Civil Engineering Reference
In-Depth Information
the cohesion
c
and the dilation angle
ψ
. The “permeability” property
k/γ
w
of the soil
10
−
6
. The current example models a cohesion-
in this example is isotropic and set to 1
×
30
◦
and no dilation. The soil is initially consolidated to an isotropic
compressive stress (
σ
3
) of -100 kN/m
2
read as
cons
. The coordinate data for
x coords
and
y coords
is followed by the familiar time stepping and output control parameters.
The current example calls for
nstep=200
calculation time steps of
dtim=0.5
secs with
theta=0.5
(Crank-Nicolson). Output is required every time step (
npri=1
) at node
nres=1
.The
nr
data indicate 20 restrained nodes, which include rollers on the left and
bottom boundaries, drainage conditions on the top and right boundaries and removal of the
third freedom at all mid-side nodes, as described in Program 9.3. The next data provide
the load weightings corresponding to a unit pressure (Appendix A), to be applied to the 5
nodes at the top of the mesh. The next line reads the tolerance and iteration ceiling (
tol
and
limit
) for plastic iterations, as was first used in Program 4.5 and again extensively
in Chapter 6. The final data define the load-time function to be applied at the top of the
mesh. In this example the deviator stress is to increase linearly with time at a “fast” rate
given by d
D/
d
t
=
less soil with
φ
=
15 kNm
−
2
s
−
1
, so just two load function coordinates are required.
The results from this analysis are listed as Figure 9.13 and plotted in Figure 9.14,
together with the results of a second analysis in which a “slow” loading rate of d
D/
d
t
=
0
.
02 kNm
−
2
s
−
1
was applied. The deviator stress at failure
D
f
was computed to be about
100 kN/m
2
for the “fast” loading rate and 200 kN/m
2
for the “slow” loading rate.
For plane strain compression of a non-dilative saturated soil, Griffiths(1985) produced
the solution,
σ
3
(K
p
−
1
)(
2
β
ps
+
1
)
D
f
=
(9.4)
(K
p
+
1
)β
ps
+
1
in which
β
ps
→∞
and
β
ps
=
0 give undrained and drained limiting conditions respec-
tively.
In this example,
K
p
=
tan
2
(
45
◦
+
φ
/
2
)
=
3, so for a consolidating stress of
σ
3
=
100 kN/m
2
, indicating that “fast” loading
in this case is giving essentially “undrained” conditions. The “slow” loading result of
100 kN/m
2
, (9.4) with
β
ps
→∞
gives
D
f
≈
There are 36 equations and the skyline storage is 466
Results at node 1
time load x-disp y-disp porepressure iters
0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
0.5000E+00 0.7500E+01 0.0000E+00 -0.5025E-03 0.0000E+00 2
0.1000E+01 0.1500E+02 0.0000E+00 -0.1009E-02 0.0000E+00 2
0.1500E+01 0.2250E+02 0.0000E+00 -0.1520E-02 0.0000E+00 2
0.2000E+01 0.3000E+02 0.0000E+00 -0.2035E-02 0.0000E+00 2
0.2500E+01 0.3750E+02 0.0000E+00 -0.2555E-02 0.0000E+00 2
0.3000E+01 0.4500E+02 0.0000E+00 -0.3078E-02 0.0000E+00 2
0.3500E+01 0.5250E+02 0.0000E+00 -0.3606E-02 0.0000E+00 2
0.4000E+01 0.6000E+02 0.0000E+00 -0.4138E-02 0.0000E+00 2
0.4500E+01 0.6750E+02 0.0000E+00 -0.4674E-02 0.0000E+00 2
0.5000E+01 0.7500E+02 0.0000E+00 -0.5213E-02 0.0000E+00 2
0.5500E+01 0.8250E+02 0.0000E+00 -0.5757E-02 0.0000E+00 2
0.6000E+01 0.9000E+02 0.0000E+00 -0.6304E-02 0.0000E+00 2
0.6500E+01 0.9750E+02 0.0000E+00 -0.6886E-02 0.0000E+00 9
0.7000E+01 0.1050E+03 0.0000E+00 -0.4326E-01 0.0000E+00 250
Figure 9.13
Results from Program 9.4 example with d
D/
d
t
=
15
