DHARM
368 GEOTECHNICAL ENGINEERING
If θ is the angle made by OA with the tangent of the periphery of the loaded area from A,
σz = q (1 – cos^3 θ) ...(Eq. 10.39)
The vertical stress at a point not lying on the vertical axis through the centre of the
loaded area may also be found; but it requires the evaluation of a more difficult integral.
Spangler (1951) gives the influence coefficients for both cases, the values for the case
where the point lies directly beneath the centre of the loaded area being those in the column
for
r
a
= 0 (Table 10.5).
Table 10.5 Influence coefficients for vertical stress due to uniform
load on a circular area (After Spangler, 1951)
r/a
z
a^0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
0.5 0.911 0.840 0.418 0.060 0.010 0.003 0.000 0.000 0.000
1.0 0.646 0.560 0.335 0.125 0.043 0.016 0.007 0.003 0.000
1.5 0.424 0.374 0.256 0.137 0.064 0.029 0.013 0.007 0.002
2.0 0.284 0.258 0.194 0.127 0.073 0.041 0.022 0.012 0.006
2.5 0.200 0.186 0.150 0.109 0.073 0.044 0.028 0.017 0.011
3.0 0.146 0.137 0.117 0.091 0.066 0.045 0.031 0.022 0.015
4.0 0.087 0.083 0.076 0.061 0.052 0.041 0.031 0.024 0.018
5.0 0.057 0.056 0.052 0.045 0.039 0.033 0.027 0.022 0.018
10.0 0.015 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.011
Pressure bulbs or isobar patterns for vertical stresses and shear stresses, as presented
by Jürgenson (1934), are shown in Figs. 10.13 (a) and (b).
If Westergaard’s theory is to be used, equation 10.14 may be integrated to obtain the
vertical stress at a point beneath the centre of a uniformly loaded circular area, which is given
by:
σz = q
az
1 1
1 2
−
+
L
N
M
M
O
Q
P
(/ )η P ...(Eq. 10.40)
where η =
12
22
−
−
ν
ν
, ν being Poisson’s ratio.
for n = 0, η =
1
2
= 0.707