2 (45o + ) exp(p tan(38.52)) tan
=
(45 +
) 52.45
2
2
Nγ =2( N q + 1) tan φ =2(52.45 + 1) tan(38.52) =
85.10
Shape factors are:
Sq =
1 + 0.2
B
1.5
=
1 + 0.2*
=
1.10
L
3
Sγ =
1 − 0.4
B
1.5
1 − 0.4*
0.8
=
=
L
3
Depth factors are:
D
dq =
dγ =
1 + 0.1 f Nff
for > 10o
B
1 + 0.1*
dq =
dγ =
0.75
38.52
tan(45 +
)=
1.10
1.5
2
Load inclination factors are:
iq= iγ= 1.0 (Because there is no load inclination)
Water table correction:
Vol. 15 [2011], Bund. J
W ' = 0.5(1 +
10944
Z
0.25
) = 0.5(1 +
) = 0.5833
B
1.5
N c , s c , d c and ic are not required because c = 0.
qn=
cN c sc d c ic + q ( N q − 1) sq d q iq +
.u
1 '
B γ Nγ sγ dγ iγ W '
2
qn.u = (20*0.75)(52.45 − 1) *1.10*1.10*1 +
1
*1.5* 20*85.10*0.8*1.10*1*0.5833
2
qn.u = 1594.57 kN/m 2
By the simplified procedure proposed by the paper:
From SPT N1)60 = 40, get Nq and Nɣ from author’s Equations (16) and (17)
For SPT N1)60 31-50
=
N q 0.04255( N1)60 )1.828=
+ 16.36 0.04255(40)1.828=
+ 16.36 52.46
=
=
+ 21.29 0.02808*(40) 2.095
=
+ 21.29 85.07
Nγ 0.02808( N1)60 ) 2.095
Shape factors are:
Sq =
1 + 0.2
B
1.5
=
1 + 0.2*
=
1.10
L
3
Sγ =
1 − 0.4
B
1.5
1 − 0.4*
0.8
=
=
L
3
Depth factors are:
(From Equation (12)) -
=
φ 26.625 + 0.3486 N1)60 − 0.001( N1)60 ) − 7 ×10−6 ( N1)60 )
2
= 26.625 + 0.3486* 40 − 0.001( 40 ) − 7 ×10−6 ( 40 )
2
= 38.52o
D
dq =
dγ =
1 + 0.1 f Nff
for > 10o
B
1 + 0.1*
dq =
dγ =
0.75
38.24
tan(45 +
)=
1.10
1.5
2
Load inclination factors are:
iq= iγ= 1.0 (Because there is no load inclination)
Water table correction:
3
3
Vol. 15 [2011], Bund. J
W ' = 0.5(1 +
10945
Z
0.25
) = 0.5(1 +
) = 0.5833
B
1.5
N c , s c , d c and ic are not required because c = 0.
qn=
cN c sc d c ic + q ( N q − 1) sq d q iq +
.u
1 '
B γ Nγ sγ dγ iγ W '
2
qn.u = (20*0.75)(52.46 − 1) *1.10*1.10*1 +
1
*1.5* 20*85.07 *0.8*1.10*1*0.5833
2
qn.u = 1569.66 kN/m 2
1569.66 − 1594.57
% error = (
) *100 = −1.6%
1594.57
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